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It used to be that if you duplicated a smart object layer using Ctrl/Cmd+J or the Duplicate Layer option in the context menu the smart object layers were linked so that changes made to one were also made to the other. That is no longer true in the current version of Ps. Does anyone know if that is a bug or is it an intentional change? Thanks.
Just to be clear, I don't think you mean Linked Smart Objects as opposed to Smart Objects created and copied inside a single document?
If you double click to open the SO in an edit window, changes made are reflected in all other copies of the SO.
If you apply effects, Transform etc. inside the document window to one SO, it will only affect that SO and not its copies.
Was that not always the case? It seems like a sensible workflow to me as it gives us both options.
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Hi @Bazsl!
I'm testing on my end, and duplicated Smart Objects remain linked to the original Smart Object using Ctrl + J. Since when do you recall this behavior began happening? Was it after a recent update to Photoshop or your operating system?
An initial thing we can try is resetting Photoshop preferences, that will help restore Photoshop to its default state and fix various issues. Here's how to do that: https://adobe.ly/415lOzW
Let me know how it goes, thakns!
Alek
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I don't know when the behavior changed since I use smart objects rarely. My test was a follows.
Open a .psd file with several layers.
Create a stamp layer (Ctrl+Shift+Alt+E) at the top of the layer stack.
Convert the stamp layer to a smart object.
Duplicate the smart object using Ctrl+J.
Select the duplicate layer.
Choose Filter > Camera raw filter from the menu.
Convert the image to black and white.
Close the camera raw filter.
The other smart object layer is still in color and does not have the camera raw filter added.
I have tried using Gaussian blur instead of the camera raw filter and I have tried adding filters to both the original smart object and the copy. In no case is the edit propagated to the other smart object. Can you provide a step-by-step example that does work? Thanks.
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@Bazsl I think they did that intentional. Changes made to one instance doesn't affect the others. This means you have to edit each instance separately. Have you tried duplicating the image as a smart objects by either copy/paste or right-click on a layer and select New Smart Object via Copy. or if you already have separate Smart Objects, you can link them. Right-click on a Smart Object layer and choose Link to Layer. You'll then be able to choose the Smart Object you want to link it to. All linked layers will share the same Smart Object content.
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New smart object via copy works as it always has. It creates a copy that is independent of the original smart object. Link Layers links the selected layers so they move together. It is not specific to smart objects and does propagate edits between the linked layers.
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Just to be clear, I don't think you mean Linked Smart Objects as opposed to Smart Objects created and copied inside a single document?
If you double click to open the SO in an edit window, changes made are reflected in all other copies of the SO.
If you apply effects, Transform etc. inside the document window to one SO, it will only affect that SO and not its copies.
Was that not always the case? It seems like a sensible workflow to me as it gives us both options.
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Yes, this is entirely logical and they way it should behave. Nothing wrong here.
When you apply a filter to a smart object layer like this, the filter is applied to the layer itself - from the outside, so to speak.
You're not going inside the smart object and editing the embedded or linked file. If you did that (by double-clicking the smart object), it would update all copied instances.
If you made a new smart object via copy, then they would be split into separate smart objects and remain independent even if you went in and edited the embedded/linked file.
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If you made a new smart object via copy, then they would be split into separate smart objects and remain independent even if you went in and edited the embedded/linked file.
By @D Fosse
Just adding to this...
The "New Smart Object Via Copy" command originated before Linked Smart Objects were available. One could argue that it needs to be renamed to include the word Embedded to clarify the intended use now that there are both Linked and Embedded Smart Object types, i.e.: "New Embedded Smart Object Via Copy".
One can copy a Linked Smart Object layer and then use the "Convert to Smart Object" command, which will Embed the previously Linked Smart Object layer. This isn't intuitive as it's already a Smart Object, so one may not think to convert a (Linked) Smart Object to an (Embedded) Smart Object. Again, perhaps the command should now have the word Embedded added to clarify what it actually does, i.e.: "Convert to Embedded Smart Object".
There isn't a native feature to create a "New Smart Object Via Copy" for a Linked Smart Object.
I created a couple of scripts to create a "New Linked Smart Object Via Copy" that breaks the association to the original file in a similar way as the native command does for Embedded Smart Objects:
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Again, perhaps the command should now have the word Embedded added to clarify what it actually does, i.e.: "Convert to Embedded Smart Object".
By @Stephen Marsh
I'd vote for that. Same terminology as Place Embedded to be consistent and avoid confusion, which was doing its best to happen in this thread.
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Again, perhaps the command should now have the word Embedded added to clarify what it actually does, i.e.: "Convert to Embedded Smart Object".
By @Stephen Marsh
I'd vote for that. Same terminology as Place Embedded to be consistent and avoid confusion, which was doing its best to happen in this thread.
By @Trevor.Dennis
You can vote here:
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Yes, @Stephen Marsh , absolutely correct and clear (I don't use linked objects a lot, mainly embedded).
Voted on your "ideas" thread - but could it perhaps be renamed in a simpler way? There may not even be room in the menu for New Embedded Smart Object Via Copy. Maybe something with "duplicate" instead of "new via copy"?
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Thanks for the clear explanation, Trevor.
Bill