Vector as ItemsData property in List Box

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Oskar Kilgus
Oskar Kilgus 2022 年 8 月 20 日
回答済み: Oskar Kilgus 2022 年 8 月 21 日
Hi all,
im coding a fairly easy app with app designer and im struggling with the input/output for List Box property "ItemsData".
What i want to do is to show the user 2 options ("Items") in a list-box. These options then lead to vectors of 2 numbers ("ItemsData"). Lets say:
With ItemsData(Option 1) = [2950 3350] and ItemsData(Option 2) = [3600 4500]
Now i want to use these 2 integers in my main function and i cant figure out how to do the indexing right. I tried using a cell array as ItemsData with combined indexing:
A = B{1,1}(1,1)
which leeds to an error because of the curly brackets.
If it is any easier with a Switch or DropDown its fine too! There are just 2 options.
Thanks for any help on this!

採用された回答

Oskar Kilgus
Oskar Kilgus 2022 年 8 月 21 日
I found a way to do it (If anyone is having trouble with this issue in the future)
1) List-Box ItemsData as 1x2 numeric vector:
[1 2]
2) value-changed function callback:
% Value changed function: choosefromListBox
function choosefromListBoxValueChanged(app, event)
app.whatever = str2num(app.choosefromListBox.Value)
end
3) reference in other function:
app.whatever(1,1) %returns 1
______________________________
Thanks @Simon Chan for your effort!

その他の回答 (1 件)

Simon Chan
Simon Chan 2022 年 8 月 21 日
Suppose the listbox is defined as follows and ItemsData are given in your example.
app.ListBox = uilistbox(app.UIFigure);
You may get the data directly via the following:
A = app.ListBox.Value
  8 件のコメント
Simon Chan
Simon Chan 2022 年 8 月 21 日
Could you please check whether the following lines gives the correct values to you or not?
app.choosefromListBox.Value(1) % gives 2950 to another function
app.choosefromListBox.Value(2) % gives 3350 to another function
Oskar Kilgus
Oskar Kilgus 2022 年 8 月 21 日
編集済み: Oskar Kilgus 2022 年 8 月 21 日
I tried exactly that (see the code i supplied) and it resulted in the error
"'Value' must be a double scalar within the range of 'Limits'."
The interesting value at the end of the function is m(o,:) = mean(p). I tried it again and now i end up with m=[0; NaN]. For the next iteration i still end up with the error message above!

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