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Distinguish uifigure from figure programmatically

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Adam
Adam 2017 年 7 月 12 日
編集済み: Adam Danz 2021 年 4 月 22 日 2:01
Is it possible to distinguish a uifigure (appdesigner) from an old-style figure programmatically by an if type test?
class( hFig )
returns
'matlab.ui.Figure'
for both of them and I can't see anything obvious in the metaclass or properties that I can use to check which it is.
I could just use a try-catch, but I'd prefer to just filter uifigures than catch the error (I have a BusyCursor class which changes the pointer on all open figures to the busy cursor and then back again when it is deleted, but it crashes on uifigures which do not support changes to the 'pointer' property).

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Ralph Coleman
Ralph Coleman 2019 年 9 月 18 日
編集済み: Ralph Coleman 2019 年 9 月 18 日
Hi
From R2018b, you can simply use:
matlab.ui.internal.isUIFigure(hFig)
  3 件のコメント
Adam Danz
Adam Danz 2021 年 4 月 21 日 17:51
A combination of the top 2 solutions here that cover all existing matlab releases to test if a figure handle h is a uifigure.
if verLessThan('Matlab','9.0') %version < 16a (release of uifigs)
isuifig = @(~)false;
elseif verLessThan('Matlab','9.5') % 16a <= version < 18b
isuifig = @(h)~isempty(matlab.ui.internal.dialog.DialogHelper.getFigureID(h));
else % version >= 18b (written in r21a)
isuifig = @(h)matlab.ui.internal.isUIFigure(h);
end
tf = isuifig(h);
...however regular figures generated in live editor are incorrectly identified as UIFigures using Ralf's method 🤨. I haven't tested it thoroughly but a potential workaround is to test for the presence of an undocumented property of figures generated in live editor.
isuifig = @(h)matlab.ui.internal.isUIFigure(h) && ~isprop(h,'LiveEditorRunTimeFigure');

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その他の回答 (4 件)

Zhengyi
Zhengyi 2019 年 2 月 18 日
編集済み: Zhengyi 2019 年 2 月 18 日
TL;DR
function val = isuifigure(h)
val = ~isempty(matlab.ui.internal.dialog.DialogHelper.getFigureID(h));
end
The slightly longer story
If you type the following into the console, you will get an error:
>> uialert(figure,'a','b')
Error using uialert (line 42)
First argument must be a figure handle created using the uifigure function.
On line 42 of uialert function, we can see it uses the following method to validate the input handle:
matlab.ui.internal.dialog.DialogHelper.validateUIfigure(hUIFigure)
By reading through this function, we can see the error is thrown when the following function returns empty:
matlab.ui.internal.dialog.DialogHelper.getFigureID(h)
and
function out = getFigureID(f)
out = f.getId();
end
However,
getId()
seems to be a private function. The only option we have is to use getFigureID and check whether it returns a non-empty value.
Finaly, if you like, you can create this utility function to help you get over the long function call:
function val = isuifigure(h)
val = ~isempty(matlab.ui.internal.dialog.DialogHelper.getFigureID(h));
end
  1 件のコメント
Adam Danz
Adam Danz 2021 年 3 月 4 日
Thanks for digging, @Zhengyi. This solution works all the way back to r2016a which is when uifigures were released. However, is breaks in r2020b (and, perhaps, prior releases) when the handle was created by figure().
matlab.ui.internal.dialog.DialogHelper.getFigureID(figure)
% Error:
% First argument must be a figure handle created using the uifigure function.
A workaround would be to use a try/catch and to test the MException for the error ID 'MATLAB:uitools:uidialogs:NotAnAppWindowHandle' but, according to your answer, that line returns an empty value in earlier releases and who knows if this error message will change in the future.

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Dev-iL
Dev-iL 2017 年 8 月 15 日
I tried comparing the outputs of:
F = struct(uifigure());
G = struct(figure());
There are several fields that are different by default. Of increased interest are:
JavaFrame, Controller, ControllerInfo, NodeChildren
I cannot guarantee that it will work 100% of the time, but you can try testing for
isstruct(struct(hFig).ControllerInfo)
... if the above returns "true" - it's likely a uifigure.

Celso Reyes
Celso Reyes 2018 年 8 月 20 日
編集済み: Celso Reyes 2018 年 8 月 21 日
The new figures do not show in groot by default -- At least as of Mac version, R2018a -- when ShowHiddenHandles is 'off'. If ShowHiddenHandles is 'on', then it will appear.
As a side note, the new-style uifigure appears when you use allchild(groot) regardless of the ShowHiddenHandles value.
So, one could non-invasive check would look like so...
function style = figtype(f)
g = groot;
oldStatus = g.ShowHiddenHandles;
g.ShowHiddenHandles = 'off';
if any(g.Children == f)
style = 'figure';
else
style = 'uifigure';
end
g.ShowHiddenHandles = oldStatus;
end
So, in practice...
>> newf = uifigure('Name','new');
>> oldf = figure('Name','old');
>> figtype(newf)
ans =
'uifigure'
>> figtype(oldf)
ans =
'figure'
I'm not sure what would make a regular figure's handle hidden, since the Visibility property doesn't do it.

Andreas Klotzek
Andreas Klotzek 2019 年 1 月 9 日
Hi
I also need an answer to this question.
The answers provided here are not certain to give the correct result. For example if I have a old-style figure with property HandleVisibility = 'off', the test using groot will give the wrong result.
Using properties is also not a stable idea. The behavior might change.
The problem is not only for figure and uifigure. I need the same check on axes/uiaxes or uipanel handles, basically on any graphics handle. One could probably use the ancestor function to redirect the problem for any graphics handle to the figure handle.

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