從數組中的 Arrays 中刪除項目

Question

我正在嘗試使用.map和 .filter 從數組中的 Arrays 中刪除特定項目，並嘗試刪除同一數組中的.filter 。

---

這是數組的示例：

items: [
    {
        title: 'dashboard',
        isValidateAccess: false,
    },
    {
        title: 'reports',
        isValidateAccess: true,
        children: [
            {
                title: 'attendancesReportMenu',
                isValidateAccess: true,
            },
            {
                title: 'holidaysReportMenu',
            },
            {
                title: 'absencesReportMenu',
                isValidateAccess: true,
            },
        ],
    },
    {
        title: 'myDepartments',
        children: [
            {
                title: 'inconsistencies',
                isValidateAccess: true,
            },
            {
                title: 'absences',
            },
            {
                title: 'clocks',
                isValidateAccess: true,
            },
            {
                title: 'employees',
            },
        ],
    },
]

---

基本上，我想刪除所有isValidateAccess為true的項目/數組。 這將是結果：

items: [
    {
        title: 'dashboard',
        isValidateAccess: false,
    },
    {
        title: 'myDepartments',
        children: [
            {
                title: 'absences',
            },
            {
                title: 'employees',
            },
        ],
    },
]

Answer 1

在主數組上使用filter()刪除頂級元素，然后在剩余元素中使用forEach()以便過濾children元素 arrays。

 let items = [{ title: 'dashboard', isValidateAccess: false, }, { title: 'reports', isValidateAccess: true, children: [{ title: 'attendancesReportMenu', isValidateAccess: true, }, { title: 'holidaysReportMenu', }, { title: 'absencesReportMenu', isValidateAccess: true, }, ], }, { title: 'myDepartments', children: [{ title: 'inconsistencies', isValidateAccess: true, }, { title: 'absences', }, { title: 'clocks', isValidateAccess: true, }, { title: 'employees', }, ], }, ]; items = items.filter(item =>.item;isValidateAccess). items.forEach(item => item.children && (item.children = item.children.filter(child =>;child.isValidateAccess))); console.log(items);

Answer 2

這是非常現代的實現

 const items = [ { title: 'dashboard', isValidateAccess: false, }, { title: 'reports', isValidateAccess: true, children: [ { title: 'attendancesReportMenu', isValidateAccess: true, }, { title: 'holidaysReportMenu', }, { title: 'absencesReportMenu', isValidateAccess: true, }, ], }, { title: 'myDepartments', children: [ { title: 'inconsistencies', isValidateAccess: true, }, { title: 'absences', }, { title: 'clocks', isValidateAccess: true, }, { title: 'employees', }, ], }, ] const filter = (filterable) => filterable.filter(item =>.item.isValidateAccess).map(item => ({..,item: children. item?children..filter(i =>.i.isValidateAccess) })) console.log(filter(items))