我如何在uitable中分配值? [英] How can l assign values in uitable?

问题描述

我试图弄清楚如何在uitable中分配值.我可以从数据矩阵中自动获取表大小，并且可以根据需要在前两列中定义值，但是我不知道如何填充第三列.可以手动填写此列吗?尝试填充第三列时，我得到警告:该位置的表数据不可编辑.

I am trying to figure out how can I assign values in uitable. I got the table size from my data matrix automatically and I am able to define values in first two columns as I want but I don't know how can I fill the third column. Is that possible to fill this column by hand? During trying to fill third column I am getting a Warning: Table data is not editable at this location.

感谢您的帮助.

a = text2(:,(1:c11));
b = in_matrix(1,(1:c11));
cnames = {'Name','Name','Value'};
rnames = 1:c11;
Data = transpose(a);
f = figure('Position', [100 100 500 500]);
t = uitable('Parent', f, 'Position', [50 100 300 400],...
        'Data',Data,...
        'ColumnName',cnames,...
        'RowName',rnames,...
        'Enable','on',...
        'Visible','on');   
set(t,'ColumnFormat',{'char','char','numeric'});
set(t,'ColumnEditable',[false,false,true]);   

推荐答案

首先，请勿将text用作变量名，因为它是内置函数，可能会引起麻烦.其次，这就是我到目前为止所掌握的:

First, don't use text as a variable name, as it is an in-built function, which can cause trouble. Second, this is what I think you have so far:

content = {'neuer Schaufeltyp'         '[]'    [     0.8]; ...
'Geometrienummer'           '[]'    [ 0.16667]; ...
'neuer Bearbeitungstyp'     '[]'    [ 0.29231]; ...
'D'                         '[]'    [0.066667]; ...
'L1'                        '[]'    [ 0.73529]; ...
'Schneiden'                 '[]'    [     0.1]; ...
'fz'                        '[]'    [     0.1]; ...
'Einstellwinkel'            '[]'    [       0]; ...
'Schneidenradius'           '[]'    [       0]; ...
'Kühlung'                   '[]'    [ 0.33333]; ...
'GleichGegen'               '[]'    [       1]; ...
'Verh. D-WZ/Eingriffsbr'    '[]'    [       0]; ...
'neuer Werkstoff'           '[]'    [ 0.11111]}.';

c11 = 13;
a = content(:,(1:c11));
cnames = {'Name','Name','Value'};
rnames = 1:c11;
Data = a.';
f = figure('Position', [100 100 500 500]);
t = uitable('Parent', f, 'Position', [50 100 300 400],...
        'Data',Data,...
        'ColumnName',cnames,...
        'RowName',rnames,...
        'Enable','on',...
        'Visible','on');

现在我有些困惑，您要填写第二栏还是第三栏?您要通过代码还是手动完成此操作?

Now I'm a little confused, do you want to fill the second column or the third? Do you want to do it by code or by hand?

要用手填充第三列，您需要使用属性ColumEditable使其可编辑，并且还要定义ColumnFormat:

To fill the third column by hand you need to make it editable using the property ColumEditable and rather define the ColumnFormat also:

set(t,'ColumnFormat',{'char','char','numeric'})
set(t,'ColumnEditable',[false,false,true])

或在开始时将其与所有其他属性一起使用.

or do it at the beginning together with all other properties.

得到表格，您可以在其中手动在第三列中插入值(我总是插入42.):

and you get your table, where you can manually insert the values in the third column (I inserted the 42 always.):

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