Remove embeddedset item by reference?

[Jonathan Rosen]

let's say i have an embedded set field "people" in record #5:12:

["bob", "jon"]

this doesnt seem to work:
update #5:12 remove people[0]

this does work however: update #5:12 remove "bob"

is there any way to remove items by reference, as in case #1?

[Luca Garulli]

Hi,

You're right, we missed this simple syntax to remove item from Lists:

update #5:12 remove 0 <- there 0 is the position.

Please could you open a new issue for that?

Lvc@

 

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[Jonathan Rosen]

Luca-

I'm confused. Are you saying that "update #5:12 remove 0" is the current syntax, or the suggested syntax?

update #5:12 remove 0 will not work, because you have to target the correct collection. For example, a single object could have an embeddedset called "people" and one called "things". The only way to remove from these collections is to reference them first, and the the position of the object to be removed.

i.e. update #5:12 remove people[0]...

i just want to get on the same page before i create the issue.

[Red-0ne]

I can't find the issue about this problem and it doesn't seem to be resolved. Any update on that ?

[Jonathan Rosen]

I asked Luca a question about this that he didn't respond to. I didn't create the issue because we never clarified how this needs to be fixed.

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[Luca Garulli]

Hi,

Sorry for the delay. May you try:

update #5:12 remove people 0

Lvc@

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