根据其属性数组 object 找到一个数组

Question

我试图根据 object 数组的元素 id 获取一个数组，结构看起来像这样

{"data": [
        {
            "id": 46,
            "name": "shsjks",
            "desc": "ehejej",
            "code": "hshsbsb",
            "activation_type": 1,
            "detail": "[{\"id\": 413, \"name\": \"A\"}, {\"id\": 416, \"name\": \"B\"}]",
        },
        {
            "id": 47,
            "name": "hhksns",
            "desc": "benemne",
            "code": "gevewk",
            "activation_type": 1,
            "detail": "[{\"id\": 419, \"name\": \"C\"}, {\"id\": 423, \"name\": \"D\"}]"
        },
    ]}

我试图根据详细信息 id 和我做了什么来获取数据

let arr = a.data.filter(x => {
    return (JSON.parse(x.detail).filter(x => x.id === 419))
});
// returned all instead of first element of the array

我希望它返回

// filter where id 419
{
  "id": 47,
  "name": "shjks",
  "detail": "[{\"id\": 419, \"name\": \"C\"}, {\"id\": 423, \"name\": \"D\"}]"
  ....
}

Answer 1

.filter不会执行 map。 其回调 function 的返回值应该只是指示是否应包含数组元素（来自顶级数组）。 因此，当您不希望包含数组时，您应该返回一个虚假值。 由于.some返回 boolean，这是用于该目的的完美候选方法。 如果您希望只有一个匹配项，那么.find比.filter更合适：

 let a = {"data": [{"id": 46,"name": "shsjks","desc": "ehejej","code": "hshsbsb","activation_type": 1,"detail": "[{\"id\": 413, \"name\": \"A\"}, {\"id\": 416, \"name\": \"B\"}]",},{"id": 47,"name": "hhksns","desc": "benemne","code": "gevewk","activation_type": 1,"detail": "[{\"id\": 419, \"name\": \"C\"}, {\"id\": 423, \"name\": \"D\"}]"},]} let result = a.data.find(x => JSON.parse(x.detail).some(x => x.id === 419)); console.log(result);

如果您希望details在结果中保持解析，则首先执行map ：

 let a = {"data": [{"id": 46,"name": "shsjks","desc": "ehejej","code": "hshsbsb","activation_type": 1,"detail": "[{\"id\": 413, \"name\": \"A\"}, {\"id\": 416, \"name\": \"B\"}]",},{"id": 47,"name": "hhksns","desc": "benemne","code": "gevewk","activation_type": 1,"detail": "[{\"id\": 419, \"name\": \"C\"}, {\"id\": 423, \"name\": \"D\"}]"},]} let result = a.data.map(x => ({...x, detail: JSON.parse(x.detail)})).find(x => x.detail.some(x => x.id === 419)); console.log(result);

如果您只对id键本身感兴趣，则执行.flatMap以首先收集所有已解析的detail arrays，这样您将获得一个包含所有详细信息的数组，然后.find您需要的元素：

 let a = {"data": [{"id": 46,"name": "shsjks","desc": "ehejej","code": "hshsbsb","activation_type": 1,"detail": "[{\"id\": 413, \"name\": \"A\"}, {\"id\": 416, \"name\": \"B\"}]",},{"id": 47,"name": "hhksns","desc": "benemne","code": "gevewk","activation_type": 1,"detail": "[{\"id\": 419, \"name\": \"C\"}, {\"id\": 423, \"name\": \"D\"}]"},]} let result = a.data.flatMap(x => JSON.parse(x.detail)).find(x => x.id === 419); console.log(result);

Answer 2

你应该做这样的事情。

let arr = a.data.filter(x => {
    return (JSON.parse(x.detail).filter(x => x.id === 419).length)
});

如果不存在值，则 Array.filter 返回[] 。 因此，当您返回数组本身时，它永远不会被视为错误。

Answer 3

您的代码中的问题是第一个过滤器需要返回 boolean 值（过滤器与否）

这对你有用：

 const test = { data: [ { id: 46, name: 'shsjks', desc: 'ehejej', code: 'hshsbsb', activation_type: 1, detail: '[{"id": 413, "name": "A"}, {"id": 416, "name": "B"}]', }, { id: 47, name: 'hhksns', desc: 'benemne', code: 'gevewk', activation_type: 1, detail: '[{"id": 419, "name": "C"}, {"id": 423, "name": "D"}]', }, ], }; const arr = test.data.find((x) => JSON.parse(x.detail).some((d) => d.id === 419)); console.log(JSON.stringify(arr, null, 2));