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Hello all,
I have an issue that I may be overthinking but the solution eludes me.
Running LRC 9.1 on Mac Mojave 10.14.6
Issue:
> Have an existing catalog that points to a now non-existing network drive.
> I have moved all relavent pictures with the same directory structure (minus the parent folder since it no longer exists) to a directly attached drive.
Can I point this existing catalog in LR to this new location and retain all the metadata, as this is my primary concern? Is there a better way of achieving this that I'm not seeing?
Much appreciated.
USING YOUR OPERATING SYSTEM:
Re-create the parent folder on the new drive, and then drag all folders containing photos into it.
THEN USING LIGHTROOM:
Use the Find Missing folder command on the parent folder, point to the new parent folder. Full details are here: http://www.computer-darkroom.com/lr2_find_folder/find-folder.htm
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USING YOUR OPERATING SYSTEM:
Re-create the parent folder on the new drive, and then drag all folders containing photos into it.
THEN USING LIGHTROOM:
Use the Find Missing folder command on the parent folder, point to the new parent folder. Full details are here: http://www.computer-darkroom.com/lr2_find_folder/find-folder.htm
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It is important to keep the relative relationships between the image folders the same. It is the single parent folder under which these are all organised, which allows you to re-address the whole lot in one go (your single point of management for where everything lives).
Depending on how you previously imported the photos, this single parent folder may or may not be showing in your current Catalog (the one that refers to photos living on this redundant network drive). You can cause it to be shown now. Even though this network drive is now not available, LR knows what the previous full path was. So once visible, you can then redirect it to point to a new parent folder that the photos now sit under. This means that you will need such a folder: it is a lot more laborious if image folders are put directly in the root of the drive; because each then becomes a separate "tree" needing separate readdressing.
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Thanks all, I knew I was overthinking this.