Pass the whole object to the cellFilter function in ng-grid

Question

In ng-grid I need one of my columns to show a value that is dependent on multiple properties of my object. So I need to pass the whole object and not just one of its properties.

In another location of the project, I solved this by wrapping my objects in another object's say, value property, so for example one column was bound to value.FirstName and another one just to value.

Is there a way to avoid this warping, for example to just pass an empty field property, or just dropping the field property? Non of these methods seem to work.

Thanks

Answer 1

Do not set a field and then use row.entity in the cellTemplate. Here is an example:

$scope.myData = [
  {
    firstName: 'Bob',
    lastName: 'Smith',
    age: 43
  },
  {
    firstName: 'Joe',
    lastName: 'Williams',
    age: 39
  }
];

$scope.gridOptions = {
  data: myData,
  colDefs: [
    {
      displayName: 'Name',
      cellTemplate: '<div>{{row.entity.firstName}} {{row.entity.lastName}}</div>'
    },
    {
      displayName: 'Age',
      field: age
    }
  ]
};

This will result in a table like this:

+--------------+-----+
| Name         | Age |
+--------------+-----+
| Bob Smith    | 43  |
+--------------+-----+
| Joe Williams | 39  |
+--------------+-----+

Answer 2

You can use row.entity object which can be referenced from any place of the cell template. It is linked to the whole object for the current row when building the grid.