How to delete the last letter from EditText with button?

Question

I have to buttons that writes A and B to an edittext. If there is something in the edittext how can I delete the last letters with the "Del" button ?

My layout:

<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
tools:context=".MainActivity" >

<Button
android:id="@+id/buttonb"
android:layout_width="100dp"
android:layout_height="100dp"
android:layout_alignParentLeft="true"
android:layout_below="@+id/buttona"
android:text="@string/buttonb"
android:textSize="50sp"
android:textStyle="bold" />

<Button
android:id="@+id/buttona"
android:layout_width="100dp"
android:layout_height="100dp"
android:layout_alignParentLeft="true"
android:layout_centerVertical="true"
android:text="@string/buttona"
android:textSize="50sp"
android:textStyle="bold" />

<Button
android:id="@+id/buttondel"
android:layout_width="100dp"
android:layout_height="100dp"
android:layout_alignParentRight="true"
android:layout_centerVertical="true"
android:text="@string/buttondel"
android:textSize="50sp"
android:textStyle="bold" />

<EditText
android:id="@+id/editText"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:layout_alignParentLeft="true"
android:layout_alignParentTop="true"
android:ems="58"
android:textSize="20sp"
android:textStyle="bold"
android:inputType="text" >

<requestFocus />
</EditText>

</RelativeLayout>

And my java:

package com.koostamas.keyboard;

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

public class MainActivity extends Activity implements OnClickListener {

Button buttona, buttondel;
EditText editText;

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
editText = (EditText) findViewById(R.id.editText);

addListenerOnButton();
}


   public void addListenerOnButton() {

buttona = (Button) findViewById(R.id.buttona);
buttona.setOnClickListener(this);

buttondel = (Button) findViewById(R.id.buttonb);
buttondel.setOnClickListener(this);


   }

public void onClick(View v) {
switch(v.getId()) {
    case R.id.buttona : 
        Button buttona = (Button)v;
    editText.setText(editText.getText().toString()+buttona.getText().toString());
        break;
    case R.id.buttondel :
        String text = editText.getText().toString();
        editText.setText(text.substring(0, text.length() - 1));
        break;



}

}


}

How can I do it? Thanks inn advance.

Solution 1

You can retrieve the text of EditText and then get the sub-string of that text and set again that text to EditText as below...

String text = editText.getText().toString();
editText.setText(text.substring(0, text.length() - 1));

You can also use following procedure....it will be more efficient.

int length = editText.getText().length();
if (length > 0) {
    editText.getText().delete(length - 1, length);
}

You should use switch-case as below...and handle your nuttondel onclick() as follows...

public class MainActivity extends Activity implements OnClickListener {

    Button buttona, buttonb;
    Button buttonDel;
    EditText editText;

    @Override
    protected void onCreate(Bundle savedInstanceState) {

        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        editText = (EditText) findViewById(R.id.editText);

        addListenerOnButton();

    }


   public void addListenerOnButton() {

    buttona = (Button) findViewById(R.id.buttona);
    buttona.setOnClickListener(this);

    buttonb = (Button) findViewById(R.id.buttonb);
    buttonb.setOnClickListener(this);

    buttonDel = (Button) findViewById(R.id.buttondel);
    buttonDel.setOnClickListener(this);


   }

    public void onClick(View v) {

       switch(v.getId()) {

          case R.id.buttona:
                      editText.setText(editText.getText().toString()+buttona.getText().toString());
              break;

          case R.id.buttonb:
                      editText.setText(editText.getText().toString()+buttonb.getText().toString());
              break;

          case R.id.buttondel:

              int length = editText.getText().length();
              if (length > 0) {
                  editText.getText().delete(length - 1, length);
              }
              break;
       }

    }

}

Solution 2

Using substring() method , you can do it,

String str = myEditText.getText().toString();
str = str.substring ( 0, str.length() - 1 );
// Now set this Text to your edit text
myEditText.setText ( str );

you need to write above lines in onClick() method.

Comments

[Tamas Koos]

But where should I put it? Could you write it into the code? Thanks.

[Tamas Koos]

But where should I put it? Could you write it into the code? Thanks.

[Android]

on button click u want na? then put it on button click event... this will remove last character

[Tamas Koos]

And what is the str variable in the 3rd line?

[user2060383]

Please tell me, which is your del button ?

[Tamas Koos]

I can't do it. It does nothing.

[Tamas Koos]

Please see my update.

[Hamid Shatu]

@user2923589...see my updated answer...I think this answer will help you lot.

[user2060383]

Good, you have done right thing.

[Android]

then again settext(str1)

[Tamas Koos]

Thanks all. It works.

[Tamas Koos]

Thanks all. It works.

[Tamas Koos]

Thanks all. It works.

[Gopal Gopi]

An efficient way would be int length = editText.getText().length();if (length > 0) {editText.getText().delete(length - 1, length);}

[Hamid Shatu]

@GopalRao...thanks for your concern. In have updated my answer. I wouldn't know that.

[Erich]

That would cause an error when the EditText is empty. Add if(str.length() > 0) { ... } to catch this case.

[Rajesh Nasit]

what if my cursor is at center and want to remove center charactor

[ingyesid]

i use the efficient way just because the cursor is placed in the correct position