Keep lesson ids of all selections #224
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hupf
requested changes
Oct 28, 2020
| .reduce((acc, lessonIds) => { | ||
| return acc.concat(lessonIds); | ||
| }, []) | ||
| ) |
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Kleine Kosmetik:
Du loopst mit dem map und dem reduce zweimal über alle Einträge. Ich würde das in diesem Fall einfach in einem Schritt machen:
map((selectedIds) =>
selectedIds
.reduce((acc, s) => [...acc, ...s.lessonIds], [])
)
Eine Alternative wäre flatten zu verwenden:
map((selectedIds) => uniq(flatten(selectedIds.map(s => s.lessonIds))))
Sollte nicht zusätzlich auch ein uniq gemacht werden, damit diese Lektionen nicht mehrfach vorkommen?
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Danke. Ich habe das nun so umgesetzt (für alle drei Fälle).
| .reduce((acc, lessonIds) => { | ||
| return acc.concat(lessonIds); | ||
| }, []) | ||
| ) |
| .reduce((acc, lessonIds) => { | ||
| return acc.concat(lessonIds); | ||
| }, []) | ||
| ) |
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Und auch hier das gleiche Thema.
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Nicht nur die erste Selection verwenden um die LessonIds zu ermitteln, sondern alle.
Würdest du das noch in eine Hilfsmethode oder so auslagern?