Comparing strings in Java

Question

Im trying to compare the values of two edittext boxes. What i would like is to just compare passw1 = passw2. As my code is now comparing two strings i have entered as i could not get to compare them.

 final EditText passw1= (EditText) findViewById(R.id.passw1);
 final EditText passw2= (EditText) findViewById(R.id.passw2);
 Button buttoks = (Button) findViewById(R.id.Ok);
      buttoks.setOnClickListener(new OnClickListener() {    

    public void onClick(View v) {       

     if (passw1.toString().equalsIgnoreCase("1234") && passw2.toString().equalsIgnoreCase("1234")){
      Toast.makeText(getApplication(),"Username and password match", Toast.LENGTH_SHORT).show();
      }
    else {
        Toast.makeText(getApplication(),"Username and password doesn't match", Toast.LENGTH_SHORT).show();
      }
     }   });

Andrew T Finnell · Accepted Answer · 2011-01-15 16:38:29Z

35

Using the == operator will compare the references to the strings not the string themselves.

Ok, you have to toString() the Editable. I loaded up some of the code I had before that dealt with this situation.

String passwd1Text = passw1.getText().toString();
String passwd2Text = passw2.getText().toString();

if (passwd1Text.equals(passwd2Text))
{
}

edited Jan 15, 2011 at 16:38

answered Jan 15, 2011 at 15:51

Andrew T Finnell

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1 Comment

M2Gd Over a year ago

i tried that and i still have else msg come up when they are matching

mrPjer · Accepted Answer · 2011-01-15 16:32:30Z

[EDIT] I made a mistake earlier, because, to get the text, you need to use .getText().toString().

Here is a full working example:

package com.psegina.passwordTest01;

import android.app.Activity;
import android.os.Bundle;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.EditText;
import android.widget.LinearLayout;
import android.widget.Toast;

public class Main extends Activity implements OnClickListener {
    LinearLayout l;
    EditText user;
    EditText pwd;
    Button btn;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);

        l = new LinearLayout(this);
        user = new EditText(this);
        pwd = new EditText(this);
        btn = new Button(this);

        l.addView(user);
        l.addView(pwd);
        l.addView(btn);
        btn.setOnClickListener(this);

        setContentView(l);
    }

    public void onClick(View v){
        String u = user.getText().toString();
        String p = pwd.getText().toString();
        if( u.equals( p ) )
            Toast.makeText(getApplicationContext(), "Matches", Toast.LENGTH_SHORT).show();
        else
            Toast.makeText(getApplicationContext(), user.getText()+" != "+pwd.getText(), Toast.LENGTH_SHORT).show();
    }
}

Original answer (Will not work because of the lack of toString())

Try using .getText() instead of .toString().

if( passw1.getText() == passw2.getText() )
#do something

.toString() returns a String representation of the whole object, meaning it won't return the text you entered in the field (see for yourself by adding a Toast which will show the output of .toString())

i tried this but all i get is they do not match even when they are
Can you post the code block with getText() and the output of Log.d("test", passw1.getText()) and Log.d("test", passw2.getText())?
Also, you can try the following: if( (passw1.getText()).equals(passw2.getText()) )
public void onClick(View v) { if( passw1.getText() == passw2.getText() ){ Toast.makeText(getApplication(),"Username and password match", Toast.LENGTH_SHORT).show(); } else{ Toast.makeText(getApplication(),"Username and password doesn't match", Toast.LENGTH_SHORT).show(); }}
I edited the answer with a full working example that should do what you are asking for.

Dinakar Prasad Maurya · Accepted Answer · 2013-08-23 10:18:05Z

6

In onclik function replace first line with this line u will definitely get right result.

if (passw1.getText().toString().equalsIgnoreCase("1234") && passw2.getText().toString().equalsIgnoreCase("1234")){

answered Aug 23, 2013 at 10:18

Dinakar Prasad Maurya

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Comments

Jason Sturges · Accepted Answer · 2012-11-03 23:36:34Z

3

You can compare the values using equals() of Java :

public void onClick(View v) {
    // TODO Auto-generated method stub

    s1=text1.getText().toString();
    s2=text2.getText().toString();

    if(s1.equals(s2))
        Show.setText("Are Equal");
    else
        Show.setText("Not Equal");
}

edited Nov 3, 2012 at 23:36

Jason Sturges

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answered Dec 20, 2011 at 5:28

Ranajit mishra

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Comments

Highland Mark · Accepted Answer · 2011-01-15 16:58:31Z

2

You need both getText() - which returns an Editable and toString() - to convert that to a String for matching. So instead of: passw1.toString().equalsIgnoreCase("1234") you need passw1.getText().toString().equalsIgnoreCase("1234").

answered Jan 15, 2011 at 16:58

Highland Mark

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Comments

Sathish · Accepted Answer · 2013-05-13 07:25:10Z

ou can use String.compareTo(String) that returns an integer that's negative (<), zero(=) or positive(>).

Use it so:

You can use String.compareTo(String) that returns an integer that's negative (<), zero(=) or positive(>).

Use it so:

  String a="myWord";
  if(a.compareTo(another_string) <0){
    //a is strictly < to another_string
  }
  else if (a.compareTo(another_string) == 0){
    //a equals to another_string
  }
else{
  // a is strictly > than another_string
}

Jave · Accepted Answer · 2013-08-23 10:22:02Z

1

Try to use .trim() first, before .equals(), or create a new String var that has these things.

edited Aug 23, 2013 at 10:22

Jave

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answered Aug 23, 2013 at 10:03

raphaelle

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Comments

A_rmas · Accepted Answer · 2014-01-06 11:16:30Z

0

did the same here needed to show "success" twice response is data from PHP

 String res=response.toString().trim;

                            Toast.makeText(sign_in.this,res,Toast.LENGTH_SHORT).show();
    if ( res.compareTo("success")==0){
    Toast.makeText(this,res,Toast.LENGTH_SHORT).show();
    }

edited Jan 6, 2014 at 11:16

answered Jan 6, 2014 at 6:05

A_rmas

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2 Comments

poohdedoo Over a year ago

This is not a proper answer to the question

A_rmas Over a year ago

try string res = response.toString().trim()

Suraj Rao · Accepted Answer · 2022-08-22 05:03:02Z

0

String password=passw1.trim();
if (password.equalsIgnoreCase("1234")){
      Toast.makeText(getApplication(),"Username and password match", Toast.LENGTH_SHORT).show();
}

You need to use the trim method in the String

edited Aug 22, 2022 at 5:03

Suraj Rao

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answered Aug 22, 2022 at 4:51

joey

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