how to change each cell color in a uitable

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I wonder how can i make a uitable such that cells at this table has different background colorsand each column has different width
thank you

采纳的回答

Friedrich
Friedrich 2012-1-3
编辑:Eric Sargent 2023-2-11
Update:
Starting in R2019b you can use the uistyle, addStyle, and removeStyle functions.
f = uifigure;
uit = uitable(f, Data = "hi");
s = uistyle('BackgroundColor','red');
addStyle(uit,s,'column',1)
Original Answer
Hi,
this is pretty tricky within an uitable since the text you display will have a red background and not the full cell in the table:
uitable('Data',{'<body bgcolor="#FF0000">Hello</body>'})
You will see that Hello has a read background but thats all.
So we can do a small trick. Instead of adding the text we add a html table which contains the text. In addition this html table is sooooo wide that it needs the full cell^^:
uitable('Data',{'<table border=0 width=400 bgcolor=#FF0000><TR><TD>Hello</TD></TR> </table>' })
So overall you can do a colored table like this:
colergen = @(color,text) ['<table border=0 width=400 bgcolor=',color,'><TR><TD>',text,'</TD></TR> </table>'];
data = { 2.7183 , colergen('#FF0000','Red')
'dummy text' , colergen('#00FF00','Green')
3.1416 , colergen('#0000FF','Blue')
}
uitable('data',data)
  14 个评论
Craig
Craig 2024-7-24
Eric Sargent, it would appear that uistyle, addStyle, and removeStyle functions only work on uitable when the uitable is made in a uifigure, but not when a uitable is made in figure. That has me stuck at the moment.
Error using matlab.ui.control.Table/addStyle (line 63)
Functionality not supported with figures created with the figure function.

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更多回答(5 个)

Yair Altman
Yair Altman 2013-6-6
This is answered in great detail here: http://undocumentedmatlab.com/blog/uitable-cell-colors/

Philip
Philip 2017-3-31
In Matlab 2017a (and at least as far back as 2013b) this has improved. The uitable now has a "BackgroundColor" property. This is an array of n rows and 3 columns, RGB values between 0 and 1. It appears that if this array has less rows than the data in the table, then the colours are repeated throughout the table.
To highlight a single row, create a colour array the same size as your data and specify the row of interest as the colour you want.
This is how the alternating row colours are created (and so those colours can be changed if you want as well).
  1 个评论
Torsion27
Torsion27 2018-4-24

Hello, that is my way to try to get the GUI uitable BAckgroundColor changed. But it is only changing the whole uitable. I want it to just change the row of the checkbox green if the checkbox is 1 and if checkbox is 0, change to red background of the row. Thanks for your help. best regards

 function pushbutton2_Callback(hObject, eventdata, handles)
 global pushbutton
 [num,txt,raw]= xlsread(uigetfile ({'.xlsx'}))
 anzahl_kriterien = size([raw],1)% --> aendern
 kriterium =  cell(anzahl_kriterien);
 tabledata = [num2cell(true(length(raw),1)),raw];
 set(handles.uitable3, 'data',tabledata)
 setappdata(handles.uitable3,'RawTableDat',raw)
 % --- Executes when entered data in editable cell(s) in uitable3.
 function uitable3_CellEditCallback(hObject, eventdata, handles)
 tabledata=get(handles.uitable3,'Data');
 spalte1=tabledata(:,1) 
 spalte2=tabledata(:,2)
 zeilenanzahl=length(tabledata); 
 % uitablehandles=findobj(handles.uitable3);
  for j = 1:1:zeilenanzahl %für zeile 1 bis ende
    if (spalte1{j,1}==1)   
       tabledata{j,2}=spalte2{j,1}
       set(handles.uitable3,'BackgroundColor',[0 1 0]) 
    end
for j = 1:1:zeilenanzahl  
    if (tabledata{j,1}==0) 
 tabledata{j,2}= set(handles.uitable3,'BackgroundColor',[1 0 0])
 end
  end 
 end
 guidata(hObject,handles);
 % --- Executes during object creation, after setting all    properties.
  function uitable3_CreateFcn(hObject, eventdata, handles)
  hObject.ColumnFormat = {'logical',[]}; 
  hObject.ColumnEditable = logical([1 0]); 

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Walter Roberson
Walter Roberson 2011-12-31
  4 个评论
Mahmoud Hassan Eldally
Sorry this is the first time to use html. when i used the statment in the code you gave me before it gives no colr difference. What I want is to form a table say 5rx6c and the last column to be with five background colors
regards
Rohail Razzaq
Rohail Razzaq 2015-4-15
what if i have to use a particular variable instead of a specific text in the following line :
html<font color="blue">'my text'</font></html>

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Torsion27
Torsion27 2018-4-22
编辑:Walter Roberson 2018-4-23

Hey, i have a similar problem with a Gui uitable and it would be great if you can help me with that. Thaks a lot :)

https://de.mathworks.com/matlabcentral/answers/393866-how-can-i-change-the-color-of-the-row-uitable-when-if-0-die-farbe-der-reihe-im-iutable-soll-sic?s_tid=prof_contriblnk


Torsion27
Torsion27 2018-4-24
Hello, that is my way to try to get the GUI uitable BAckgroundColor changed. But it is only changing the whole uitable. I want it to just change the row of the checkbox green if the checkbox is 1 and if checkbox is 0, change to red background of the row. Thanks for your help. best regards
function pushbutton2_Callback(hObject, eventdata, handles)
global pushbutton
[num,txt,raw]= xlsread(uigetfile ({'.xlsx'}))
anzahl_kriterien = size([raw],1)% --> aendern
kriterium = cell(anzahl_kriterien);
tabledata = [num2cell(true(length(raw),1)),raw];%das nur einmal checkbox spalte da ist und am anfang
set(handles.uitable3, 'data',tabledata)
setappdata(handles.uitable3,'RawTableDat',raw)
% --- Executes when entered data in editable cell(s) in uitable3.
function uitable3_CellEditCallback(hObject, eventdata, handles)
tabledata=get(handles.uitable3,'Data');
spalte1=tabledata(:,1)
spalte2=tabledata(:,2)
zeilenanzahl=length(tabledata);
for j = 1:1:zeilenanzahl %für zeile 1 bis ende
if (spalte1{j,1}==1)
tabledata{j,2}=spalte2{j,1}
set(handles.uitable3,'BackgroundColor',[0 1 0])
end
for j = 1:1:zeilenanzahl
if (tabledata{j,1}==0)
tabledata{j,2}= set(handles.uitable3,'BackgroundColor',[1 0 0])
end
end
end
guidata(hObject,handles);
% --- Executes during object creation, after setting all properties.
function uitable3_CreateFcn(hObject, eventdata, handles)
hObject.ColumnFormat = {'logical',[]};
hObject.ColumnEditable = logical([1 0]);
  1 个评论
Himanshu Verma
Himanshu Verma 2020-5-19
Hi there...
Did you find any solution to this problem? I'm also looking for the same.

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