如何在 Javascript/AngularJS 中按 Json 对象分组答案

【问题标题】：How to Group By Json object in Javascript/AngularJS如何在 Javascript/AngularJS 中按 Json 对象分组
【发布时间】：2021-03-26 19:39:19
【问题描述】：

我有一个包含 zip 条目的 json 对象。在下面的 json 对象中，“Zipcode”键下有 zip 条目，基本上我想按 Zip code 对其进行分组以显示哪个邮政编码属于所有区域 ID。

例如 - 邮政编码 7 属于 zoneIds 12005,12008,12006,12009，邮政编码 12 属于 zoneIds 12004,11001。最终目标是在一些错误消息中显示这一点。

var Json = [{
City: "ABC",
ZipCode: "7",
ZoneID: 12008,
ZoneName: "test_TP41"},
{City: "ABC",
ZipCode: "7",
ZoneID: 12005,
ZoneName: "test_TP4"
},
{City: "ABC",
ZipCode: "7",
ZoneID: 12007,
ZoneName: "test_TP456"},
{City: "ABC",
ZipCode: "7",
ZoneID: 12006,
ZoneName: "test_TP5"},
{City: "ABC",
ZipCode: "7",
ZoneID: 12009,
ZoneName: "testgrp16"},
{City: "CDE",
ZipCode: "12",
ZoneID: 12004,
ZoneName: "test_TP2"},
{City: "CDE",
ZipCode: "12",
ZoneID: 11001,
ZoneName: "test 20201"
}]

【问题讨论】：

标签： javascript json angularjs

【解决方案1】：

您可以执行以下操作来获得这样的结构，

obj = {
  ZipCode: [all the objects with this ZipCode],
}

Json = [{
City: "ABC",
ZipCode: "7",
ZoneID: 12008,
ZoneName: "test_TP41"},
{City: "ABC",
ZipCode: "7",
ZoneID: 12005,
ZoneName: "test_TP4"
},
{City: "ABC",
ZipCode: "7",
ZoneID: 12007,
ZoneName: "test_TP456"},
{City: "ABC",
ZipCode: "7",
ZoneID: 12006,
ZoneName: "test_TP5"},
{City: "ABC",
ZipCode: "7",
ZoneID: 12009,
ZoneName: "testgrp16"},
{City: "CDE",
ZipCode: "12",
ZoneID: 12004,
ZoneName: "test_TP2"},
{City: "CDE",
ZipCode: "12",
ZoneID: 11001,
ZoneName: "test 20201"
}]

const res = Json.reduce((acc, curr) => {
  if(acc.hasOwnProperty(curr.ZipCode)) {
    acc[curr.ZipCode].push(curr);
  } else {
    acc[curr.ZipCode] = [curr];
  }
  return acc;
}, {});

console.log(res);

如果您愿意，您可以通过这样做仅在数组中插入 ZoneID，

acc[curr.ZipCode] = [curr.ZoneID];

和

acc[curr.ZipCode].push(curr.ZoneID);

【讨论】：

感谢您的回答，但现在如何从新构建的对象中获取特定 zip 的信息，例如 7？
obj[ZipCode] 就足够了，您会在一个数组中找到所有区域。

【解决方案2】：

您可以关注this answer了解更多分类

使用 ES5+，

var grouped = _.mapValues(_.groupBy(Json , 'ZipCode'),
                          singleObject=> singleObject.map(item=> _.omit(Json , 'ZipCode')));

console.log(grouped);

希望对你有帮助

【讨论】：

【解决方案3】：

您可以循环创建一个以 ZipCode 为键、ZoneID 为值的对象

const Json = [{
City: "ABC",
ZipCode: "7",
ZoneID: 12008,
ZoneName: "test_TP41"},
{City: "ABC",
ZipCode: "7",
ZoneID: 12005,
ZoneName: "test_TP4"
},
{City: "ABC",
ZipCode: "7",
ZoneID: 12007,
ZoneName: "test_TP456"},
{City: "ABC",
ZipCode: "7",
ZoneID: 12006,
ZoneName: "test_TP5"},
{City: "ABC",
ZipCode: "7",
ZoneID: 12009,
ZoneName: "testgrp16"},
{City: "CDE",
ZipCode: "12",
ZoneID: 12004,
ZoneName: "test_TP2"},
{City: "CDE",
ZipCode: "12",
ZoneID: 11001,
ZoneName: "test 20201"
}];

const ZipMappedObj = {}

Json.forEach((k)=>{
if(ZipMappedObj[k.ZipCode]){
  ZipMappedObj[k.ZipCode].push(k.ZoneID)
} else {
ZipMappedObj[k.ZipCode] = [k.ZoneID]
}
})
console.log(ZipMappedObj)

【讨论】：