Angular UI中的弹出模板

Question

我在当前项目中使用angular-ui-bootstrap,并且我需要在显示行特定信息的网格中的每一行中打开一个popover.弹出框应该有自己的HTML模板,并且有多个字段可以绑定在HTML模板中.这是我为实现同样目的而编写的代码.我确实试图通过html模板没有运气.任何帮助表示赞赏.

HTML代码:

<div ng-app="helloAngular" ng-controller="casesCntrl">
  <table class="table table-bordered">
    <thead>
      <tr>
        <th class="">Date</th>
        <th class="">Case</th>                    
        <th class="">Severity</th>                
        <th class="">Status</th>
        <th class="">Site</th>
      </tr>
    </thead>
    <tbody>
      <tr ng-repeat="case in cases">
        <td>{{case.casedate | date:'MM/dd/yyyy'}}</td>  
        <td><a ng-href="#/site/{{case.id}}">{{case.name}}</a></td>
        <td ng-class="{'case-critical':case.severity==1, 'case-urgent':case.severity==2, 'case-normal':case.severity==3}"  class="case-none"></td>  
        <td>{{case.status}}</td>    
        <td>{{case.sitename}}&nbsp;<button popover="{{case.sitename}}" popover-title="{{case.sitedescription}}"  data-placement="bottom" data-trigger="focus" class="btn btn-default" popover-unsafe-html="This is a Help but please <b> focus </b> on this">V</button></td>    
      </tr>
    </tbody>
  </table>      
</div>

JS代码:

function CasesController($scope) {    

var casesData = [
  {

    "name": "Case -1",
    "casedate":"2013-06-26T08:02:00-0700",
    "caseid":1,
    "severity": "1",
    "status":"New",
    "siteid":1,
    "sitename":"Merchant Demo 1"
  },
  {
    "name": "Case -2",
    "casedate":"2013-01-26T08:02:00-0700",
    "caseid":2,
    "severity": "1",
    "status":"New",
    "siteid":2,
    "sitename":"Merchant Demo 2"
  },
  {
    "name": "Case -3",
    "casedate":"2013-02-26T08:02:00-0700",
    "caseid":3,
    "severity": "1",
    "status":"Accepted",
    "siteid":1,
    "sitename":"Merchant Demo 3"
  } ,
  {
    "name": "Case -4",
    "casedate":"2013-05-26T08:02:00-0700",
    "caseid":4,
    "severity": "2",
    "status":"New",
    "siteid":1,
    "sitename":"Merchant Demo 4"
  } ,
  {
    "name": "Case -5",
    "casedate":"2013-09-26T08:02:00-0700",
    "caseid":5,
    "severity": "3",
    "status":"New",
    "siteid":1,
    "sitename":"Merchant Demo 5"
  } ,
  {
    "name": "Case -6",
    "casedate":"2013-04-26T08:02:00-0700",
    "caseid":6,
    "severity": "1",
    "status":"Sent to billing",
    "siteid":1,
    "sitename":"Merchant Demo 6"
  },
  {
    "name": "Case -7",
    "casedate":"2013-10-26T08:02:00-0700",
    "caseid":7,
    "severity": "3",
    "status":"New",
    "siteid":1,
    "sitename":"Merchant Demo 7"
  }
 ];
$scope.cases = casesData;
}

这是与上述代码相同的小提琴:http: //jsfiddle.net/anirbankundu/YyK5s/4/

我确实通过传递popover-unsafe-html来尝试该选项,如https://github.com/angular-ui/bootstrap/pull/641中所述

Answer 1

现在无法完成,请查看以下问题以获取更多详细信息:https: //github.com/angular-ui/bootstrap/issues/220

您可以如何使用角度带来实现这一目标.详情请访问:http://mgcrea.github.io/angular-strap/##popovers

Answer 2

使用popover -template属性在角度UI bootsrap版本0.13.0中引入了为弹出框提供模板的功能,该属性获取用于弹出框体的模板的位置. http://angular-ui.github.io/bootstrap/#/popover