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java – Converting a string to an integer on Android

Posted by: admin March 10, 2020 Leave a comment

Questions:

How do I convert a string into an integer?

I have a textbox I have the user enter a number into:

EditText et = (EditText) findViewById(R.id.entry1);
String hello = et.getText().toString();

And the value is assigned to the string hello.

I want to convert it to a integer so I can get the number they typed; it will be used later on in code.

Is there a way to get the EditText to a integer? That would skip the middle man. If not, string to integer will be just fine.

How to&Answers:

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);
} 

Answer:

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

Answer:

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;

Answer:

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;

Answer:

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]", ""));
}

Answer:

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);
} 

Answer:

Best way to convert your string into int is :

 EditText et = (EditText) findViewById(R.id.entry1);
 String hello = et.getText().toString();
 int converted=Integer.parseInt(hello);

Answer:

You should covert String to float. It is working.

float result = 0;
 if (TextUtils.isEmpty(et.getText().toString()) {
  return;
}

result = Float.parseFloat(et.getText().toString());

tv.setText(result); 

Answer:

You can use the following to parse a string to an integer:

int value=Integer.parseInt(textView.getText().toString());

(1) input: 12 then it will work.. because textview has taken this 12 number as “12” string.

(2) input: “abdul” then it will throw an exception that is NumberFormatException.
So to solve this we need to use try catch as I have mention below:

  int tax_amount=20;
  EditText edit=(EditText)findViewById(R.id.editText1);
     try
       {

        int value=Integer.parseInt(edit.getText().toString());
        value=value+tax_amount;
        edit.setText(String.valueOf(value));// to convert integer to string 

       }catch(NumberFormatException ee){
       Log.e(ee.toString());
       }

You may also want to refer to the following link for more information:
http://developer.android.com/reference/java/lang/Integer.html

Answer:

You can also do it one line:

int hello = Integer.parseInt(((Button)findViewById(R.id.button1)).getText().toString().replaceAll("[\D]", ""));

Reading from order of execution

  1. grab the view using findViewById(R.id.button1)
  2. use ((Button)______) to cast the View as a Button
  3. Call .GetText() to get the text entry from Button
  4. Call .toString() to convert the Character Varying to a String
  5. Call .ReplaceAll() with "[\\D]" to replace all Non Digit Characters with “” (nothing)
  6. Call Integer.parseInt() grab and return an integer out of the Digit-only string.

Answer:

The much simpler method is to use the decode method of Integer so for example:

int helloInt = Integer.decode(hello);

Answer:

Kotlin

There are available Extension methods to parse them into other primitive types.

Java

String num = "10";
Integer.parseInt(num );

Answer:

There are five ways to convert
The First Way :

String str = " 123" ;
int i = Integer.parse(str); 
output : 123

The second way :

String str = "hello123world";
int i = Integer.parse(str.replaceAll("[\D]" , "" ) );
output : 123

The Third Way :

String str"123";
int i = new Integer(str);
output "123 

The Fourth Way :

String str"123";
int i = Integer.valueOf(Str);
output "123 

The Fifth Way :

String str"123";
int i = Integer.decode(str);
output "123 

There could be other ways
But that’s what I remember now