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刪除深度嵌套一維數組中的所有嵌套子節點

首頁猿問刪除深度嵌套一維數組中的所有嵌套子節點

刪除深度嵌套一維數組中的所有嵌套子節點

JavaScript

溫溫醬 2023-07-29 11:10:02

我有一個一維深度嵌套數組：nestedObj: [ { id: 1, parentId: null, taskCode: '12', taskName: 'Parent one', duration: 0, assignee: '', crewCount: 0, startDate: null, endDate: null, dependencies: []}, { id: 2, parentId: 1, taskCode: '12100', taskName: 'Child one', duration: 0, assignee: '', crewCount: 0, startDate: null, endDate: null, dependencies: []}, { id: 3, parentId: 2, taskCode: '12200', taskName: 'SubChild one', duration: 0, assignee: '', crewCount: 0, startDate: null, endDate: null, dependencies: []}, { id: 4, parentId: 3, taskCode: '122001', taskName: 'Sub-dub-Child one', duration: 0, assignee: '', crewCount: 0, startDate: null, endDate: null, dependencies: []}, { id: 5, parentId: null, taskCode: '13', taskName: 'Parent two', duration: 0, assignee: '', crewCount: 0, startDate: null, endDate: null, dependencies: []}]如上面的數據結構，以taskName為樹形視圖如下所示-> Parent one -> Child one -> SubChild one ->Sub-sub-Child one-> Parent two 在這里，如果我刪除一個節點（比如父節點），那么它的所有嵌套子節點（直到子子節點）都應該被刪除。那么我怎樣才能使用遞歸來做到這一點呢？

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1 回答

長風秋雁

TA貢獻1757條經驗獲得超7個贊

const nestedObj = [

{ id: 1, parentId: null, taskCode: '12', taskName: 'Parent one', duration: 0, assignee: '', crewCount: 0, startDate: null, endDate: null, dependencies: []},

{ id: 2, parentId: 1, taskCode: '12100', taskName: 'Child one', duration: 0, assignee: '', crewCount: 0, startDate: null, endDate: null, dependencies: []},

{ id: 3, parentId: 2, taskCode: '12200', taskName: 'SubChild one', duration: 0, assignee: '', crewCount: 0, startDate: null, endDate: null, dependencies: []},

{ id: 4, parentId: 3, taskCode: '122001', taskName: 'Sub-dub-Child one', duration: 0, assignee: '', crewCount: 0, startDate: null, endDate: null, dependencies: []},

{ id: 5, parentId: null, taskCode: '13', taskName: 'Parent two', duration: 0, assignee: '', crewCount: 0, startDate: null, endDate: null, dependencies: []}

];

function deepDelete(id) {

const index = nestedObj.findIndex((element) => element.id === id);

if (index === -1) return;

nestedObj.splice(index, 1);

const childElements = nestedObj.filter((element) => element.parentId === id);

for (const element of childElements) {

deepDelete(element.id);

}

deepDelete(2);

console.log(nestedObj);

反對回復 2023-07-29

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刪除深度嵌套一維數組中的所有嵌套子節點

刪除深度嵌套一維數組中的所有嵌套子節點

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