CODEWITHSUNDEEP
androidbuttonandroid-alertdialog
Rahul
Rahul

Reputation: 1667

Dialog open twice on fast click of button

I have Button, which on clicking, displays a Dialog. Everything works like a charm, but if I double click the button or click the button fast, the Dialog opens two or three times. I have to click the back button twice or thrice to dismiss the Dialog.

I have searched for related questions on SO, but most of the answers suggest disabling the button or to use a variable and setting it to true and false, which is not my requirement.

If anyone knows how to solve this problem, please help me.

Code I have used

// Delete item on click of delete button
holder.butDelete.setOnClickListener(new OnClickListener() {         
@Override
    public void onClick(View v) {
        Dialog passwordDialog = new Dialog(SettingsActivity.this);      
        passwordDialog.show();
    }
});

Upvotes: 14

Views: 18744

Answers (10)

ucMax
ucMax

Reputation: 5428

When calling the show() method it takes a little while to isShowing() returns the true.
At first, it returns false because the dialog entrance animation doesn't finish yet.
My solution is about building a new implementation of the dialog class.

public class ObedientDialog extends Dialog{

    private boolean showing;

    public ObedientDialog(Context context) {

        setOnCancelListener(new OnCancelListener() {
            @Override
            public void onCancel(DialogInterface dialog) {
                showing=false;
            }});

        setOnDismissListener(new OnDismissListener() {
            @Override
            public void onDismiss(DialogInterface dialog) {
                showing=false;
            }});
    }

    @Override
    public void show() {
        if (!showing) super.show();
        showing=true;
    }

    public boolean isShowing(){
        return showing;
    }
}

Upvotes: 1

Sarmad Sohail
Sarmad Sohail

Reputation: 275

I faced the same issue. That's how I fixed it

private AlertDialog.Builder alertDialogBuilder;
private AlertDialog alertDialog;

Declare the above variables globally

alertDialogBuilder = new AlertDialog.Builder(this);

Then like above create alertDialogBuilder in the OnCreate method

    private void AlertDialogMessage(){
        alertDialogBuilder.setMessage("Internet disconnected!");
        alertDialogBuilder.setCancelable(true);
        alertDialogBuilder.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
            @Override
            public void onClick(DialogInterface dialog, int which) {
                dialog.dismiss();
            }
        });
        alertDialog = alertDialogBuilder.create();
   }

We have just baked our alert dialog in this function. Call this function in the OnCreate method.
It should be called after the alertDialogBuilder creation statement. After that in your button OnClickListener put that below code.

   if (!alertDialog.isShowing()){
        alertDialog.show();
   }

That's it. It fixed my issue. Hopefully, it will fix yours too.

Upvotes: 1

Saloni
Saloni

Reputation: 46

You can also use this code given below

 private long mLastClickTime = 0;

public boolean isSingleClick() {
            if (SystemClock.elapsedRealtime() - mLastClickTime < 1000) {
                return false;
            }
            mLastClickTime = SystemClock.elapsedRealtime();
            return true;
        }

    if(isSingleClick){
    openDialog();
    }

Upvotes: 0

Huy Tran
Huy Tran

Reputation: 69

Put this code where you want to show dialog, it checks exist dialog base on its tag name.

FragmentTransaction ft = getFragmentManager().beginTransaction();
Fragment prevFragment = getFragmentManager().findFragmentByTag("dialog");
if (prevFragment != null) {
    return;
}

MyDialog dialog = MyDialog.newInstance();
dialog.show(ft, "dialog");

Upvotes: 4

user370305
user370305

Reputation: 109237

You have to just check whether your Dialog is already shown or not:

Dialog passwordDialog = new Dialog(SettingsActivity.this);
holder.butDelete.setOnClickListener(new OnClickListener() {           
    @Override
    public void onClick(View v) {                  
        if(!passwordDialog.isShowing()) {
            passwordDialog.show();
        }
    }
});

Update:

If this doesn't work in your case, then in your activity declare globally:

Dialog passwordDialog = null;

and on Button's click:

holder.butDelete.setOnClickListener(new OnClickListener() {           
    @Override
    public void onClick(View v) {                  
        if(passwordDialog == null) {
            passwordDialog = new Dialog(SettingsActivity.this); 
            passwordDialog.show(); 
        }
    }
});

Upvotes: 13

Chirag Solanki
Chirag Solanki

Reputation: 1277

try this.

Step 1. Declare dialog object as instance variable or global variable

  private MediaPlayerDialog dialog;

Step 2. now put below code in you btton click.

 if (dialog == null || (!dialog.isVisible())) {
        dialog = new MediaPlayerDialog();
        dialog.setData(songListDatas, selectPosition);
        dialog.show(getSupportFragmentManager(), dialog.getClass().getName().toString());
    }

Note : MediaPlayer is my custom DialogFragment you can change it according to your requirement.

Best of Luck

Upvotes: -1

Priyank Patel
Priyank Patel

Reputation: 12382

try like this...

Dialog passwordDialog = new Dialog(SettingsActivity.this);


holder.butDelete.setOnClickListener(new OnClickListener()
{           
            @Override
            public void onClick(View v)
            {                         
                if(!passwordDialog.isShowing()) {   
                   passwordDialog.show();
                }
            }
});

Upvotes: -2

Versa
Versa

Reputation: 625

Declared globally:

private Boolean dialogShownOnce = false;
private mDialog dia;

Where your dialog.show(); is called:

dia = new mDialog(getContext());

if (!dia.isShowing() && !dialogShownOnce) {
    dia.show();
    dialogShownOnce = true;
}

dia.setOnDismissListener(new DialogInterface.OnDismissListener() {
    @Override
    public void onDismiss(DialogInterface dialog) {
        dialogShownOnce = false;
    }
});

mDialog does not have to be global, but I was calling mDialog.dismiss() in some out-of-local-scope interfaces.

Still uses a Boolean, but I don´t understand why it can´t be used.

Upvotes: 9

sdabet
sdabet

Reputation: 18670

You should probably have your dialogs managed by your activity, by overriding onCreateDialog() method and then calling showDialog(). That would solve your problem.

See http://developer.android.com/reference/android/app/Activity.html#onCreateDialog(int, android.os.Bundle)

Callback for creating dialogs that are managed (saved and restored) for you by the activity. The default implementation calls through to onCreateDialog(int) for compatibility. If you are targeting HONEYCOMB or later, consider instead using a DialogFragment instead.

Example:

public class TestActivity extends Activity {

private static final int TEST_DIALOG_ID = 1;

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

    showDialog(TEST_DIALOG_ID);
}

@Override
protected Dialog onCreateDialog(int id) {
    if(id == TEST_DIALOG_ID) {
        Dialog passwordDialog = new Dialog(this);      
        return passwordDialog;
    }
    return super.onCreateDialog(id);
}

}

Upvotes: -1

Sankar
Sankar

Reputation: 1691

disable the button once you clicked it and enable again once you cancel the dialog. like below

 holder.butDelete.setOnClickListener(new OnClickListener()
            {           
                @Override
                public void onClick(View v)
                {   
                    holder.butDelete.setEnabled(false);
                    Dialog passwordDialog = new Dialog(SettingsActivity.this);      
                    passwordDialog.show();
                }
            });

if it didn't work you have to take one boolean variable and use that to show and cancel the dialog.

Upvotes: 1

Related Questions