I'm coming from C#, so typically I try to relate everything that i'm doing.
I cannot figure out why the below statement doesn't work. Basically String val = "admin". Then an I have an if statement, however the if statement is always false. I'm sure it's something simple.
Thanks!
EditText edt = (EditText) findViewById(R.id.email);
//String val = edt.getText().toString();
String val = "admin";
EditText edt2 = (EditText) findViewById(R.id.password);
String val2 = edt2.getText().toString();
if(val.toString() == "admin") {
String hero = val;
}
You should use
if (val.equals("admin")) {
String hero = val;
}
instead of using an equal sign. Using an equal sign in java is asking if they're the same object, which will be false even if the strings are the same.
Also, be careful with what you're doing inside of the if statement, because the variable "hero" won't be accessible outside of that block.
In Java you can't compare strings using ==
You need to change your if statment like this
if(val.equals("admin")){}
First of all you have never changed the value of String val to anything so there is no need to try convert it to a string in your if statement.
String val = "admin";
if (val == "admin") {
//code here
}else{
//code here
}
Hope this helps
In java, == operator check the address of each value, and equals() method check the value.
So If you want to compare the value of each string, you should use the equals() method.
Please search for the concept of 'call by reference' and 'call by value'.
And you already declare val to String, so it didn't need toString().
if(val.equals("admin")) {
String hero = val;
}
I'm surprised no one mentioned the difference between .matches() and .equals() depending on your needs, what you could also be looking for is .matches()
if(val.toString().matches("admin")) {
String hero = val;
}
Matches checks the match of a String to a regular expression pattern, not the same string.
For example:
"hello".equals(".*e.*"); // false
"hello".matches(".*e.*"); // true
use .equals() instead of ==.
for example:
if (val.equals("admin")) ...
Related
if condition doesn't validate the value inside string mentioned below.
//getting input box value
newText = input.getText().toString();
//using if condition newText got value "a"
if (newText=="a")
{
//do something
}
but the above if condition doesn't work i check string got right value which is a.
thanks in advance
In java you should use equals method to compare the values of strings:
if ("a".equals(newText)) {
//do something
}
To compare the references you can use ==. For more information check this: https://stackoverflow.com/a/513839/3225458
I want to split string got from bluetooth. i'm using
StringTokenizer splitStr = new StringTokenizer(readMessage, "\\|");
String numberSpeed = splitStr.nextToken(); //splitStr[0].replaceAll("\\D+","");
String numberTorque = splitStr.nextToken(); //splitStr[1].replaceAll("\\D+","");
numberSpeed = numberSpeed.replaceAll("\\D+","");
numberTorque = numberTorque.replaceAll("\\D+","");
Did it with split string before.
If i get corupted data without delimiter the app crashes while trying to do impossible.
How to check if there is delimiter or not and then proceed split or skip it?
you can check for delimeter in string by contains() method
if(str.contains("_your_delimiter")) { //for safe side convert your delimeter and search string to lower case using method toLowerCase()
//do your work here
}
Try this, I use it in my app.
String container = numberSpeed ;
String content = "\\D+";
boolean containerContainsContent = StringUtils.containsIgnoreCase(container, content);
It will return true if it has delimiter, and false it not.
Use that with an if statement.
ex.
if(containerContainsContent){
//split it
} else {
//skip it
}
This is quote from tokenizer docs: StringTokenizer is a legacy class that is retained for compatibility reasons although its use is discouraged in new code.
Try to user String.split() instead.
if(str.contains(DEILIMITER)) {
String tab[] = str.split(DEILIMITER);
//enter code here
}
This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 9 years ago.
I can't seem to get the code within the second if statement to execute. I have logged both values being compared and also ran the debugger and checked them. They are both "a". It always shows the incorrectPasswordDialog. This question seems difficult because it seems as though it should just work but any help is appreciated.
private void logUserIn(AppUser user) {
if (user != null){
Log.d("mPassword: ", mPassword);
Log.d("user.getPassword(): ", user.getPassword());
String userPassword = user.getPassword().toString();
String formPassword = mPassword.toString();
if ( userPassword == formPassword ){
Intent welcomePage = new Intent(this, StartScreenActivity.class);
welcomePage.putExtra("name", mName);
startActivity(welcomePage);
}
else {
showIncorrectPasswordDialog();
}
}else {
showIncorrectUserNameDialog();
}
}
You are comparing the object identity. Use string.equals() to check for equivalence.
if(userPassword.equals(formPassword)){
}
Change
if ( userPassword == formPassword ){
to
if ( userPassword.equals(formPassword) ){
== compares object references, whereas .equals compared String values.
You can not compare string using ==
try like this
Compares the specified object to this string and returns true if they are equal. The object must be an instance of string with the same characters in the same order.
if ( userPassword.equals(formPassword)){
// do some stuff
}
or
Compares the specified string to this string ignoring the case of the characters and returns true if they are equal.
if(userPassword.equalsIgnoreCase(formPassword))
{
//do some stuff
}
In JAVA to compare between strings, you should use:
if(userPassword.equals(formPassword)) {
// they are equal
}
Change
if ( userPassword == formPassword ){
to
if ( userPassword.equals(formPassword)){
In Java, String comparison is done with .equals(). Using == compares the object reference and not their values.
Java String comparison
In JAVA you should use equals to judge equal of value of different String.
== is used to judge object pointer for string, so it would only return false in your case.
I've a string value String str="Success"
How to check the "str" value is not equal to "Success". If I try if(str!="Success"). It is not working properly.
I've a integer value int a=250 If I use If(a!=250) this is also not working properly.
How to code for these conditional statements ? I'm using Android and Eclipse version 2.1.
Any help would be appreciable.
a!=250 should work.
For string try str.equals("Success")
As every one said, String values should compare with .equals method.
like
if(str.equals("Success"))
{
System.out.println("My String is Success");
}
Next coming to integer comparison.
As you have said, you have the integer value
int a=250
if(a!=250)
{
// Some code
}
Here already your a value is 250. then you are executing some lines of code if a value is not equal to 250. Then how the condition will execute. if you want to test then
change the a value and then check again as
int a=50
if(a!=250)
{
// Some code===============> Now this code will execute
}
For string comparison, use the standard String.equals() method.
str.equals("Success"); - For Equals
if(!str.equals("Success")) - For Not Equals(your case)
And for int, what is being used, is proper. a!=250 will return false in your case. Hence, it'll not enter the if block.
You should try to study Java first ;)
This is not the way to do a String comparison. For String comparison, you must use .equals(String) method.
String str1 = "str1";
String str2 = "str2";
if(str1.equals(str2)) {
//do something
}
You should not compare strings with != or =. That won't work in most cases.
new String("test").equals("test")
Should be used for comparison.
!= compares the value and is true if they are NOT equal.
you could also use something like if(a == 250){...}
In Java you cannot use == to compare Strings, you must use:
if(string.equals("example"))
So let's use equals() in your conditional and optimize it:
if(!str.equals("Success"))
this will work
I am trying to validate my string with the regular expression. Here is what I am trying to do
EditText serialText = (EditText) findViewById(R.id.pinText);
serialText.setVisibility(View.VISIBLE);
serialNumber = serialText.getText().toString();
I am storing the serial number in serialNumber
I have the following method to match the regular expression
boolean isRegularSerialNumber(String pinNumber) {
// regular expression to be matched against
String regularString = "[0-9]{4}-[0-9]{4}-[0-9]{4}-[0-9]{4}";
Pattern pattern = Pattern.compile(regularString);
Matcher matcher = pattern.matcher(pinNumber);
boolean isRegularSerialNumberValid ;
if (pinNumber.matches(regularString))
isRegularSerialNumberValid = true;
else
isRegularSerialNumberValid = false;
return isRegularSerialNumberValid;
}
But I am not able to match this.
Any answer for this? Hope Pattern and Matcher are the right one for this.
What I am trying to do is this, this matched serialNumber I am validating against serial number stored in the database. If match found, it returns success or else failure. And i have entered the exact serial number which is stored in the database but even then it returns failure.
I followed the method what #Stevehb said and i got the match true in that case.
This is how I am sending my data
parameter.add(new BasicNameValuePair("validate", serialNumber));
Breaking my head on this.
The built in String functions should work by themselves. isRegularSerialNumber() could just be
boolean isRegularSerialNumber(String pinNumber) {
String regularString = "[0-9]{4}-[0-9]{4}-[0-9]{4}-[0-9]{4}";
return pinNumber.matches(regularString);
}
This works for me when I tested 1234-5678-9012-1324 (true) and 12-1234-123-1324 (false).
Also, it looks like you're maybe grabbing the input string from serialText right after you make it visible. Could your problem be in grabbing the text before the user has made any input?
looks much alike .net regex code.
instead of
if (pinNumber.matches(regularString))
try
if (matcher.matches())