Lodash：如何使用ID从数组中删除项或嵌套项？

Question

我有一个数组，它长这样：

[
    {
        "id": "denuzi",
        "sub":
        [
            {"id": "s4p58o"},
            {"id": "xzbqi"},
        ],
    },
    {
        "id": "4rcodm",
        "sub": [],
    }
]

我想做的是使用提供的id从数组或嵌套数组中删除一个项目。

例如，如果提供了denuzi，则将删除整个第一个对象；如果提供了s4p58o，则只会删除第一个对象的子数组中的该对象。

我知道可以使用以下代码删除整个对象：

_.remove(this.items, function(item) {
    return item.id !== id;
});

但我不确定如何进行检查以删除子项？

- crotanite

Answer 1

您可以通过检查id或再次调用sub采用迭代和递归的方法。如果找到了，退出。

function remove(array, id) {
    return array.some((o, i, a) => o.id === id
        ? a.splice(i, 1)
        : remove(o.sub || [], id)
    );
}

var array = [{ id: "denuzi", sub: [{ id: "s4p58o" }, { id: "xzbqi" }] }, { id: "4rcodm", sub: [] }];

remove(array, 's4p58o');
console.log(array);

remove(array, 'denuzi');
console.log(array);

.as-console-wrapper { max-height: 100% !important; top: 0; }

- Nina Scholz

Answer 2

只需使用 Array.prototype.filter

var list = [
    {
        "id": "denuzi",
        "sub":
        [
            {"id": "s4p58o"},
            {"id": "xzbqi"},
        ],
    },
    {
        "id": "4rcodm",
        "sub": [],
    }
]

const result = list.filter(a => a.id !== 'denuzi');

console.log(result);

- Cristian S.

---

它不适用于嵌套结构，例如，您的解决方案无法删除s4p58o。 - asifsaho

Answer 3

不使用 lodash

let data = [
    {
        "id": "denuzi",
        "sub":
        [
            {"id": "s4p58o"},
            {"id": "xzbqi"},
        ],
    },
    {
        "id": "4rcodm",
        "sub": [],
    }
]
function remove(data, removeId){
  return data
    .filter(({id}) => id!==removeId) // removes parent object if its id matches removeId
    .map(({id, sub}) => ({id, sub: sub.filter(({id}) => id!==removeId)})) // replaces sub array with new sub array with subobject missing
}

console.log(remove(data, "s4p58o")); 
console.log(remove(data, "denuzi"))

使用 lodash

const data = [
    {
        "id": "denuzi",
        "sub":
        [
            {"id": "s4p58o"},
            {"id": "xzbqi"},
        ],
    },
    {
        "id": "4rcodm",
        "sub": [],
    }
]


function remove(data, removeId){
  return _.remove(data, (item) => item.id !== removeId)  // removes parent object if its id matches removeId
    .map(({id, sub}) => ({id, sub: _.remove(sub, (item) => item.id !== removeId)})) // replaces sub array with new sub array with subobject missing
}

// i parse data to do a deep copy for showing the property
console.log(remove(JSON.parse(JSON.stringify(data)), "s4p58o")); 
console.log(remove(JSON.parse(JSON.stringify(data)), "denuzi"))

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

- Krzysztof Krzeszewski