过滤对象数组在javascript中具有相同的嵌套属性

Question

[{"code":"96","value":"RemoveTest","parentLabelCode":1,"color":null,"level":1,"children":[{"code":"97","value":"R1","parentLabelCode":96,"color":null,"level":2,"children":[],"visible":true,"retired":false,"systemLabel":true,"removeOnArchive":true,"displayNumber":1,"labelTranslations":null},{"code":"98","value":"R2","parentLabelCode":96,"color":null,"level":2,"children":[],"visible":true,"retired":true,"systemLabel":true,"removeOnArchive":true,"displayNumber":2,"labelTranslations":null},{"code":"99","value":"R3","parentLabelCode":96,"color":null,"level":2,"children":[],"visible":true,"retired":true,"systemLabel":true,"removeOnArchive":true,"displayNumber":2,"labelTranslations":null}],"visible":true,"retired":false,"systemLabel":true,"removeOnArchive":true,"displayNumber":0,"labelTranslations":null}]

我想要这个输出：基于退休的过滤器：假，退休：真，不应该显示并且有多个对象并且有深度嵌套的孩子我正在递归地做，但没有得到预期的输出

 var list = label_lists.filter(function f(item) {
            return item.retired==false ||
            item.children && (item.children = item.children.filter(f)).length
          })

[{"code":"96","value":"RemoveTest","parentLabelCode":1,"color":null,"level":1,"children":[{"code":"97","value":"R1","parentLabelCode":96,"color":null,"level":2,"children":[],"visible":true,"retired":false,"systemLabel":true,"removeOnArchive":true,"displayNumber":1,"labelTranslations":null}],"visible":true,"retired":false,"systemLabel":true,"removeOnArchive":true,"displayNumber":0,"labelTranslations":null}]

Answer 1

这是一个递归解决方案，它也复制而不是修改源对象：

function notRetired(arr) {
    return arr.filter(item => !item.retired)
        .map(item => Object.assign({}, item, {
            children: notRetired(item.children)
        }));
}

这是一个适用于循环结构的（略显老套的）变体：

function notRetired(arr, cache = null) {
    return arr.filter(item => !item.retired)
        .map((item, ret) => (cache || (cache = new Map())).get(item) ||
            (cache.set(item, ret = {}) && Object.assign(ret, item, {
                children: notRetired(item.children, cache)
            })));
}

Answer 2

您正在过滤以下人员：未退休 (item.retired==false) 或 (||) 有孩子 (item.children ) 和 (&&) 他们的孩子都没有退休 (item.children = item.children.filter(f)).length)

我认为您希望包括儿童在内的所有人给我所有未退休的。对吗？

var label_lists = [{
    "code":"96","value":"RemoveTest","parentLabelCode":1,"color":null,"level":1,
    "children":[{
        "code":"97","value":"R1","parentLabelCode":96,"color":null,"level":2,"children":[],"visible":true,"retired":false,"systemLabel":true,"removeOnArchive":true,"displayNumber":1,"labelTranslations":null
    }],
    "visible":true,"retired":false,"systemLabel":true,"removeOnArchive":true,"displayNumber":0,"labelTranslations":null}];

function filterActive(dic) {
    return dic.retired==false ? dic : [];
}

function findAll(array) {
    var res = [];
    for (item of array) {
        res.push(filterActive(item));
        res = res.concat(findAll(item.children));
    }
    return res;
}

console.log(findAll(label_lists));

--->

[ { code: '96',
    value: 'RemoveTest',
    parentLabelCode: 1,
    color: null,
    level: 1,
    children: [ [Object] ],
    visible: true,
    retired: false,
    systemLabel: true,
    removeOnArchive: true,
    displayNumber: 0,
    labelTranslations: null },
  { code: '97',
    value: 'R1',
    parentLabelCode: 96,
    color: null,
    level: 2,
    children: [],
    visible: true,
    retired: false,
    systemLabel: true,
    removeOnArchive: true,
    displayNumber: 1,
    labelTranslations: null } ]

关于函数式编程的注意事项。你的意图确实不错。但是根据您的问题，您想更改列表的结构。请注意，过滤器在任何情况下都只能减小列表大小。 reduce 是适合您的激光剑。

先把它写成循环和函数，然后数组上的任何循环都可以改成map、filter或reduce。

顺便欢迎SO！