我有一个ng-repeat,它返回对象数组,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,以便将它们格式化为:
[
{"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
}
最佳答案
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)
==>将为您提供属性Year
中的月份到数组中Array.prototype.map
==>您传入months数组,然后从所有月份中返回二维的days数组。[].concat
==>返回级联数组的数组。它可能需要多个数组作为参数,因此我们方便地使用function.apply
将2D转换为1D。Bin
其他简单且性能有效的方法始终是循环添加。