Loopback custom method call from Android - android

I am looking for example where I can call loopback's custom method from Android. To explain more, lets say I have a method on server side with name "greet(name)" that will greet someone. I want to invoke that from Android. Any example, or link is ok.
Thanks in advance.
Jahid

In the examples below, I'll assume your model is called Greeter and the static method Greeter.greet is invoked via GET /greeters/greet?name=Alex.
First of all, you need to describe the REST mapping of your method. Then you can call the method using invokeMethod.
public class GreeterRepository extends ModelRepository<Greeter> {
public RestContract createContract() {
RestContract contract = super.createContract();
contract.addItem(new RestContractItem("/" + getNameForRestUrl() + "/greet", "POST"),
getClassName() + ".greet");
return contract;
}
public void greet(name, final VoidCallback callback) {
invokeStaticMethod("greet", ImmutableMap.of("name", name), new Adapter.Callback() {
#Override
public void onError(Throwable t) {
callback.onError(t);
}
#Override
public void onSuccess(String response) {
callback.onSuccess();
}
});
}
}
See ModelRepository.java and Model.java for examples of methods that parse the response body.
Disclaimer: I am one of the developers of LoopBack, loopback-sdk-android is one of my specialisations.

Related

EventBus calling two different handlers for a single post

I am having problems with EventBus 3.0.0 where I post a single event like this:
Call<List<SessionSpec>> call = httpService.getSessionSpecs();
call.enqueue(new Callback<List<SessionSpec>>() {
#Override
public void onResponse(Call<List<SessionSpec>> call, Response<List<SessionSpec>> response) {
if (response.isSuccessful()) {
List<SessionSpec> specs = response.body();
EventBus.getDefault().post((List<SessionSpec>)specs);
}
else Log.e(TAG, "sendSessionSpecs(): request NOT successful");
}
#Override
public void onFailure(Call<List<SessionSpec>> call, Throwable t) {
Log.e(TAG, "sendSessionsSpecs(): failed");
}
});
I have two subscribers in the same fragment, each with different signatures, but they are both getting called from a single post:
#Subscribe
public void onSessionSpec(List<SessionSpec> specs) {
Log.d(TAG, "onSessionSpec(): entered");
Log.d(TAG, " : number of session specs: " + specs.size());
}
The second subscriber is defined as:
#Subscribe
public void onOverlayType(List<OverlayType> types) {
Log.d(TAG, "onOverlayType(): entered");
Log.d(TAG, " : number of overlay types: " + types.size());
}
Both of these callbacks are in a single fragment which is active when the post is done and I have verified that the post is only called once. When the single SessionSpec event is posted, both the onSessionSpec and the onOverlayType callbacks are dispatched by EventBus with the event type of List> so the onOverlayType callback receives the wrong type in its callback argument. The class OverlayType is a simple POJO class with 2 members, a int "sid" and a String "name". The class SessionSpec is more complex; it does have a member String "name" but other than that, nothing else is common between these 2 classes. I have verified that there is nothing closely resembling "OverlayType" in the SessionSpec class.
The interface definition is this:
public interface VcapHttpInterface {
#GET("overlay/types")
Call<List<OverlayType>> getOverlayTypes();
#GET("session/list")
Call<List<SessionSpec>> getSessionSpecs();
#GET("session/{id}")
Call<Session> getSession(#Path("id") int sid);
}
The getSession event post/callback has no problems.
I have spend all day trying to figure what is going wrong so I am clueless at this point. Anybody know what might be wrong with my code?
Thanks,
-Andres
Edit: How does EventBus know which handler to call for a particular response? Some posts I have read said that EventBus does not use the handler signature, but how else would it know how to map a response to the right subscribed handler routine? Is there a way to explicitly define the handler callback for a given event?
EventBus checks the class of the object that you are posting, and calls the methods that expect that class in their parameters. In your case you are posting an object which is a List. In both your listeners you expect an object of type List. It doesn't matter what generic you put in OverlayType or SessionSpec, eventbus will call both. In order to make it work you gotta define to models as events.
public class OverlayTypeEvent {
public List<OverlayType> types;
public OverlayTypeEvent(List<OverlayType> types) {
this.types = types;
}
}
and
public class SessionSpecEvent {
public List<SessionSpec> types;
public SessionSpecEvent(List<SessionSpec> types) {
this.types = types;
}
}
And listen on them seperatley. Then post events with the specific type.
#Subscribe
public void onSessionSpec(OverlayTypeEvent event) {
List<OverlayType> overlayTypes = event.overlayType;
}
If you don't want to create new class as a container everytime you send a list data, you can you Pair as simple container, it has two generic fields (first and second) to contain variables.
You can use first as a key to check the type of class, second contains the actually data.
List<SessionSpec> specs = response.body();
EventBus.getDefault().post(new Pair<>(SessionSpec.class.getSimpleName(), specs));
Receive data:
#Subscribe
public void onSessionSpec(Pair<String, List<SessionSpec>> specContainer){
if (SessionSpec.class.getSimpleName().equals(specContainer.first)) {
List<SessionSpec> sessionSpecs = specContainer.second;
}
}
#Subscribe
public void onOverlayType(Pair<String, List<OverlayType>> overlayContainer) {
if (OverlayType.class.getSimpleName().equals(overlayContainer.first)) {
List<OverlayType> overlayTypes = overlayContainer.second;
}
}
Advantage of this solution: Reduce creating unneeded classes.
Disadvantage: both onSessionSpec and onOverlayType get called.

Is it possible to skip onPause() method of BaseActivity in MainActivity when latter extends former? [duplicate]

I read this question and thought that would easily be solved (not that it isn't solvable without) if one could write:
#Override
public String toString() {
return super.super.toString();
}
I'm not sure if it is useful in many cases, but I wonder why it isn't and if something like this exists in other languages.
What do you guys think?
EDIT:
To clarify: yes I know, that's impossible in Java and I don't really miss it. This is nothing I expected to work and was surprised getting a compiler error. I just had the idea and like to discuss it.
It violates encapsulation. You shouldn't be able to bypass the parent class's behaviour. It makes sense to sometimes be able to bypass your own class's behaviour (particularly from within the same method) but not your parent's. For example, suppose we have a base "collection of items", a subclass representing "a collection of red items" and a subclass of that representing "a collection of big red items". It makes sense to have:
public class Items
{
public void add(Item item) { ... }
}
public class RedItems extends Items
{
#Override
public void add(Item item)
{
if (!item.isRed())
{
throw new NotRedItemException();
}
super.add(item);
}
}
public class BigRedItems extends RedItems
{
#Override
public void add(Item item)
{
if (!item.isBig())
{
throw new NotBigItemException();
}
super.add(item);
}
}
That's fine - RedItems can always be confident that the items it contains are all red. Now suppose we were able to call super.super.add():
public class NaughtyItems extends RedItems
{
#Override
public void add(Item item)
{
// I don't care if it's red or not. Take that, RedItems!
super.super.add(item);
}
}
Now we could add whatever we like, and the invariant in RedItems is broken.
Does that make sense?
I think Jon Skeet has the correct answer. I'd just like to add that you can access shadowed variables from superclasses of superclasses by casting this:
interface I { int x = 0; }
class T1 implements I { int x = 1; }
class T2 extends T1 { int x = 2; }
class T3 extends T2 {
int x = 3;
void test() {
System.out.println("x=\t\t" + x);
System.out.println("super.x=\t\t" + super.x);
System.out.println("((T2)this).x=\t" + ((T2)this).x);
System.out.println("((T1)this).x=\t" + ((T1)this).x);
System.out.println("((I)this).x=\t" + ((I)this).x);
}
}
class Test {
public static void main(String[] args) {
new T3().test();
}
}
which produces the output:
x= 3
super.x= 2
((T2)this).x= 2
((T1)this).x= 1
((I)this).x= 0
(example from the JLS)
However, this doesn't work for method calls because method calls are determined based on the runtime type of the object.
I think the following code allow to use super.super...super.method() in most case.
(even if it's uggly to do that)
In short
create temporary instance of ancestor type
copy values of fields from original object to temporary one
invoke target method on temporary object
copy modified values back to original object
Usage :
public class A {
public void doThat() { ... }
}
public class B extends A {
public void doThat() { /* don't call super.doThat() */ }
}
public class C extends B {
public void doThat() {
Magic.exec(A.class, this, "doThat");
}
}
public class Magic {
public static <Type, ChieldType extends Type> void exec(Class<Type> oneSuperType, ChieldType instance,
String methodOfParentToExec) {
try {
Type type = oneSuperType.newInstance();
shareVars(oneSuperType, instance, type);
oneSuperType.getMethod(methodOfParentToExec).invoke(type);
shareVars(oneSuperType, type, instance);
} catch (Exception e) {
throw new RuntimeException(e);
}
}
private static <Type, SourceType extends Type, TargetType extends Type> void shareVars(Class<Type> clazz,
SourceType source, TargetType target) throws IllegalArgumentException, IllegalAccessException {
Class<?> loop = clazz;
do {
for (Field f : loop.getDeclaredFields()) {
if (!f.isAccessible()) {
f.setAccessible(true);
}
f.set(target, f.get(source));
}
loop = loop.getSuperclass();
} while (loop != Object.class);
}
}
I don't have enough reputation to comment so I will add this to the other answers.
Jon Skeet answers excellently, with a beautiful example. Matt B has a point: not all superclasses have supers. Your code would break if you called a super of a super that had no super.
Object oriented programming (which Java is) is all about objects, not functions. If you want task oriented programming, choose C++ or something else. If your object doesn't fit in it's super class, then you need to add it to the "grandparent class", create a new class, or find another super it does fit into.
Personally, I have found this limitation to be one of Java's greatest strengths. Code is somewhat rigid compared to other languages I've used, but I always know what to expect. This helps with the "simple and familiar" goal of Java. In my mind, calling super.super is not simple or familiar. Perhaps the developers felt the same?
There's some good reasons to do this. You might have a subclass which has a method which is implemented incorrectly, but the parent method is implemented correctly. Because it belongs to a third party library, you might be unable/unwilling to change the source. In this case, you want to create a subclass but override one method to call the super.super method.
As shown by some other posters, it is possible to do this through reflection, but it should be possible to do something like
(SuperSuperClass this).theMethod();
I'm dealing with this problem right now - the quick fix is to copy and paste the superclass method into the subsubclass method :)
In addition to the very good points that others have made, I think there's another reason: what if the superclass does not have a superclass?
Since every class naturally extends (at least) Object, super.whatever() will always refer to a method in the superclass. But what if your class only extends Object - what would super.super refer to then? How should that behavior be handled - a compiler error, a NullPointer, etc?
I think the primary reason why this is not allowed is that it violates encapsulation, but this might be a small reason too.
I think if you overwrite a method and want to all the super-class version of it (like, say for equals), then you virtually always want to call the direct superclass version first, which one will call its superclass version in turn if it wants.
I think it only makes rarely sense (if at all. i can't think of a case where it does) to call some arbitrary superclass' version of a method. I don't know if that is possible at all in Java. It can be done in C++:
this->ReallyTheBase::foo();
At a guess, because it's not used that often. The only reason I could see using it is if your direct parent has overridden some functionality and you're trying to restore it back to the original.
Which seems to me to be against OO principles, since the class's direct parent should be more closely related to your class than the grandparent is.
Calling of super.super.method() make sense when you can't change code of base class. This often happens when you are extending an existing library.
Ask yourself first, why are you extending that class? If answer is "because I can't change it" then you can create exact package and class in your application, and rewrite naughty method or create delegate:
package com.company.application;
public class OneYouWantExtend extends OneThatContainsDesiredMethod {
// one way is to rewrite method() to call super.method() only or
// to doStuff() and then call super.method()
public void method() {
if (isDoStuff()) {
// do stuff
}
super.method();
}
protected abstract boolean isDoStuff();
// second way is to define methodDelegate() that will call hidden super.method()
public void methodDelegate() {
super.method();
}
...
}
public class OneThatContainsDesiredMethod {
public void method() {...}
...
}
For instance, you can create org.springframework.test.context.junit4.SpringJUnit4ClassRunner class in your application so this class should be loaded before the real one from jar. Then rewrite methods or constructors.
Attention: This is absolute hack, and it is highly NOT recommended to use but it's WORKING! Using of this approach is dangerous because of possible issues with class loaders. Also this may cause issues each time you will update library that contains overwritten class.
#Jon Skeet Nice explanation.
IMO if some one wants to call super.super method then one must be want to ignore the behavior of immediate parent, but want to access the grand parent behavior.
This can be achieved through instance Of. As below code
public class A {
protected void printClass() {
System.out.println("In A Class");
}
}
public class B extends A {
#Override
protected void printClass() {
if (!(this instanceof C)) {
System.out.println("In B Class");
}
super.printClass();
}
}
public class C extends B {
#Override
protected void printClass() {
System.out.println("In C Class");
super.printClass();
}
}
Here is driver class,
public class Driver {
public static void main(String[] args) {
C c = new C();
c.printClass();
}
}
Output of this will be
In C Class
In A Class
Class B printClass behavior will be ignored in this case.
I am not sure about is this a ideal or good practice to achieve super.super, but still it is working.
Look at this Github project, especially the objectHandle variable. This project shows how to actually and accurately call the grandparent method on a grandchild.
Just in case the link gets broken, here is the code:
import lombok.val;
import org.junit.Assert;
import org.junit.Test;
import java.lang.invoke.*;
/*
Your scientists were so preoccupied with whether or not they could, they didn’t stop to think if they should.
Please don't actually do this... :P
*/
public class ImplLookupTest {
private MethodHandles.Lookup getImplLookup() throws NoSuchFieldException, IllegalAccessException {
val field = MethodHandles.Lookup.class.getDeclaredField("IMPL_LOOKUP");
field.setAccessible(true);
return (MethodHandles.Lookup) field.get(null);
}
#Test
public void test() throws Throwable {
val lookup = getImplLookup();
val baseHandle = lookup.findSpecial(Base.class, "toString",
MethodType.methodType(String.class),
Sub.class);
val objectHandle = lookup.findSpecial(Object.class, "toString",
MethodType.methodType(String.class),
// Must use Base.class here for this reference to call Object's toString
Base.class);
val sub = new Sub();
Assert.assertEquals("Sub", sub.toString());
Assert.assertEquals("Base", baseHandle.invoke(sub));
Assert.assertEquals(toString(sub), objectHandle.invoke(sub));
}
private static String toString(Object o) {
return o.getClass().getName() + "#" + Integer.toHexString(o.hashCode());
}
public class Sub extends Base {
#Override
public String toString() {
return "Sub";
}
}
public class Base {
#Override
public String toString() {
return "Base";
}
}
}
Happy Coding!!!!
I would put the super.super method body in another method, if possible
class SuperSuperClass {
public String toString() {
return DescribeMe();
}
protected String DescribeMe() {
return "I am super super";
}
}
class SuperClass extends SuperSuperClass {
public String toString() {
return "I am super";
}
}
class ChildClass extends SuperClass {
public String toString() {
return DescribeMe();
}
}
Or if you cannot change the super-super class, you can try this:
class SuperSuperClass {
public String toString() {
return "I am super super";
}
}
class SuperClass extends SuperSuperClass {
public String toString() {
return DescribeMe(super.toString());
}
protected String DescribeMe(string fromSuper) {
return "I am super";
}
}
class ChildClass extends SuperClass {
protected String DescribeMe(string fromSuper) {
return fromSuper;
}
}
In both cases, the
new ChildClass().toString();
results to "I am super super"
It would seem to be possible to at least get the class of the superclass's superclass, though not necessarily the instance of it, using reflection; if this might be useful, please consider the Javadoc at http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Class.html#getSuperclass()
public class A {
#Override
public String toString() {
return "A";
}
}
public class B extends A {
#Override
public String toString() {
return "B";
}
}
public class C extends B {
#Override
public String toString() {
return "C";
}
}
public class D extends C {
#Override
public String toString() {
String result = "";
try {
result = this.getClass().getSuperclass().getSuperclass().getSuperclass().newInstance().toString();
} catch (InstantiationException ex) {
Logger.getLogger(D.class.getName()).log(Level.SEVERE, null, ex);
} catch (IllegalAccessException ex) {
Logger.getLogger(D.class.getName()).log(Level.SEVERE, null, ex);
}
return result;
}
}
public class Main {
public static void main(String... args) {
D d = new D();
System.out.println(d);
}
}
run:
A
BUILD SUCCESSFUL (total time: 0 seconds)
I have had situations like these when the architecture is to build common functionality in a common CustomBaseClass which implements on behalf of several derived classes.
However, we need to circumvent common logic for specific method for a specific derived class. In such cases, we must use a super.super.methodX implementation.
We achieve this by introducing a boolean member in the CustomBaseClass, which can be used to selectively defer custom implementation and yield to default framework implementation where desirable.
...
FrameworkBaseClass (....) extends...
{
methodA(...){...}
methodB(...){...}
...
methodX(...)
...
methodN(...){...}
}
/* CustomBaseClass overrides default framework functionality for benefit of several derived classes.*/
CustomBaseClass(...) extends FrameworkBaseClass
{
private boolean skipMethodX=false;
/* implement accessors isSkipMethodX() and setSkipMethodX(boolean)*/
methodA(...){...}
methodB(...){...}
...
methodN(...){...}
methodX(...){
if (isSkipMethodX()) {
setSKipMethodX(false);
super.methodX(...);
return;
}
... //common method logic
}
}
DerivedClass1(...) extends CustomBaseClass
DerivedClass2(...) extends CustomBaseClass
...
DerivedClassN(...) extends CustomBaseClass...
DerivedClassX(...) extends CustomBaseClass...
{
methodX(...){
super.setSKipMethodX(true);
super.methodX(...);
}
}
However, with good architecture principles followed in framework as well as app, we could avoid such situations easily, by using hasA approach, instead of isA approach. But at all times it is not very practical to expect well designed architecture in place, and hence the need to get away from solid design principles and introduce hacks like this.
Just my 2 cents...
IMO, it's a clean way to achieve super.super.sayYourName() behavior in Java.
public class GrandMa {
public void sayYourName(){
System.out.println("Grandma Fedora");
}
}
public class Mama extends GrandMa {
public void sayYourName(boolean lie){
if(lie){
super.sayYourName();
}else {
System.out.println("Mama Stephanida");
}
}
}
public class Daughter extends Mama {
public void sayYourName(boolean lie){
if(lie){
super.sayYourName(lie);
}else {
System.out.println("Little girl Masha");
}
}
}
public class TestDaughter {
public static void main(String[] args){
Daughter d = new Daughter();
System.out.print("Request to lie: d.sayYourName(true) returns ");
d.sayYourName(true);
System.out.print("Request not to lie: d.sayYourName(false) returns ");
d.sayYourName(false);
}
}
Output:
Request to lie: d.sayYourName(true) returns Grandma Fedora
Request not to lie: d.sayYourName(false) returns Little girl Masha
I think this is a problem that breaks the inheritance agreement.
By extending a class you obey / agree its behavior, features
Whilst when calling super.super.method(), you want to break your own obedience agreement.
You just cannot cherry pick from the super class.
However, there may happen situations when you feel the need to call super.super.method() - usually a bad design sign, in your code or in the code you inherit !
If the super and super super classes cannot be refactored (some legacy code), then opt for composition over inheritance.
Encapsulation breaking is when you #Override some methods by breaking the encapsulated code.
The methods designed not to be overridden are marked
final.
In C# you can call a method of any ancestor like this:
public class A
internal virtual void foo()
...
public class B : A
public new void foo()
...
public class C : B
public new void foo() {
(this as A).foo();
}
Also you can do this in Delphi:
type
A=class
procedure foo;
...
B=class(A)
procedure foo; override;
...
C=class(B)
procedure foo; override;
...
A(objC).foo();
But in Java you can do such focus only by some gear. One possible way is:
class A {
int y=10;
void foo(Class X) throws Exception {
if(X!=A.class)
throw new Exception("Incorrect parameter of "+this.getClass().getName()+".foo("+X.getName()+")");
y++;
System.out.printf("A.foo(%s): y=%d\n",X.getName(),y);
}
void foo() throws Exception {
System.out.printf("A.foo()\n");
this.foo(this.getClass());
}
}
class B extends A {
int y=20;
#Override
void foo(Class X) throws Exception {
if(X==B.class) {
y++;
System.out.printf("B.foo(%s): y=%d\n",X.getName(),y);
} else {
System.out.printf("B.foo(%s) calls B.super.foo(%s)\n",X.getName(),X.getName());
super.foo(X);
}
}
}
class C extends B {
int y=30;
#Override
void foo(Class X) throws Exception {
if(X==C.class) {
y++;
System.out.printf("C.foo(%s): y=%d\n",X.getName(),y);
} else {
System.out.printf("C.foo(%s) calls C.super.foo(%s)\n",X.getName(),X.getName());
super.foo(X);
}
}
void DoIt() {
try {
System.out.printf("DoIt: foo():\n");
foo();
Show();
System.out.printf("DoIt: foo(B):\n");
foo(B.class);
Show();
System.out.printf("DoIt: foo(A):\n");
foo(A.class);
Show();
} catch(Exception e) {
//...
}
}
void Show() {
System.out.printf("Show: A.y=%d, B.y=%d, C.y=%d\n\n", ((A)this).y, ((B)this).y, ((C)this).y);
}
}
objC.DoIt() result output:
DoIt: foo():
A.foo()
C.foo(C): y=31
Show: A.y=10, B.y=20, C.y=31
DoIt: foo(B):
C.foo(B) calls C.super.foo(B)
B.foo(B): y=21
Show: A.y=10, B.y=21, C.y=31
DoIt: foo(A):
C.foo(A) calls C.super.foo(A)
B.foo(A) calls B.super.foo(A)
A.foo(A): y=11
Show: A.y=11, B.y=21, C.y=31
It is simply easy to do. For instance:
C subclass of B and B subclass of A. Both of three have method methodName() for example.
public abstract class A {
public void methodName() {
System.out.println("Class A");
}
}
public class B extends A {
public void methodName() {
super.methodName();
System.out.println("Class B");
}
// Will call the super methodName
public void hackSuper() {
super.methodName();
}
}
public class C extends B {
public static void main(String[] args) {
A a = new C();
a.methodName();
}
#Override
public void methodName() {
/*super.methodName();*/
hackSuper();
System.out.println("Class C");
}
}
Run class C Output will be:
Class A
Class C
Instead of output:
Class A
Class B
Class C
If you think you are going to be needing the superclass, you could reference it in a variable for that class. For example:
public class Foo
{
public int getNumber()
{
return 0;
}
}
public class SuperFoo extends Foo
{
public static Foo superClass = new Foo();
public int getNumber()
{
return 1;
}
}
public class UltraFoo extends Foo
{
public static void main(String[] args)
{
System.out.println(new UltraFoo.getNumber());
System.out.println(new SuperFoo().getNumber());
System.out.println(new SuperFoo().superClass.getNumber());
}
public int getNumber()
{
return 2;
}
}
Should print out:
2
1
0
public class SubSubClass extends SubClass {
#Override
public void print() {
super.superPrint();
}
public static void main(String[] args) {
new SubSubClass().print();
}
}
class SuperClass {
public void print() {
System.out.println("Printed in the GrandDad");
}
}
class SubClass extends SuperClass {
public void superPrint() {
super.print();
}
}
Output: Printed in the GrandDad
The keyword super is just a way to invoke the method in the superclass.
In the Java tutorial:https://docs.oracle.com/javase/tutorial/java/IandI/super.html
If your method overrides one of its superclass's methods, you can invoke the overridden method through the use of the keyword super.
Don't believe that it's a reference of the super object!!! No, it's just a keyword to invoke methods in the superclass.
Here is an example:
class Animal {
public void doSth() {
System.out.println(this); // It's a Cat! Not an animal!
System.out.println("Animal do sth.");
}
}
class Cat extends Animal {
public void doSth() {
System.out.println(this);
System.out.println("Cat do sth.");
super.doSth();
}
}
When you call cat.doSth(), the method doSth() in class Animal will print this and it is a cat.

RxJava/Retrofit - How can I force users to use a specific subclass of Subscriber?

Before using rx.Observable, I used a custom callback with retrofit so I can add some specific logic for handling response/error and not have to do that inside the callback for every request as boilerplate code.
I force users to use the custom callback by putting it in the method signature like this:
#GET("/user_endpoint/")
void getUser(CustomCallback<User> callback);
#GET("/profile_endpoint/")
void getProfile(CustomCallback<Profile> callback);
but now that I'm returning an Observable:
#GET("/user_endpoint/")
Observable<User> getUser();
#GET("/profile_endpoint/")
Observable<Profile> getProfile();
I can't figure out a way to make sure that a custom callback always proxies the error/response.
Also, with retrofit2.0, how can I force the user to use a custom callback with the returned Call object?
CustomCallback for reference:
public abstract class CustomCallback<T> implements Callback<T> {
#Override public final void success(T t, Response response) {
// do some logic
onSuccess(t);
}
#Override public final void failure(RetrofitError error) {
// do something with the error here such as show a Toast
Toast.makeText(Application.getInstance(), error.getLocalizedMessage(), Toast.LENGTH_SHORT).show();
onFailure(error);
}
public abstract void onSuccess(T response);
public abstract void onFailure(Throwable error);
}
Stop. You're thinking this the wrong way.
Instead consider this: You have the normal Retrofit interface:
interface Foo {
#GET("/user_endpoint/")
Observable<User> getUser();
}
And then you have your decorator class:
public class FooDecorator implements Foo {
private Foo delegate = ...; // inject or create the Retrofit instance.
#Override
public Observable<User> getUser() {
return delegate.getUser().doOnNext(...).doOnError(...);
}
}
Then you use only the second class everywhere in your code (preferably just let the DI system use that) and you're set.
If you're feeling adventurous, you could even adapt the RxJavaCallAdapterFactory so that it modifies the returned observables without the need of a custom class.

Here-api offline maps installation

In MapEngine initialization I want to install all packages but I am stuck here installMapPackages(List packageIdList) from where can I find List packageIdList.
You should use the MapLoader#getMapPackages() API to retrieve the root MapPackage object. You can then use the MapPackage#getId() method to find the Id's of the countries/regions you wish to install. Note that the MapPackage object is not returned directly from the MapLoader#getMapPackages() call, but instead through a Listener object. You must provide your own MapLoader.Listener implementation and set it by way of the MapLoader#addListener(MapLoader.Listener listener) method before calling getMapPackages().
For Example:
MapLoader.Listener mapLoaderListener = new MapLoader.Listener() {
public void onUninstallMapPackagesComplete(MapPackage rootMapPackage,
MapLoader.ResultCode mapLoaderResultCode) {
}
public void onProgress(int progressPercentage) {
}
public void onPerformMapDataUpdateComplete(MapPackage rootMapPackage,
MapLoader.ResultCode mapLoaderResultCode) {
}
public void onInstallationSize(long diskSize, long networkSize) {
}
public void onInstallMapPackagesComplete(MapPackage rootMapPackage,
MapLoader.ResultCode mapLoaderResultCode) {
}
public void onGetMapPackagesComplete(MapPackage rootMapPackage,
MapLoader.ResultCode mapLoaderResultCode) {
// This method will be called after MapLoader#getMapPackages()
// is called
// You can use the rootMapPackage object to find the Id's to
// pass to installMapPackages()
}
public void onCheckForUpdateComplete(boolean updateAvailable,
String currentMapVersion,String newestMapVersion,
MapLoader.ResultCode mapLoaderResultCode) {
}
};
MapLoader mapLoader = MapLoader.getInstance();
mapLoader.addListener(mapLoaderListener);
mapLoader.getMapPackages();
Further details here:
Developer Guide
https://developer.here.com/mobile-sdks/documentation/android-hybrid-plus/topics/maps-offline.html
API Reference
https://developer.here.com/mobile-sdks/documentation/android-hybrid-plus/topics_api_nlp_hybrid_plus/com-here-android-mpa-odml-maploader.html
https://developer.here.com/mobile-sdks/documentation/android-hybrid-plus/topics_api_nlp_hybrid_plus/com-here-android-mpa-odml-maploader-listener.html
https://developer.here.com/mobile-sdks/documentation/android-hybrid-plus/topics_api_nlp_hybrid_plus/com-here-android-mpa-odml-mappackage.html
The only way (I think), is to recursively call getChildren() on MapPackage, then check getTitle() for each of the children package to find the region you need.
For example, to get the id of the "Bretagne" region in france, you need to go through rootMapPackage.getChildren().get(2/Europe/).getChildren().get(1/France/).getChildren().get(3/Bretagne/).getId()
Not very convenient. A method "search(String title)" on the root package would be handy.

android interface make callback function optional

I have a simple class with an interface enabled and works proper when used.
interface interfacename{
void function1();
void function2();
}
public class asyncfunction(){
public interfacename listener;
...
onasyncStart( ... ){
listener.function1();
}
...
...
onasyncComplete( ... ){
listener.function2();
}
}
public myclass(){
....
....
methodcall(new interfacename(){
#Override
public void function1(){
// executes proper
}
#Override
public void function2(){
// executes proper
}
}
}
So the above method works proper.
But i want to call only the function1() sometimes and only function2() when needed.
I don't want both methods to be implemented always. The code looks big and im not sure if it slows down code or not ( not on the milli second level btw ) but it would be really nice if there was another way to have the option to execute particular call backs when needed.
It sounds like you're really looking at splitting up your interface into multiple interfaces, and change the method that accepts this interface as a parameter, so that it will instead accept the interface that it requires (e.g. InterfaceOne) in order to call a method in that interface (e.g. function1()). Another method might want to call function2(), in which case it will accept an argument of type InterfaceTwo.
If however you need to always call both methods of the interface in your method, but you don't always need to call any code in the methods of that interface, what you're looking at instead is the following.
Instead of creating a new anonymous class of type interfacename, you could use a base class with empty method bodies, and simply override the ones you need. Methods implemented by the abstract base class are essentially optional, while those that are not implemented are required methods.
This is a very common pattern in Java development.
public interface InterfaceName {
void function1();
void function2();
}
public abstract class BaseInterfaceName implements InterfaceName {
public void function1() {
}
public void function2() {
}
}
public class MyClass {
public void myMethod() {
myMethodWithInterface(new BaseInterfaceName() {
#Override
public void function2() {
System.out.println("function2");
}
})
}
public void myMethodWithInterface(InterfaceName intf) {
intf.function1();
intf.function2();
}
}
A possible solution is the one explained by #Nicklas.
But, if you use Java 8, you can use the default method. So you can declare your interface in this way:
public interface InterfaceName {
default void function1(){ /* do nothing */}
default void function2(){ /* do nothing */}
}
So, you can avoid implementing the methods, since you are providing a default implementation. In my example the default is to do nothing, but of course, you can personalize them.

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