How can I get the Appsheet url for an image?

[dalebyers]

I am using an external app to annotate appsheet images in certain cases.  I have this working with a drive look up to load the image, but it is slow.  If I can get the appsheet image URL that appsheet uses to display the original image, vs the simple file path stored in the data source, it makes the loading of the image almost instant.  I can trick the app into showing this to me if I set a virtual column equal to the og image col and set the type to text, then instead of displaying the image, it shows the appsheet URL, which is how I have tested its speed.  But I do not know how to extract it as a value in an appsheet field.  Anyone know of a way to extract the appsheet url from appsheet through an expression?  I am so close to making this work flawlessly.

[Suvrutt_Gurjar]

Your requirement description is not exactly clear. However it sounds that you are probably looking for creating a URL from AppSheet images column?

If it is so, please take a look at the following article and carefully read the "Turn on external URLs for image files" in the help article below

Display images, documents, and audio files - AppSheet Help

 

[Kirk_Masden · Accepted solution]

I'm not sure I understand what you are asking for but I wonder if this is it?

https://support.google.com/appsheet/answer/10107317?hl=en&sjid=17802089530411919826-NC

[PauloExpert]

You're so close — and you're on the right track by using a virtual column to expose the underlying AppSheet-hosted image URL.

Here’s the thing: AppSheet does not officially expose the hosted image URL directly through expressions, but there's a widely used trick that works for most cases.

The Trick: Reconstruct the AppSheet Image URL

AppSheet stores uploaded images in your cloud drive (e.g., Google Drive), but it serves them via a CDN when used in the app. You can construct this URL manually, and it’ll work just like the one AppSheet uses to display images fast.

Here’s the format:

https://www.appsheet.com/template/gettablefileurl?appName=YOUR_APP_ID_HERE&tableName=YOUR_TABLE_NAME...

To do this dynamically in a virtual column:

"https://www.appsheet.com/template/gettablefileurl?appName=" & ENCODEURL(CONTEXT("AppName")) & "&tableName=YOUR_TABLE_NAME&fileName=" & ENCODEURL([ImageColumn])

Example:

Let’s say your app name is MyApp-123456 and your table is Inspections, and you have an image column [Photo].

Then your virtual column expression becomes:

"https://www.appsheet.com/template/gettablefileurl?appName=" & ENCODEURL(CONTEXT("AppName")) & "&tableName=Inspections&fileName=" & ENCODEURL([Photo])

This will give you the same fast-loading AppSheet-hosted URL that the platform uses internally.

Important Notes:

Make sure [Photo] contains the file path (e.g., folder/image_1234.jpg), which is what AppSheet typically stores.

This works even if the image is stored in a subfolder.

ENCODEURL() helps to ensure the image path is properly escaped for URLs.

If you want, I can help you turn this into a reusable snippet or logic inside your app. What’s your app/table/image column setup?