Modify cell value of a row based on other cell value in same row uitable

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Cristian Martin
Cristian Martin 2022-6-8
回答： Simon Chan 2022-6-8
Hi,
function uitable4_CellEditCallback(hObject, eventdata, handles)
a=get(handles.uitable4,'Data');
a = cell2table(a);
a.a6 = (10 / a.a5)*1^4; %where a.a6 is the column to be auto modify after editing column a.a5
a = table2cell(a);
set(handles.uitable4, 'Data',a);
Practically, I want every time I modify the value in a cell from a.a5 column, the a.a6 to be modified automaticaly, for the first row it's working but when I add another row i get the error:
To assign to or create a variable in a table, the number of rows must match the height of the table.
Practically, I unerstand the error but how to make the code working for every row? Thanks!
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Simon Chan
Simon Chan 2022-6-8
编辑：Simon Chan 2022-6-8
This is another issue, you need to make uitable editable as follows:
Assume your uitable have 6 columns and if you would like to allow user to edit column 5 only, do the following.
uitable4.ColumnEditable=[false,false,false,false,true,false]
OR only update column 6 when column 5 is being edited.
function uitable4_CellEditCallback(hObject, eventdata, handles)
% Get the row number of your edited cell
rownum = eventdata.Indices(1);
colnum = eventdata.Indices(2);
% Update column 6 when column 5 has a new value
if isequal(colnum,5)
hObject.Data(rownum,6)=hObject.Data(rownum,colnum)+1000; % Add 1000 to the new value
end
end
Cristian Martin
Cristian Martin 2022-6-8
if colnum == 5;
hObject.Data{rownum,6}=hObject.Data{rownum,colnum}+1000; % Add 1000 to the new value
end
that resolve the problem
Thanks @Simon Chan for your patience, and answers !

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采纳的回答

Simon Chan
Simon Chan 2022-6-8
Just move the last comment as an answer.
function uitable4_CellEditCallback(hObject, eventdata, handles)
% Get the row number of your edited cell
rownum = eventdata.Indices(1);
colnum = eventdata.Indices(2);
% Update column 6 when column 5 has a new value
if isequal(colnum,5)
hObject.Data(rownum,6)=hObject.Data(rownum,colnum)+1000; % Add 1000 to the new value
end
end
Please accept the answer if it is useful for you.

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