Develop Reference

How to get last 4 digit of number in android?

I have an app which contain mobile number edit text in which user can edit mobile number and I have to send two request to server like:- mobile number and mssdn,mobile number(which is full lenghth ) and mssdn(which contain mobile number last 4 digit).How can I do that

Try this. Check for length greater than 4 before calling subString to avoid IndexOutOfBounds Exception.
EditText mEdtPhoneNumber = (EditText) findViewById(R.id.edtPhoneNumber);
String phoneNumber = mEdtPhoneNumber.getText().toString().trim();
String strLastFourDi = phoneNumber.length() >= 4 ? phoneNumber.substring(phoneNumber.length() - 4): "";
Also what is mssdn?? Is it msisdn??

Use the modulus (%) operator:
To get the last digit: use number % 10
To get the last 2 digits: use number % 100
and so on
For example:
42455%10000 = 2455

You could do something like this:
EditText phoneNumberEditText = (EditText) findViewById(R.id.phoneNumberEditText);
String phoneNumber = phoneNumberEditText.getText().toString();
String lastFourDigits = phoneNumber.substring(phoneNumber.length() - 4);

you should use regex because this will only give you result if the last four letters are actually numbers on the other hand the substring function simply give you last four letters no matter they are numbers or characters. e.g 4344sdsdss4 will give you dss4 which is clearly not a part of phone number
String str="4444ntrjntkr555566";
Pattern p = Pattern.compile("(\\d{4})$");
Matcher m = p.matcher(str);
if (m.find()) {
System.out.println(m.group(m.groupCount()));
}
this will produce 5566
Working
//d mean digits
{4} for fix length as 4
$ mean at the end

List<Integer> f(String str){
ArrayList<Integer> digits = new ArrayList<>();
if (null == str || str.length() < 4){
Log.i(LOG_TAG, "there are less than 4 digits");
return digits;
}
String digitsStr = str.substring(str.length() - 4);
for (char c : digitsStr.toCharArray()){
try {
Integer digit = Integer.parseInt(String.valueOf(c));
digits.add(digit);
} catch (Exception e){
continue;
}
}
return digits;
}

We can also use a new method introduced in kotlin: takeLast(n)
fun getLastDigit(data: String, n:Int): String {
return if(data.length > n){
data.takeLast(n)
}else {
""
}
}

How to Validate Phone Number format

i am about to create a validation for phone number format..The format is 10 digit including the plus sign eg:+0133999504. Even though I have declare the pattern which is I try to disallow the "-" symbol or any other characters, but the validation is not working. Any other Idea or solution?
1st I declared the string regex:
String PhoneNo;
String PhoneNo_PATTERN ="[\\+]\\d{3}\\d{7}";
2nd I make a if..else statement:
{
Pattern pattern = Pattern.compile(PhoneNo_PATTERN);
Matcher matcher = pattern.matcher(PhoneNo);
if (!matcher.matches())
{
inputemergencyContactNo.setError("Please enter Emergency Contact No");
}
else{
Toast.makeText(RiderProfile.this, "Please filled in All field", Toast.LENGTH_SHORT).show();
}

Why not remove all non-digits and then count the digits left and put the plus back in later? This allows users the freedom to fill out their phone number anyway they want...
String PhoneNo = "+123-456 7890";
String Regex = "[^\\d]";
String PhoneDigits = PhoneNo.replaceAll(Regex, "");
if (PhoneDigits.length()!=10)
{
// error message
}
else
{
PhoneNo = "+";
PhoneNo = PhoneNo.concat(PhoneDigits); // adding the plus sign
// validation successful
}
If your app is intended for international use replace
if (!PhoneDigits.length()!=10)
with
if(PhoneDigits.length() < 6 || PhoneDigits.length() > 13)
as Fatti Khan suggested.
To apply this in the code you posted at Android EditText Validation and Regex first include this method in your public class or the class containing onClick():
public boolean validateNumber(String S) {
String Regex = "[^\\d]";
String PhoneDigits = S.replaceAll(Regex, "");
return (PhoneDigits.length()!=10);
}
And include this method in the CreateNewRider class:
protected String tidyNumber(String S) {
String Regex = "[^\\d]";
String PhoneDigits = S.replaceAll(Regex, "");
String Plus = "+";
return Plus.concat(PhoneDigits);
}
This is where the validation happens...
#Override
public void onClick(View view) {
Boolean b = false;
if(inputfullname.getText().toString().equals("")) b = true;
else if(... // do this for all fields
else if(inputmobileNo.getText().toString().equals("")) b=true;
else if(inputemergencyContactNo.getText().toString().equals("")) b=true;
else {
if(validateNumber( inputmobileNo.getText().toString() )
Toast.makeText(RiderProfile.this, "Invalid mobile number", Toast.LENGTH_SHORT).show();
else if(validateNumber( inputemergencyContactNo.getText().toString() )
Toast.makeText(RiderProfile.this, "Invalid emergency contact number", Toast.LENGTH_SHORT).show();
else {
// Validation succesful
new CreateNewRider().execute();
}
}
if(b) Toast.makeText(RiderProfile.this, "Please filled in All field", Toast.LENGTH_SHORT).show();
}
And then use tidyNumber() in the CreateNewRider class:
protected String doInBackground(String... args) {
String fullname= inputfullname.getText().toString();
String IC= inputIC.getText().toString();
String mobileNo= tidyNumber( inputmobileNo.getText().toString() );
String emergencyContactName= inputemergencyContactName.getText().toString() );
String emergencyContactNo= tidyNumber( inputemergencyContactNo.getText().toString() );
...

Given the rules you specified:
upto length 13 and including character + infront.
(and also incorporating the min length of 10 in your code)
You're going to want a regex that looks like this:
^\+[0-9]{10,13}$
With the min and max lengths encoded in the regex, you can drop those conditions from your if() block.
Off topic: I'd suggest that a range of 10 - 13 is too limiting for an international phone number field; you're almost certain to find valid numbers that are both longer and shorter than this. I'd suggest a range of 8 - 20 to be safe.
[EDIT] OP states the above regex doesn't work due to the escape sequence. Not sure why, but an alternative would be:
^[+][0-9]{10,13}$
[EDIT 2] OP now adds that the + sign should be optional. In this case, the regex needs a question mark after the +, so the example above would now look like this:
^[+]?[0-9]{10,13}$
For Valid Mobile You need to consider 7 digit to 13 digit because some country have 7 digit mobile number . Also we can not check like mobile number must starts with 9 or 8 or anything..
For mobile number I used this this Function
private boolean isValidMobile(String phone2)
{
boolean check;
if(phone2.length() < 6 || phone2.length() > 13)
{
check = false;
txtPhone.setError("Not Valid Number");
}
else
{
check = true;
}
return check;
}

^[\\+]\\d{3}\\d{7}$
Use anchors to limit the match.
^ => start of match
$=> end of match

To validate India's mobile number.
Your edit text input
edt_mobile.text.toString().trim()
Number validation method
fun isValidMobile(phone: String): Boolean {
return phone.matches(Constants.REGEX_MOBILE.toRegex()) && phone.trim().length == 10
}
Regression expression
const val REGEX_MOBILE = "^[6-9]{1}[0-9]{9}\$"

Android: Converting a set of functions into an equation

I am 99% sure that this cannot be done, however I thought I would ask to be certain.
I am attempting to create an application that calculates the required dice roll for an action in a popular tabletop war game.
The following is this calculation in Java
int x = ((WSattacker * 2) - WSdefender);
int y = (WSattacker - WSdefender);
String result;
// Calculation for a +5
if (x <= -1) {
result = "5+";
}
// Calculation for a +4
else if (x >= 0 && y <= 0) {
result = "4+";
}
// Calculation for a +3
else if (y > 0) {
result = "3+";
} else {
result = "Error";
}
return result;
Now my issue is that to avoid copywriter infringement I cannot mention the name of the game in my application, and probably cannot hard code the above calculation in the app.
This means that it is difficult to tell a potential user what the app will do.
The only solution I can think of is to make the application generic and allow the user to input the calculation required in the form of an equation.
An equation that I can place anonymously on a public board or similar.
Therefore my questions are as follows.
Is there another way of going about this?
If no, is it possible to condense the above code into a single expression/ equationi.e. one that removes the if and else statements

To answer question 2:
result = test_condition_1 ? result2_if_true : (test_condition_2 ? result2_if_true : test3_or_result2);
You can then build up 'compound' test conditions this way, and it's based upon ternary operators.
EDIT
Ternary operators are a short-hand way of writing if..then..else statments, and more information can be found in the wiki-link above. An example of its use is below, which you can compile and run:
public class TernaryTest {
public static void main(String [] args){
int x = 14;
int y = 5;
String result = ( x <= 10 ) ? "Less than 10" : "More than 10";
System.out.println("Result is: " + result);
}
}
Try running it and see the result as you change the value of x to understand how it works. Then it's possible to extend it to include and else by replacing the "more than 10" string.

Loop structure gives wrong result

I am trying to compare items out of my DB to the value of an EditText (user input). The answer can have multiple answers, seperated by a ','. I first put them into a stringarray and then compare them to the answer. The LevenshteinDistance checks if the answer is more or les good (http://en.wikipedia.org/wiki/Levenshtein_distance#Computing_Levenshtein_distance).
userAnswer = etUserAnswer.getText().toString().toLowerCase();
String[] answers = qAnswer.split(",");
for (String answer : answers) {
if (answer.equals(userAnswer)) {
Toast.makeText(getApplicationContext(), ("Answer Correct"),
Toast.LENGTH_SHORT).show();
tvMessage.setText("You smartass!");
} else {
Toast.makeText(getApplicationContext(), ("Wrong"),
Toast.LENGTH_SHORT).show();
points = points - 4;
String answerGood = answer.toLowerCase();
LevenshteinDistance lDistance = new LevenshteinDistance();
int comparisonCheck = lDistance.computeLevenshteinDistance(
userAnswer, answerGood);
if (comparisonCheck == 1) {
tvMessage.setText("Almost there, but not quite yet!");
} else if (comparisonCheck > 1) {
tvMessage.setText("Are you serious, totally wrong?!");
}
}
}
Suppose I am having the answers for a question in the DB as follows: tree,test,radio
I am having two problems:
1. When I type "radi" it gives me 'Almost there...' which is good. It should also give me this if I enter "tes", but instead it gives me the 'Are you serious,...' line. I guess it keeps comparing to the last one.
2. Every time I type in something which is not correct, I get -12 instead of -4. I suppose this is due to the fact I am having three answers and it loops three times.. but I don't know how I can make it count only once..
Anyone can help me on the way? Thanks!

Assuming you don't need to know the word which gives the least Levenshtein distance, you could modify your loop to find smallest distance only;
userAnswer = etUserAnswer.getText().toString().toLowerCase();
String[] answers = qAnswer.split(",");
LevenshteinDistance lDistance = new LevenshteinDistance();
int minDistance = lDistance.computeLevenshteinDistance(
userAnswer, answers[0].toLowerCase());
for (int i = 1; i < answers.length; ++i) {
minDistance = Math.min(minDistance, lDistance.computeLevenshteinDistance(
userAnswer, answers[i].toLowerCase()));
}
if (minDistance == 0) {
// Correct answer...
} else {
// Wrong answer...
points -= 4;
// etc etc...
}

Get phone number without country code for the purpose of comparing numbers

I can obtain the phone number from an incoming call or from a sms message. unfortunately, in case of the SMS there might be the country code in it. So, basically I need to obtain the plain phone number, without country code, in order to compare it with existing numbers in Contacts.

If you want to compare phone numbers you can always use the
PhoneNumberUtils.compare(number1, number2);
or
PhoneNumberUtils.compare(context, number1, number2);
Then you don't have to worry about the country code, it will just compare the numbers from the reversed order and see if they match (enough for callerID purposes at least).

fast untested approach (AFAIK phone numbers have 10 digits):
// As I said, AFAIK phone numbers have 10 digits... (at least here in Mexico this is true)
int digits = 10;
// the char + is always at first.
int plus_sign_pos = 0;
// Always send the number to this function to remove the first n digits (+1,+52, +520, etc)
private String removeCountryCode(String number) {
if (hasCountryCode(number)) {
// +52 for MEX +526441122345, 13-10 = 3, so we need to remove 3 characters
int country_digits = number.length() - digits;
number = number.substring(country_digits);
}
return number;
}
// Every country code starts with + right?
private boolean hasCountryCode(String number) {
return number.charAt(plus_sign_pos) == '+'; // Didn't String had contains() method?...
}
then you just call these functions