如何在Javascript中迭代嵌套对象数组答案

【问题标题】：How to iterate over a nested object array in Javascript如何在Javascript中迭代嵌套对象数组
【发布时间】：2019-04-03 08:33:52
【问题描述】：

如何遍历 Javascipt 中的嵌套对象数组？我有一个名为obj 的对象。我想检索in 是credit 和out 是bank 的对象。

// I have tried using filter but returns empty array
const s = obj.filter(function(t){
  return t.in == "credit" && t.out == "bank";
})
console.log(s);

这是数据：

var obj = [{
  "btob": [{
    "id": "trans",
    "in": "bank",
    "out": "bank",
    "value": 10
  }],
  "ctob": [{
    "id": "trans",
    "in": "credit",
    "out": "bank",
    "value": 20
  }],
  "dtob": [{
    "id": "trans",
    "in": "debit",
    "out": "bank",
    "value": 30
  }]
}, {
  "btob": [{
    "id": "fund",
    "in": "bank",
    "out": "bank",
    "value": 10
  }],
  "ctob": [{
    "id": "fund",
    "in": "credit",
    "out": "bank",
    "value": 10
  }],
  "dtob": [{
    "id": "fund",
    "in": "debit",
    "out": "bank",
    "value": 30
  }]
}]

预期输出：

  [{
    "id": "trans",
    "in": "credit",
    "out": "bank",
    "value": 20
  },
  {
    "id": "fund",
    "in": "credit",
    "out": "bank",
    "value": 10
  }]

【问题讨论】：

根据您当前的预期输出，您不能在对象中使用相同的键
输出无效，因为在同一个对象中不能有相同的键。也许您想要一个对象数组？此外，您正在考虑将 t 作为内部数组，尽管 t 是一个包含动态键 (ctob, btob) 的对象，并且对于该键，是一个对象数组。您可能希望改为过滤该数组。
另外，您的对象在某些行的末尾缺少“，”。
@briosheje 是的，更新了输出
@manuman94 道歉和更新的变化

标签： javascript jquery arrays object

【解决方案1】：

这是一个函数式风格的解决方案：

data.flatMap(obj => Object.values(obj).flatMap(arr => 
    arr.filter(t => t.in === "credit" && t.out === "bank")
));

const data = [{"btob": [{"id": "trans","in": "bank","out": "bank","value": 10}],"ctob": [{"id": "trans","in": "credit","out": "bank","value": 20}],"dtob": [{"id": "trans","in": "debit","out": "bank","value": 30}]}, {"btob": [{"id": "fund","in": "bank","out": "bank","value": 10}],"ctob": [{"id": "fund","in": "credit","out": "bank","value": 10}],"dtob": [{"id": "fund","in": "debit","out": "bank","value": 30}]}];

const result = data.flatMap(obj => Object.values(obj).flatMap(arr => arr.filter(t => t.in === "credit" && t.out === "bank")));

console.log(result);

但是就像被评论的那样，如果你的对象键“ctob”意味着“creedit to bank "，那么就不需要测试嵌套的 "credit" 和 "bank" 属性值了。

【讨论】：

【解决方案2】：

由于ctob 键引用的对象符合您的选择要求，您可以简单地这样做：

const output = obj.map(entry => {
  return entry.ctob[0];
});

var obj = [{
  "btob": [{
    "id": "trans",
    "in": "bank",
    "out": "bank",
    "value": 10
  }],
  "ctob": [{
    "id": "trans",
    "in": "credit",
    "out": "bank",
    "value": 20
  }],
  "dtob": [{
    "id": "trans",
    "in": "debit",
    "out": "bank",
    "value": 30
  }]
}, {
  "btob": [{
    "id": "fund",
    "in": "bank",
    "out": "bank",
    "value": 10
  }],
  "ctob": [{
    "id": "fund",
    "in": "credit",
    "out": "bank",
    "value": 10
  }],
  "dtob": [{
    "id": "fund",
    "in": "debit",
    "out": "bank",
    "value": 30
  }]
}];

const output = obj.map(entry => {
  return entry.ctob[0];
});

console.log(output);

当然，如果您想绝对确定，则必须遍历每个嵌套对象。请记住，由于每个嵌套数组的长度为 1（它是对象的单长度数组），因此您需要在比较其键之前使用 [0] 访问正确的对象：

const output = [];
obj.forEach(entry => {
  Object.keys(entry).forEach(key => {
    const entity = entry[key][0];
    if (entity.in === 'credit' && entity.out === 'bank') {
      output.push(entity);
    }
  });
});

var obj = [{
  "btob": [{
    "id": "trans",
    "in": "bank",
    "out": "bank",
    "value": 10
  }],
  "ctob": [{
    "id": "trans",
    "in": "credit",
    "out": "bank",
    "value": 20
  }],
  "dtob": [{
    "id": "trans",
    "in": "debit",
    "out": "bank",
    "value": 30
  }]
}, {
  "btob": [{
    "id": "fund",
    "in": "bank",
    "out": "bank",
    "value": 10
  }],
  "ctob": [{
    "id": "fund",
    "in": "credit",
    "out": "bank",
    "value": 10
  }],
  "dtob": [{
    "id": "fund",
    "in": "debit",
    "out": "bank",
    "value": 30
  }]
}];

const output = [];
obj.forEach(entry => {
  Object.keys(entry).forEach(key => {
    const entity = entry[key][0];
    if (entity.in === 'credit' && entity.out === 'bank') {
      output.push(entity);
    }
  });
});

console.log(output);

【讨论】：

【解决方案3】：

您必须遍历数组和单个数组项的每个属性；我以更易读的方式编写代码，添加了一些 cmets：

var searched = [];

// iterate on each array elements
for(var i = 0; i < obj.length; i++){

    // take the array element as an object
    var element = obj[i];

    // iterate to all the properties of that object
    for (var property in element) {
      if (element.hasOwnProperty(property)) {
          // take the property as an object
          var propObj = element[property][0];

          // verify if the property has searched value, if so, add to the result array
          if(propObj.in == "credit" && propObj.out == "bank")
              searched.push(propObj)
      }
    }
}

// print searched array
console.log(searched);

【讨论】：

【解决方案4】：

这是您的解决方案：

x=[];
obj.forEach((t)=>{
   for(key in t){
     if ((t[key][0].in == "credit") && (t[key][0].out == "bank")){
       x.push(t[key][0])
     }
  }
});
console.log(x);

【讨论】：