Javascript 使用更新的日期对ng repeat中的项目进行分组
我想显示如下项目:Javascript 使用更新的日期对ng repeat中的项目进行分组,javascript,angularjs,ionic-framework,Javascript,Angularjs,Ionic Framework,我想显示如下项目: date:14 march 2016 2042 2000 date:15 march 2016 1500 1501 1600 我的代码是: <div ng-repeat="(key, value) in oc.mileage_data | groupBy: 'updated_at' "> <div class="div_day" > <p>{{key * 1000 | date: 'd MMMM y' }
date:14 march 2016
2042
2000
date:15 march 2016
1500
1501
1600
我的代码是:
<div ng-repeat="(key, value) in oc.mileage_data | groupBy: 'updated_at' ">
<div class="div_day" >
<p>{{key * 1000 | date: 'd MMMM y' }}</p>
</div>
<ion-list can-swipe="listCanSwipe">
<ion-item ng-repeat="mileage in value" class="" style="padding: 0px;">
<div class="listing_div">
<i class="icon-mileage milage_color"></i>
<h2>{{mileage.mileage}}</h2>
<span></span>
</div>
<ion-option-button class="button-positive button_delete" ng-click="share(item)">
<i class="ion-ios-trash"></i> Delete
</ion-option-button>
</ion-item>
</ion-list>
</div>
您可以创建多个ng repeat,并仅在数据与ng匹配时显示它们,如下所示:
date:14 march 2016
2042
date:14 march 2016
2000
date:15 march 2016
1500
date:15 march 2016
1501
date:15 march 2016
1600
//here only 2014
<div ng-repeat="item in data | groupBy: 'datetime' ">
<span ng-if="item.datetime.indexOf(2014) > -1">{{item.datetime }}</span>
</div>
//here only 2015
<div ng-repeat="item in data | groupBy: 'datetime' ">
<span ng-if="item.datetime.indexOf(2015) > -1"> {{item.datetime}}</span>
</div>
//这里只有2014年
{{item.datetime}
//这里只有2015年
{{item.datetime}
//等等。json数据:
var data = [
{
date: '14 march 2016',
value: 1
},
{
date: '14 march 2016',
value: 2
},{
date: '15 march 2016',
value: 3
},
{
date: '15 march 2016',
value: 4
}
];
按日期分组:
$scope.uniq = [];
angular.forEach(data,function(val,index) {
if(angular.isUndefined($scope.uniq[val['date']])){
$scope.uniq[val['date']] = [val['value']];
}else{
$scope.uniq[val['date']].push(val['value']);
}
});
//输出:
['14 March 2016' : [1,2]........
ng repeat中的
键是否与updated_at
属性相同?键包含与'updated_at'相同的值,如果日期字段包含时间戳日期,则会是什么solution@KhushbooMahajan只要修改这段代码就行了,我不会写你的代码,我只是给你一个解决方案。