How do I open an image in a new window

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I am writing an executable that has a list box that contains a data template. The data is pulled from an XML file via Linq2XML.

Within the data in the list box is an image that is set to 200px wide. I have code to open a new window when when the image is clicked and show a larger version of the same image. My problem is getting the image that was clicked to open in the new window.

I assume I must have to store the image from {Binding Path=Artwork} into some sort of image variable, but I don't know how to go about it.

Posted 13-Apr-15 10:03am

Sean Donnahoe

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Sergey Alexandrovich Kryukov 13-Apr-15 16:42pm

Not clear what is the problem here. You are showing the XAML with fixed source, but how is it related to the case when you have to change the source on some event?
—SA

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Sergey Alexandrovich Kryukov · Answer 1 · 2015-04-13T10:50:00

Probably you are using System.Windows.Controls.Image, which is System.Windows.UIElement and can be used in XAML: https://msdn.microsoft.com/en-us/library/system.windows.controls.image%28v=vs.110%29.aspx.

All you need is to change its ImageSource during runtime:
https://msdn.microsoft.com/en-us/library/system.windows.controls.image.source%28v=vs.110%29.aspx (see the code sample here),
https://msdn.microsoft.com/en-us/library/system.windows.media.imagesource%28v=vs.110%29.aspx.

Roughly speaking, the "real image" is the instance of image source, and the instance of Image is just the UIElement which renders some image. Therefore, it gives your additional level of flexibility: you don't need to remove the instance of Image image from the logical tree (see also https://msdn.microsoft.com/en-us/library/ms753391%28v=vs.110%29.aspx) of your window and insert a new instance; instead, you just change the image to be rendered — the property referenced above is read/write.

—SA