Can’t delete singular KeyCommand in Cubase 13

I agree. What I meant by this is that it’s the most satisfying solution while waiting for an eventual fix.

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Now I get it. I just didn’t get it yesterday. I kept reading it over and over and it just didn’t make sense. Today, it does. Effects of sleep deprivation I guess. I apologize for the needless back and forth. However, this only works if the KeyCommand hasn’t already been deleted/changed, so it’s not a complete workaround. If a change has already been saved, this method is unusable…

Yes. MacBooks don’t come with numeric keypads. Even Apple’s standard desktop keyboard doesn’t (it used to in the past), hence why even their virtual accessibility keyboard doesn’t include one. They do make one with a numeric keypad but it’s not as available as the standard one, depending on one’s location.

Agreed, which is why I’ll leave it unsolved. That being said, both m.c.’s and Nico5’s solutions are feasible workarounds. I prefer m.c.’s as it requires less fiddling, no moving between actions, so less chance of making a mistake, and most importantly, it allowed me to re-input the original KeyCommand which had already been deleted.

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Yep, the @m.c solution is perfect if you just want to retrieve default keystrokes for a given command. But I was putting myself as someone who would try to keep custom keystrokes involving the numeric pad, would they be ‘default’ or not. In this case, @Nico5 is a better solution, IMO (Sorry, m.c … :slightly_smiling_face:). After this…

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No problem at all :smiley:

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It can also make mole hills look like mountains…

This thread contains 3 approaches, which facilitate pretty much every use case I can imagine plus a bonus utility for overcoming a MacOS virtual keyboard (accessibility) limitation.

  • If the goal is to return to factory defaults, regardless of the keyboard I have, the solution was offered by @m.c
  • If the goal is to eliminate and/or move a key command, that I can actually type on my system, then a solution was offered by me
  • If the goal is anything else, then a solution was offered by @cubic13
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