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Jan 8th, 2009, 08:05 AM
#1
Thread Starter
New Member
Re: Getting the output image path to display
Ok, I'll try to explain it better :-)
My interface has two picturebox and two buttons. The first button opens a OpenFileDialog which allows the user to choose the input image, then the image selected is displayed on picturebox1 (until here no problem). The second button calls my executable which modifies the input image chosen and creates an output one in the same directory as the input one with the same name + "_out". Now, what I want to do is display this output image in picturebox2 automatically, I mean without using any more dialogs (just clicking the button, calling .exe and displaying result).
By now I've been looking for functions which do that (find the output file path) but I've only found the ones which I mentioned on post 1.
I hope now it's clearer :-S
If somebody knows how to do it or if it is possible please tell me.
Thank you!! :-)
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