Converting a string to an integer on Android

Question

See the Integer class and the static parseInt() method:

http://developer.android.com/reference/java/lang/Integer.html

Integer.parseInt(et.getText().toString());

You will need to catch NumberFormatException though in case of problems whilst parsing, so:

int myNum = 0;

try {
    myNum = Integer.parseInt(et.getText().toString());
} catch(NumberFormatException nfe) {
   System.out.println("Could not parse " + nfe);
} 

int in = Integer.valueOf(et.getText().toString());
//or
int in2 = new Integer(et.getText().toString());

Use regular expression:

String s="your1string2contain3with4number";
int i=Integer.parseInt(s.replaceAll("[\\D]", ""));

output: i=1234;

If you need first number combination then you should try below code:

String s="abc123xyz456";
int i=NumberFormat.getInstance().parse(s).intValue();

output: i=123;

Use regular expression:

int i=Integer.parseInt("hello123".replaceAll("[\\D]",""));
int j=Integer.parseInt("123hello".replaceAll("[\\D]",""));
int k=Integer.parseInt("1h2el3lo".replaceAll("[\\D]",""));

output:

i=123;
j=123;
k=123;

Use regular expression is best way to doing this as already mentioned by ashish sahu

public int getInt(String s){
return Integer.parseInt(s.replaceAll("[\\D]", ""));
}

Try this code it's really working.

int number = 0;
try {
    number = Integer.parseInt(YourEditTextName.getText().toString());
} catch(NumberFormatException e) {
   System.out.println("parse value is not valid : " + e);
}