如何遍歷JavaScript中的嵌套數組對象

Question

我想知道如何遍歷Javascipt中嵌套的對象數組嗎？ 我有一個名為obj的示例對象。 我想基於條件“輸入”為“字符串”和“輸出”為“數字”來檢索對象。

// tried got stuck
const output = [];
    myList.forEach(entry => {
      Object.keys(entry).forEach(key => {
        const entity = entry[key][0];
        if (entity.in === "string" && entity.out === "number") {
          output.push(entity);
        }
      });
    });

var obj = [{
    "ston": [{
      "id": "identity1",
      "in": "string",
      "out": "number",
      "value": 10
    },{
        "id": "identity2",
        "in": "string",
        "out": "number",
        "value": 10
    }],
    "nton": [{
      "id": "identity1",
      "in": "number",
      "out": "number",
      "value": 20
    },{
      "id": "identity2",
      "in": "number",
      "out": "number",
      "value": 30
    }]
  }]

預期產量

   [{
      "id": "identity1",
      "in": "string",
      "out": "number",
      "value": 10
    },{
        "id": "identity2",
        "in": "string",
        "out": "number",
        "value": 10
    }]

Answer 1

簡單的遞歸函數：

 var obj = [{ "ston": [{ "id": "identity1", "in": "string", "out": "number", "value": 10 }, { "id": "identity2", "in": "string", "out": "number", "value": 10 }], "nton": [{ "id": "identity1", "in": "number", "out": "number", "value": 20 }, { "id": "identity2", "in": "number", "out": "number", "value": 30 }] }]; function getObjects(inVal, outVal) { var matches = []; obj.forEach(child => { Object.values(child).forEach(arr => { if (arr.some(e => e.in == inVal && e.out == outVal)) { matches.push([...arr.filter(e => e => e.in == inVal && e.out == outVal)]); } }); }); return matches.flat(); } console.log(getObjects("string", "number")); 

Answer 2

在這里，您有一個解決方案，該解決方案主要使用Array.reduce（）遍歷數組的外部對象，從每個外部對象獲取值的展平數組，以創建具有內部對象的數組，然后過濾滿足條件的對象，同時保存他們在一個新的數組：

 var obj = [ { "ston": [ {"id": "identity1", "in": "string", "out": "number", "value": 10}, {"id": "identity2", "in": "string", "out": "number", "value": 10} ], "nton": [ {"id": "identity1", "in": "number", "out": "number", "value": 20}, {"id": "identity2", "in": "number", "out": "number", "value": 30} ] } ]; let res = obj.reduce((acc, o) => { acc = acc.concat(Object.values(o).flat().filter( o => o.in === "string" && o.out === "number" )); return acc; }, []); console.log(res); 

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其他有關使用方法的文檔：

• Array.concat（）
• Array.flat（）
• Array.filter（）
• Object.values（）

或者，經過一段時間的思考，您可以將下一個簡化版本與Array.map（）一起使用

 var obj = [ { "ston": [ {"id": "identity1", "in": "string", "out": "number", "value": 10}, {"id": "identity2", "in": "string", "out": "number", "value": 10} ], "nton": [ {"id": "identity1", "in": "number", "out": "number", "value": 20}, {"id": "identity2", "in": "number", "out": "number", "value": 30} ] } ]; let res = obj.map(Object.values).flat(2).filter( o => o.in === "string" && o.out === "number" ); console.log(res); 

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Answer 3

您可以使用將該對象重建為嵌套數組，然后展平並最終進行filter 。

 var obj = [ { "ston": [ {"id": "identity1", "in": "string", "out": "number", "value": 10}, {"id": "identity2", "in": "string", "out": "number", "value": 10} ], "nton": [ {"id": "identity1", "in": "number", "out": "number", "value": 20}, {"id": "identity2", "in": "number", "out": "number", "value": 30} ] } ]; const tmp = obj.map(e => Object.entries(e).map(([k, v]) => v)).flat(3) const rs = tmp.filter(e => e.out==='number' && e.in ==='string') console.log(rs)