Angular ui grid ui网格angularjs中的条件单元格模板类

如何在下面的ui grid cellTemplate中显示数据时添加条件样式：我想将格式改为显示为链接或无链接。 类似于

var myTemplate='{{getExternalScopes（）.showName（row.entity）}'
$scope.gridOptions={
columnDefs：[
{字段：'代码'}，
{字段：“名称”，单元格模板：myTemplate}，
//字段链接
]
};

预期结果应该是名称为link或基于参数link的文本

---

提前感谢。

我制作了条件模板：

cellContainerNameLinkTemplate: 
'
<div class="ui-grid-cell-contents" 
   ng-if = "row.entity.IsSample"
   ng-class = "col.colIndex()">
   <div class = "gridnolinkpointer"> {{grid.getCellValue(row, col)}}</div>
</div>
<div ng-if = "!row.entity.IsSample">
   <div class = "nav nav-pills" ng-class = "col.colIndex()" > 
      <a class = "gridlinkpointer" ng-click = "getExternalScopes().followlink({event:$event,row:row})"> 
      {{grid.getCellValue(row, col)}}</a>
   </div>
</div>
', cellW

CellContainerNameLink模板：
'
{{grid.getCellValue（行，列）}
{{grid.getCellValue（行，列）}
’，塞尔夫

cellContainerNameLinkTemplate: 
'
<div class="ui-grid-cell-contents" 
   ng-if = "row.entity.IsSample"
   ng-class = "col.colIndex()">
   <div class = "gridnolinkpointer"> {{grid.getCellValue(row, col)}}</div>
</div>
<div ng-if = "!row.entity.IsSample">
   <div class = "nav nav-pills" ng-class = "col.colIndex()" > 
      <a class = "gridlinkpointer" ng-click = "getExternalScopes().followlink({event:$event,row:row})"> 
      {{grid.getCellValue(row, col)}}</a>
   </div>
</div>
', cellW