使用角度foreach循环重新排列JSON

一尘不染

使用角度foreach循环重新排列JSON

angularjs

我有一个ng-repeat返回对象数组，如下所示：

[{"day":"10","title":"day","summary":"summary","description":"ok","_id":"53f25185bffedb83d8348b22"}]
[{"day":"3","title":"day","summary":"summary","description":"ok","_id":"53f25185bffedb83d8348b22"}]

我想拉出对象并将其推入另一个数组，以便将其格式化为：

[
{"day":"10","title":"day","summary":"summary","description":"ok","_id":"53f25185bffedb83d8348b22"},
    {"day":"3","title":"day","summary":"summary","description":"ok","_id":"53f25185bffedb83d8348b22"
}]

目标是在数组上使用orderBy。是否可以将JSON重组为这种格式，然后访问数据？

这是我的观点供参考：

<div class="calDynamic" data-ng-repeat="n in [] | range:100">
<div ng-repeat="cal in calendar[n].year | filterKey:month">

    <p>{{cal}}</p>
</div>
</div>

我的JSON格式：

{
"_id" : ObjectId("53f252537d343a9ad862866c"),
"year" : {
    "December" : [],
    "November" : [],
    "October" : [],
    "September" : [],
    "August" : [],
    "July" : [ 
        {
            "day" : "21",
            "title" : "u",
            "summary" : "u",
            "description" : "ok",
            "_id" : ObjectId("53f252537d343a9ad862866d")
        }
    ],
    "June" : [],
    "May" : [],
    "April" : [],
    "March" : [],
    "February" : [],
    "January" : []
},
"__v" : 0
},
{
    "_id" : ObjectId("53f252537d343a9ad862866c"),
    "year" : {
        "December" : [],
        "November" : [],
        "October" : [],
        "September" : [],
        "August" : [],
        "July" : [ 
            {
                "day" : "3",
                "title" : "u",
                "summary" : "u",
                "description" : "ok",
                "_id" : ObjectId("53f252537d343a9ad862866d")
            }
        ],
        "June" : [],
        "May" : [],
        "April" : [],
        "March" : [],
        "February" : [],
        "January" : []
    },
    "__v" : 0
    }

阅读 174

2020-07-04

共1个答案

一尘不染

只是详细说明我的回答：-

您可以通过这种方式将各个月分散的阵列合并为1个阵列。

//obj has the the array result that is the input
var temp = [];
var result = temp.concat.apply(temp,obj.map(function(itm){ //Get each object in the source array that hold the year.
  return temp.concat.apply(temp, Object.keys(itm.year).map(function(key){ //Get Month for each yeah and flatten it out
       return itm.year[key]; //Get day array for each of the month in each year
  }));
}));

Object.keys(obj.year)==>将给您属性中的月份Year到一个数组
Array.prototype.map==>您传入月份数组，然后从所有月份中获得2D天数组。
[].concat==>返回级联数组的数组。它可以接受多个数组作为参数，因此我们可以function.apply方便地将2D转换为1D。

箱子

其他简单且性能有效的方法始终是循环添加。

2020-07-04