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Author Topic: Question on display order  (Read 267 times)

Dennis in FL

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Question on display order
« on: December 27, 2024, 11:45:52 am »

Is there a way to rearrange the order of folders under "Audio?"

Or is it always alphabetical?

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Richard Martin

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Re: Question on display order
« Reply #1 on: December 27, 2024, 06:31:35 pm »

I think you have to edit the names to add numbers at the beginning
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HPBEME

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Re: Question on display order
« Reply #2 on: December 30, 2024, 07:01:52 am »

The primary tree nodes (Audio, Images, Video, CloudPlay, etc.) follow a fixed order set by MC that isn't alphabetical or user changeable.  However, MC does alphabetically order by default all the user created sub-views underneath.  As Richard Martin noted, you can prefix view names with numbers to force a desired order (for user created views).

That said, you may have noticed in the pic below that my Genre named views are ordered above my Artist views, and there are no numbers. So how did I do that? In the customize view dialog (2nd pic) note there is an option for position. For all items forced to the "top", MC arranges them in alphabetical order as a subgroup.  All custom views set to "automatic" (the default) are alphabetized as a subgroup that follows the "top" subgroup.  And finally, all "bottom" forced views are alphabetized as the last subgroup (which is how my album view is displayed near the bottom of the view list). 

It's a bit tedious to do and will likely require some trial and error to get it exactly how you want, especially if you have many views.  But if you dislike the idea of prefixing view names with numbers, which I totally get, this top/automatic/bottom subgroup method, combined with well-thought-out view names, may get you what you want.

All that said, it would be nice if you could just drag views into the desired order. This has been brought up many times over the years, so I assume from a coding perspective that adding this must be too difficult.  Well… that, and probably a general unfamiliarity with MC's built-in subgroup method I described above.
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