CODEWITHSUNDEEP
javaandroiddialogandroid-edittext
Jens Bergvall
Jens Bergvall

Reputation: 1627

Android dialog and edittext values?

I'm having trouble saving String-values that resides within the edittext.

What happens is, the dialog shows, with edittext, an ok and a cancel-button.

When the OK button is pushed, what i want to happen is for the bar-variable to get the string-value from the edittext. 

public void dialog(){

    final Dialog dialog = new Dialog(myClass.this);
    dialog.setContentView(R.layout.mydialog);
    dialog.setTitle("I'm soo smart. S-M-R-T. Smart.");
    dialog.setCancelable(true);
    dialog.show();
    Button okButton = (Button) dialog.findViewById(R.id.dialog_OK_BUTTON);
    okButton.setOnClickListener(new OnClickListener() {
        public void onClick(View v) {
            try{

                LayoutInflater factory = LayoutInflater.from(Inloggning.this);
                final View textEntryView = factory.inflate(R.layout.myDialog, null);
                final EditText barText= (EditText) textEntryView.findViewById(R.id.dialog_FOO);


                // this gets returned empty.
                bar= barText.getText().toString();


                System.out.println("foo: "+bar);


                //call();
                dialog.hide();

            }
            catch(Exception e){

                // do whatever nessesary.
            }

        } 
    });

    Button cancelButton = (Button) dialog.findViewById(R.id.dialogbtn_cancel);
    cancelButton.setOnClickListener(new OnClickListener() {
        public void onClick(View v) {
            dialog.dismiss();
        }
    });



}

Can someone shed some light onto this please?

EDIT: This is sample code. Actual code does not have duplicates names on variables.

2ND EDIT: removed duplicates..

Upvotes: 4

Views: 2674

Answers (4)

KnowIT
KnowIT

Reputation: 2502

I am assuming this bar is a String type variable, the problem here seem that you are declaring two edittexts by same name, And String variables by same name too, I suggest you to change the names and try again. 

 final EditText barText=(EditText) textEntryView.findViewById(R.id.dialog_FOO);
 final EditText barText1=(EditText) textEntryView.findViewById(R.id.dialog_FOO1);

 // these gets returned empty.
 bar= barText.getText().toString();
 bar1= barText.getText().toString();

 System.out.println("foo: "+bar);
 System.out.println("foo: "+bar1);

Upvotes: -2

degratnik
degratnik

Reputation: 830

I had similar problems. I made subclass.

   public class InputDialog extends Dialog{

  private String result = null;

  private Context context = null;

  private EditText keyEdit = null;

  public InputDialog(Context _context, String _title, String _message) {
    super(_context);
    context = _context;

    setContentView(R.layout.input_dialog);
    setTitle(_title);

    keyEdit = ((EditText) findViewById(R.id.inputEditText));





  }

  public void onBackPressed() {
    cancel();
  }


  public InputDialog setOkListener(View.OnClickListener _onOk) {
    findViewById(R.id.okButton).setOnClickListener(_onOk);
    return this;
  }

  public InputDialog setCancelListener(View.OnClickListener _onCancel) {
    findViewById(R.id.cancelButton).setOnClickListener(_onCancel);
    return this;
  }

  public String getResult() {    
    return keyEdit.getText().toString();
  }

  public EditText getKeyEdit() {
    return keyEdit;
  }


}

Using

inputDialog = new InputDialog(context, getString(R.string.encription_dialog_title), getString(R.string.encription_dialog_message));
inputDialog.setOkListener(new OnClickListener(){
  public void onClick(View v) {        
    model.setEncriptionKey(inputDialog.getResult());
    listRefresh();
    if (inputDialog.getResult() == null || inputDialog.getResult().equals("")) {
      AppHelper.showMessage(FileManagerActivity.this, getString(R.string.encription_dialog_message));
    } else {
      inputDialog.dismiss();
      inputDialog.getKeyEdit().setText("");
    }

  }
});
inputDialog.setCancelListener(new OnClickListener(){
  public void onClick(View v) {      
    inputDialog.dismiss();
    inputDialog.getKeyEdit().setText("");
    onBackPressed();
  }
});
inputDialog.setOnCancelListener(new DialogInterface.OnCancelListener(){

  public void onCancel(DialogInterface dialog) {
     inputDialog.dismiss();
     inputDialog.getKeyEdit().setText("");
     onBackPressed();

  }
});

EditText keyEdit = inputDialog.getKeyEdit();



});

Upvotes: 0

Ricky
Ricky

Reputation: 7879

Check if barText is null.

What happens if you declare it out of the onClick listener?

Maybe change:

final EditText barText= (EditText) textEntryView.findViewById(R.id.dialog_FOO);

to:

final EditText barText= (EditText) dialog.findViewById(R.id.dialog_FOO);

Upvotes: 3

Anu
Anu

Reputation: 89

firstly two different edittext boxes must have different id. secondly there will two string variables to store them.

final EditText barFirstText= (EditText) textEntryView.findViewById(R.id.dialog1_FOO); final EditText barSecondText= (EditText) textEntryView.findViewById(R.id.dialog2_FOO);

            // these gets returned empty.
            barFirst= barText.getText().toString();
            barSecond= barText.getText().toString();

            System.out.println("foo: "+barFirst);
            System.out.println("foo: "+barSecond);

Upvotes: 0

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