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'.all()' is not needed when '.filter()' is used #879

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johnmhoran opened this issue Aug 25, 2022 · 0 comments
Open

'.all()' is not needed when '.filter()' is used #879

johnmhoran opened this issue Aug 25, 2022 · 0 comments
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From @tdruez 's comments in #875 .

.all() is not needed if you call .filter() after.
@johnmhoran johnmhoran added the ui label Aug 25, 2022
@johnmhoran johnmhoran self-assigned this Aug 25, 2022
@johnmhoran johnmhoran mentioned this issue Aug 25, 2022
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