Conditional cell template class in ui-grid angularjs

Question

How to add conditional styling when showing data in ui-grid cellTemplate below: I want to mofify format to be shown as link or no link. Similar to Conditional cell template in ui-grid angularjs.

var myTemplate = '<div>{{getExternalScopes().showName(row.entity)}}</div>'

$scope.gridOptions = {
    columnDefs: [
            { field: 'code' },
            { field: 'name', cellTemplate: myTemplate},
            // field 'link
  ]
};

The expected result should be that name is link or text based on parameter link

Thanks in advance.

LBA · Accepted Answer · 2016-10-19 17:37:24Z

0

I made conditional template:

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

edited Oct 19, 2016 at 17:37

LBA

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answered Jan 22, 2015 at 16:00

vertti

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