Question

我第一次做键值对映射而无法接近。我有一个键值对，如：

trips= {
    date1: [ 
      {
        "id": 1, 
        "Place": "Delhi", 
        "Number": "001",
        "Vehicle": {"id":"veh1", "number": "AN01001"} 
      },
      { 
        "id": 2, 
        "Place": "Bangalore", 
        "Number": "002",
        "Vehicle": {"id":"veh2", "number": "AN01002"}
      },
      { 
        "id": 3, 
        "Place": "Pune", 
        "Number": "003",
        "Vehicle": {"id":"veh3", "number": "AN01003"}
      } 
    ],
    date2: [ 
      {
        "id": 1, 
        "Place": "Lucknow", 
        "Number": "001",
        "Vehicle": {"id":"veh1", "number": "AN01002"} 
      },
      { 
        "id": 3, 
        "Place": "Pune", 
        "Number": "003",
        "Vehicle": {"id":"veh3", "number": "AN01003"}
      } 
    ],
    date3: [ 
      {
        "id": 1, 
        "Place": "Delhi", 
        "Number": "001",
        "Vehicle": {"id":"veh1", "number": "AN01001"} 
      },
      { 
        "id": 2, 
        "Place": "Bangalore", 
        "Number": "002",
        "Vehicle": {"id":"veh2", "number": "AN01002"}
      }
   ]
}

for (date in trips) {
  var places = trips[date]
  for (var i = 0; i < places.length; ++i) {
    var place = places[i]
    console.log('place', place)
    console.log('Vehicle', place.Vehicle)
  }
}

内部日期数据以具有键值对的数组形式存储。我需要打印所有车辆ID为“veh2”的日期。我试图循环数据。但是在阵列开始的某个点之后找不到正确的方法。

我已经能够遍历一个嵌套的键值对

for (key in trips){
    var value= trips[key]
    for (k in value)
        {
            //further nested logic
        }

}

Answer 1

我认为你的主要问题是错误的结构化数据，正如@Nina Scholz已经提到的那样，当纠正这个问题时，很容易贯穿所有内容：

trips= {
    date1: [ 
      {
        "id": 1, 
        "Place": "Delhi", 
        "Number": "001",
        "Vehicle": {"id":"veh1", "number": "AN01001"} 
      },
      { 
        "id": 2, 
        "Place": "Bangalore", 
        "Number": "002",
        "Vehicle": {"id":"veh2", "number": "AN01002"}
      },
      { 
        "id": 3, 
        "Place": "Pune", 
        "Number": "003",
        "Vehicle": {"id":"veh3", "number": "AN01003"}
      } 
    ],
    date2: [ 
      {
        "id": 1, 
        "Place": "Lucknow", 
        "Number": "001",
        "Vehicle": {"id":"veh1", "number": "AN01002"} 
      },
      { 
        "id": 3, 
        "Place": "Pune", 
        "Number": "003",
        "Vehicle": {"id":"veh3", "number": "AN01003"}
      } 
    ],
    date3: [ 
      {
        "id": 1, 
        "Place": "Delhi", 
        "Number": "001",
        "Vehicle": {"id":"veh1", "number": "AN01001"} 
      },
      { 
        "id": 2, 
        "Place": "Bangalore", 
        "Number": "002",
        "Vehicle": {"id":"veh2", "number": "AN01002"}
      }
   ]
}

for (date in trips) {
  var places = trips[date]
  for (var i = 0; i < places.length; ++i) {
    var place = places[i]
    console.log('place', place)
    console.log('Vehicle', place.Vehicle)
  }
}

Answer 2

使用poper格式化的对象和数组，您可以过滤id匹配的单个位置。

＆＃13;

var trips = { date1: [{ id: 1, Place: "Delhi", Number: "001", Vehicle: { id: "veh1", number: "AN01001" } }, { id: 2, Place: "Bangalore", Number: "002", Vehicle: { id: "veh2", number: "AN01002" } }, { id: 3, Place: "Pune", Number: "003", Vehicle: { id: "veh3", number: "AN01003" } }], date2: [{ id: 1, Place: "Lucknow", Number: "001", Vehicle: { id: "veh1", number: "AN01002" } }, { id: 3, Place: "Pune", Number: "003", Vehicle: { id: "veh3", number: "AN01003" } }], date3: [{ id: 1, Place: "Delhi", Number: "001", Vehicle: { id: "veh1", number: "AN01001" } }, { id: 2, Place: "Bangalore", Number: "002", Vehicle: { id: "veh2", number: "AN01002" } }] },
    id = "veh2",
    result = Object.keys(trips).reduce(function (r, k) {
        return r.concat(trips[k].filter(function (place) {
            return place.Vehicle.id === id;
        }));
    }, []);

console.log(result);

＆＃13;

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＆＃13;

Answer 3

您可以使用以下代码执行此操作。

＆＃13;

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＆＃13;

Answer 4

以下是获取包含您的车辆的所有日期的一种方法，ID === veh2：

const t = Object.entries(trips);
const res = t.filter((dates) => {
  return dates[1].some((d, e) => {
    return d.Vehicle.id == 'veh2'; 
  });
});
console.log(res);

这将返回date1和date3（包含veh2）的数组

编辑（版本2）：

这会使日期更清晰：

let res = [];
for(let i in trips) {
  const found = trips[i].filter((dates) => dates.Vehicle.id == 'veh2');
  found.length && res.push(trips[i]);
}
console.log(res);

Answer 5

您可以像这样使用Heap：

Static Area

}）;

根据特定值过滤嵌套键值对

5 个答案: