如何删除此代码 Javascript 上的突变？

Question

我设法编写了我需要的代码，但我想让它干净并删除realtiveGroups.push()的突变，但不知道如何实现它。 如何从此代码中删除突变？

export interface OfferCategory {
  id: string;
  name: string;
}

export interface OfferGroup {
  groupId: string;
  dependsOn: OfferCategory;
  name: string;
  rule: Rule;
  products: PosProducts[];
}

function relativesFromSubscription(groups: OfferGroup[], dependingGroups: OfferGroup[]): OfferGroup[] {
    const relativeGroups: OfferGroup[] = [];
    groups.forEach(group => {
      if (dependingGroups.some(dependingGroup => group?.dependsOn?.id === dependingGroup.groupId)) {
        relativeGroups.push(group);
      }
      if (relativeGroups.some(relativeGroup => group?.dependsOn?.id === relativeGroup.groupId)) {
        relativeGroups.push(group);
      }
    });
    return relativeGroups;
  }

Answer 1

与其在一个周期内完成所有事情，不如尝试将其分成几个：

function relativesFromSubscription(groups: OfferGroup[], dependingGroups: OfferGroup[]): OfferGroup[] {
    const groups1 = groups.filter(group => dependingGroups.some(dependingGroup => group?.dependsOn?.id === dependingGroup.groupId));
    const groups2 = groups.filter(group => groups1.some(relGroup=> group?.dependsOn?.id === relGroup.groupId)); 
    return [...groups1, ...groups2];
}

Answer 2

使用您的代码和Array.filter

const relativeGroups: OfferGroup[] = groups.filter(group => {
    return dependingGroups.some(dependingGroup => group?.dependsOn?.id === dependingGroup.groupId) || relativeGroups.some(relativeGroup => group?.dependsOn?.id === relativeGroup.groupId)
});

或者，如果您希望代码更具可读性，您可以添加描述性变量：

const relativeGroups: OfferGroup[] = groups.filter(group => {
  const hasDependingGroups = dependingGroups.some(dependingGroup => group?.dependsOn?.id === dependingGroup.groupId);

  const hasRelativeGroups = relativeGroups.some(relativeGroup => group?.dependsOn?.id === relativeGroup.groupId)

  return hasDependingGroups || hasRelativeGroups
});