javascript - Angular-UI grid : adding custom field in columnDefs and access it from header template - Stack Overflow

I want to add some custom field in columnDefs and want to access it from header template. As an example , i want a field let say showFile

$scope.gridOptions.columnDefs = [
       {                
                name: 'ServiceID',
                displayName: 'Service',
                showFile: some data
      }]

and want to access showFile inside header template ...

<div class="ui-grid-top-panel"  style="text-align: center">
      {{ want to access 'showFile' }}
</div>

What's the best possible way to do this. As i have tried it using custom method as

<div class="ui-grid-top-panel"  style="text-align: center">
{{grid.appScope.letter()}}
</div>

(plnkr link , ) but , the function (grid.appScope.letter()) is being called for infinite number of times . I have raised the issue but didn't got any reply .. , . Can anyone please suggest the best possible way to achieve the above mentioned task.

I want to add some custom field in columnDefs and want to access it from header template. As an example , i want a field let say showFile

$scope.gridOptions.columnDefs = [
       {                
                name: 'ServiceID',
                displayName: 'Service',
                showFile: some data
      }]

and want to access showFile inside header template ...

<div class="ui-grid-top-panel"  style="text-align: center">
      {{ want to access 'showFile' }}
</div>

What's the best possible way to do this. As i have tried it using custom method as

<div class="ui-grid-top-panel"  style="text-align: center">
{{grid.appScope.letter()}}
</div>

(plnkr link http://plnkr.co/edit/ZW43LsiLY7GdnX6XEOgG?p=preview , http://plnkr.co/edit/3E8HTz4Z2daGqRh1WHtx?p=preview) but , the function (grid.appScope.letter()) is being called for infinite number of times . I have raised the issue but didn't got any reply .. https://github./angular-ui/ui-grid/issues/4250 , https://github./angular-ui/ui-grid/issues/4314. Can anyone please suggest the best possible way to achieve the above mentioned task.

• javascript
• angularjs
• angular-ui-grid

Share Follow edited Dec 14, 2016 at 10:58 surya singh asked Sep 22, 2015 at 16:42 surya singhsurya singh 27766 silver badges2323 bronze badges

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3 Answers 3

Sorted by: Reset to default 1

Try using renderIndex. This will give you the column's index.

{{grid.appScope.gridOptions.columnDefs[renderIndex].customField}}

To access showFile in there, have you tried...

{{grid.appScope.gridOptions.columnDefs[0].showFile}}

I edited the plnkr: http://plnkr.co/edit/kdU59pZYQT0B76vYBQC8?p=preview.

I'm not sure if that's what you want to do, i used the headerCellTemplate in the columnDefs object instead of the headerTemplate, then you access with: {{col.colDef.showFile}}

columnDefs: [{
    field: 'name',
    displayName: 'First Name',
    width: 90
  }, {
    field: 'title',
    displayName: 'Last Name',
    width: 80
  }, {
    field: 'test',
    displayName: 'test',
    width: 80,
    showFile: 'FILE',
    headerCellTemplate: '<div ng-class="{ \'sortable\': sortable }">' +
      '<div class="ui-grid-cell-contents" col-index="renderIndex" title="TOOLTIP">' +
      '<span>{{ col.displayName CUSTOM_FILTERS }}</span><br /><span>{{col.colDef.showFile}}</span>' +
      '<span ui-grid-visible="col.sort.direction" ng-class="{ \'ui-grid-icon-up-dir\': col.sort.direction == asc, \'ui-grid-icon-down-dir\': col.sort.direction == desc, \'ui-grid-icon-blank\': !col.sort.direction }">' +
      '&nbsp;' +
      '</span>' +
      '</div>' +
      '<div class="ui-grid-column-menu-button" ng-if="grid.options.enableColumnMenus && !col.isRowHeader  && col.colDef.enableColumnMenu !== false" ng-click="toggleMenu($event)" ng-class="{\'ui-grid-column-menu-button-last-col\': isLastCol}">' +
      '<i class="ui-grid-icon-angle-down">&nbsp;</i>' +
      '</div>' +
      '<div ui-grid-filter></div>' +
      '</div>'
  }