Errors using imwrite "Unable to determine the file format from the filename" I am trying to splice together the name of a new image file but dont understand why imwrite isnt working.

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Marshall on 21 Dec 2015
Edited: Stephen23 on 21 Dec 2015
% --- Executes on button press in Save.
function Save_Callback(hObject, eventdata, handles)
% hObject handle to Save (see GCBO)
% eventdata reserved - to be defined in a future version of MATLAB
% handles structure with handles and user data (see GUIDATA)
global y
newname = get(handles.newname,'String')
tif = '.tif';
filename = [newname tif];
Heres hat I have so far. Not really sure how to solve this issue with the way I am going about it. I though thtat if the filename had the .tif written in it it may recognize the fileformat, but I guess not. Any suggestions? Also on another note, how can I create somesort of dropdown menu to specify the pathway for this image file? Thanks
  1 Comment
Stephen23 on 21 Dec 2015
This time I fixed your code formatting for you. In future please do it yourself by first selecting your code text, and then clicking the {} Code button. The default text that clicking the button produces is not markup.

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Accepted Answer

Stephen23 on 21 Dec 2015
Edited: Stephen23 on 21 Dec 2015
You could try reading the imwrite documentation. At the top of the page there is a list of the accepted syntaxes:
Whereas what you are trying to use is:
What does 'default' mean? It does not seem to fit any of the permitted syntaxes. Where did you get this from? I don't see 'default' listed as any of the permitted input arguments. When I try your code with an image it produces the same error:
>> X = imread('parula_0.png');
>> newname = 'test';
>> imwrite(X,'default',newname,'tif')
??? Error using ==> imwrite at 435
Unable to determine the file format from the filename.
but when I remove 'default' from your code it works fine:
>> imwrite(X,newname,'tif')
with a new file created correctly. This is why reading the documentation is useful.
Also note that your code creates a new name string in the variable filename, but that you do not use this variable. Perhaps you really meant to call this:

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