UITable dropdown option in column not working
10 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
I am trying to create a dropdown option in the third column of my table. If I use the documentation data (variable Data) with my dropdown choices, it works. If I try using my data (variable myData), it does not work. I have also tried the categorical replacement for the specific column in the table and that did not work for me. Thanks for your insight.
% Generate Data
material_list = {'material1','material2','material3'};
num = [1 2 3]';
type = {'A', 'B', 'C'}';
% Generate "Choose" for last column
choose = repmat({'Choose'},length(num),1);
% Build table
my_table = table(num,type,choose);
myData = table2cell(my_table);
fig = uifigure;
% Matlab documentation data
Data = {'Andrew' 31 'Male' 'Choose'; ...
'Bob' 41 'Male' 'Choose'; ...
'Anne' 20 'Female' 'Choose';};
% Build uitable
uit = uitable('Parent', fig, ...
'Position', [100 150 380 100], ...
'ColumnFormat',({[] [] [] material_list}), ...
'ColumnEditable',true, ...
'Data',myData); % myData does not work, % Data works.
0 Kommentare
Akzeptierte Antwort
Ameer Hamza
am 9 Apr. 2020
Change the value to the property 'ColumnFormat'
% Build uitable
uit = uitable('Parent', fig, ...
'Position', [100 150 380 100], ...
'ColumnFormat',({[] [] material_list}), ... % number of elements should be same as number of columns
'ColumnEditable',true, ...
'Data',myData); % myData does not work, % Data works.
2 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Develop Apps Using App Designer finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!