kathelia.dokguOriginal post
Rank 3: Container
I have a file ID and I only want to call the getFilePreview() if the file is a photo. If it's a video file, I'll handle it differently and won't bother getting a preview image and simply show a video tag.
Summary
The user wants to determine if a file from a bucket is a photo or a video. The solution is to use the `getFile` function which returns a file object with a `mimeType` property. If the mimeType is a photo, call `getFilePreview()`, otherwise handle it differently for video files.