我有一个Person对象,它具有Reports列表。使用ng-repeat创建显示列表信息的ng表
我有这些人从我的控制器列表 - vm.Persons和vm.PersonParams。
我的意图是为每个Person生成一个单独的ng表,每个表应显示其报表中的信息。
我现在使用的方法产生一个空表:人的
<div class="table-responsive">
<table ng-table="vm.PersonParams" class="table table-hover">
<tr ng-repeat="row in $data">
<td data-title="'Report Number'" >{{row.Reports.ReportNum }} </td>
<td data-title="'Date'" >{{row.Reports.ReceivedDate }} </td>
<td data-title="'Subject'" >{{row.Reports.Subject }} </td>
</tr>
</table>
</div>
价值观和PersonParams:
vm.Persons = [{"IdNum ":"23713","LastName ":"Smith","LocFlag":0,"Reports":[{"ReportNum":321231,"ReceivedDate":"2010-09-16T15:25:00","Subject":"Tax",}]},{"IdNum":"32552","LastName":"Xavier","LocFlag":1,"Reports":[{"ReportNum":324342,"ReceivedDate":"2013-09-11T07:50:00","Subject":"Filing Request"}]}];
vm.PersonParams = {"data":[]};
你可以发布'vm.Persons'和'vm.PersonParams'值。我的意思是这些对象是如何格式化的? – Minato
[Documentation](http://ng-table.com/#/loading/overview)表示,您必须用'NgTableParams()'构造函数创建的数据填充'ng-table'。顺便说一句,你的'vm.PersonParams'是空的? –