android: how do I check if dialogfragment is showing

Question

I launch my dialog fragment using

FragmentTransaction ft = 
getFragmentManager().beginTransaction();
MyDialogFragment dialog = new MyDialogFragment()
dialog.show(ft, "dialog");

then to get a handle on it I do

Fragment prev = getFragmentManager().findFragmentByTag("dialog");

but once I get prev, how do I check if it is showing?

Back Story

My problem is that my looping code keeps launching the dialog again and again. But if the dialog is already showing, I don't want it to launch again. This back story is just for context. The answer I seek is not: "move it out of the loop."

Call prev.isVisible(). See: https://developer.android.com/reference/android/app/Fragment.html#isVisible() — Luis, Jan 25 '14 at 15:46

j2emanue · Answer 1 · 2021-11-16T02:00:09.923

152

 if (dialogFragment != null
     && dialogFragment.getDialog() != null
     && dialogFragment.getDialog().isShowing()
     && !dialogFragment.isRemoving()) {
            //dialog is showing so do something 
 } else {
     //dialog is not showing
 }

UPDATE: team can you also try the below call, i think it will work logically:

dialogFragment.isResumed

that should mean its in the foreground displaying if im not mistaken.

edited Nov 16 '21 at 02:00

answered Jun 25 '14 at 15:36

j2emanue

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Wonderful! Worked! `dialog.isVisible()` is the real culprit which, sometimes says `false` even when dialog is visible!! So, `dialogFragment.getDialog().isShowing()` worked..like a charm! – sud007 Aug 29 '16 at 09:33
2

This works as intended, and should be the accepted answer, but gets placed last... – Sakamiai Oct 20 '16 at 09:32
If dialogFragment happens to be null, dialogFragment.getDialog() will give a null pointer exception. – ʕ ᵔᴥᵔ ʔ Apr 18 '18 at 20:38
3

If you call DialogFragment's creation several times in one moment, `dialogFragment = getSupportFragmentManager().findFragmentByTag("dialog");` will return null, and all dialogs will be shown. – CoolMind Aug 20 '18 at 09:14
1

Would it help if `dialogFragment.isRemoving() == false` is also added to this check? – rpattabi Oct 16 '18 at 16:04
@rpattabi that seems like a god idea. i updated the post. – j2emanue Oct 17 '18 at 02:53
@j2emanue Your decision to keep `isRemoving` check at the end is interesting and makes sense. – rpattabi Oct 17 '18 at 05:05
is removing is not exist more? – luke cross May 25 '20 at 20:20
dialogFragment != null is always false – luke cross May 25 '20 at 20:22
But if you click the button too fast, the dilagFragment will come up twice. – Abdulmalek Feb 02 '21 at 15:25
Just use a throttle on the click event. Eg. https://stackoverflow.com/a/22221410/835883 – j2emanue Feb 02 '21 at 16:22
Also by calling dialogFragment.getShowsDialog() – Lloen Oct 06 '21 at 16:31
Yea this still doesn't work as a check to see if the dialog is already showing if you call .show() twice one right after the other. – Flyview Dec 05 '22 at 22:39

score 57 · Answer 2 · edited Dec 14 '16 at 12:25

57

I added this to be inside my custom dialog fragment, so I don't have to worry about any logic on the outside. Override the show() and onDismiss() methods, with a boolean shown field:

  private static boolean shown = false;

    @Override
    public void show(FragmentManager manager, String tag) {
        if (shown) return;

        super.show(manager, tag);
        shown = true;
    }

    @Override
    public void onDismiss(DialogInterface dialog) {
        shown = false;
        super.onDismiss(dialog);
    }

If you want to check whether it is shown or not, you can create a getter for the shown boolean.

edited Dec 14 '16 at 12:25

Arpit Patel

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answered Mar 14 '14 at 19:44

John Leehey

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Not the best answer, If user receive call while dialog is showing the dialog will added twice because variable `show` will created again with value `false`, You must to make `show` variable `static`, `private static boolean shown = false;` – Ibrahim Disouki Apr 21 '15 at 13:47
1

+1 for the previous comment and also: this approach is not applicable if you create a new instance from the outside for each dialog invocation. One could argue that you could make "shown" static but I think this would have other implications – avalancha Oct 02 '15 at 12:21
1

See https://stackoverflow.com/a/51934699/2914140: in `void show(…)` change line: `if (shown || manager.isStateSaved()) return;`. – CoolMind Aug 20 '18 at 16:14
You're also going to want to override onCancel and set shown = false there or dismissing by tapping outside of dialog won't call onDismiss and "shown" will remain stuck on true. I tried to edit the answer but the edit queue is full. – Flyview Dec 06 '22 at 01:00

nstosic · Accepted Answer · 2014-01-25T15:47:39.403

35

simply check if it's null

if(prev == null)
    //There is no active fragment with tag "dialog"
else
    //There is an active fragment with tag "dialog" and "prev" variable holds a reference to it.

Alternatively, you could check the activity the fragment prev is currently associated with, however, make sure you ask that after you make sure it's not null or you'll get a NullPointerException. Like this:

if(prev == null)
    //There is no active fragment with tag "dialog"
else
    if(prev.getActivity() != this) //additional check
        //There is a fragment with tag "dialog", but it is not active (shown) which means it was found on device's back stack.
    else
        //There is an active fragment with tag "dialog"

edited Jan 25 '14 at 15:47

answered Jan 25 '14 at 15:42

nstosic

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This doesn't always work, for example if the dialog has just been dismissed. I ran into this problem when I sometimes wanted to show a dialog immediately after it was dismissed (I reused the same fragment to show a "different" dialog). The solution I found to be working is the same as by @j2emanue below. `if (fragment.getDialog() == null || !fragment.getDialog().isShowing()) { fragment.show(fragmentManager, tag); }` (in my case I already know that my fragment is not null). – LoPoBo Nov 22 '16 at 09:25

score 3 · Answer 4 · edited May 24 '22 at 20:55

100% Uptime

class ProgressDialogFragment : DialogFragment() {

 companion object {
    private val TAG = this::class.simpleName
 }

 lateinit var dialogFragmentManager: FragmentManager
 
 fun showDialog() {
    dialogFragmentManager.apply {
       if (findFragmentByTag(TAG) == null) show(this, TAG) else return
    }
 }

}

lateinit var progressDialog: ProgressDialogFragment

Init:

class MainActivity : AppCompatActivity() {

override fun onCreate(savedInstanceState: Bundle?) {
    super.onCreate(savedInstanceState)
    setContentView(R.layout.activity_main)
   
    initProgressDialog(supportFragmentManager)
}

fun initProgressDialog(fragmentManager: FragmentManager){
    progressDialog = ProgressDialogFragment().apply {
        dialogFragmentManager = fragmentManager
    }
}

Use { progressDialog.showDialog() } everywhere

This is gonna be disappointing to you "great" finding but just call `progressDialog.showDialog()` twice back-to-back and you will get two dialogs. Because `show` is asynchronous and your `findFragmentByTag(TAG) == null` check will be true until dialog is actually added by system. — Farid, Dec 21 '21 at 05:25

drindt · Answer 5 · 2020-06-22T10:30:39.973

-1

Kotlin style:

private fun showDialog(dialogFragment: DialogFragment, tag: String) {
    supportFragmentManager.findFragmentByTag(tag).let { fragment ->
        fragment ?: let {
            supportFragmentManager.beginTransaction().let { transition ->
                dialogFragment.show(transition, tag)
            }
        }
    }
}

edited Jun 22 '20 at 10:30

answered Jun 22 '20 at 09:56

drindt

901
10
15

Sorry, this problem is not even solved technically so there is no need to jam in your Kotlin "solution" – Farid Dec 21 '21 at 05:28

android: how do I check if dialogfragment is showing

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