根據其屬性數組 object 找到一個數組

Question

我試圖根據 object 數組的元素 id 獲取一個數組，結構看起來像這樣

{"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\"}]"
        },
    ]}

我試圖根據詳細信息 id 和我做了什么來獲取數據

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不會執行 map。 其回調 function 的返回值應該只是指示是否應包含數組元素（來自頂級數組）。 因此，當您不希望包含數組時，您應該返回一個虛假值。 由於.some返回 boolean，這是用於該目的的完美候選方法。 如果您希望只有一個匹配項，那么.find比.filter更合適：

 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在結果中保持解析，則首先執行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);

如果您只對id鍵本身感興趣，則執行.flatMap以首先收集所有已解析的detail arrays，這樣您將獲得一個包含所有詳細信息的數組，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

你應該做這樣的事情。

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

如果不存在值，則 Array.filter 返回[] 。 因此，當您返回數組本身時，它永遠不會被視為錯誤。

Answer 3

您的代碼中的問題是第一個過濾器需要返回 boolean 值（過濾器與否）

這對你有用：

 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));