Conditional cell template in ui-grid angularjs

问题

How to add conditional when showing data in ui-grid cellTemplate below:

$scope.status = ['Active', 'Non Active', 'Deleted'];
$scope.gridOptions = {
    columnDefs: [{
        field: 'code'
    }, {
        field: 'name'
    }, {
        field: 'status',
        cellTemplate: '<div>{{status[row.entity.status]}}</div>'
    }]
};

The expected result should be row status show Active/NonActive/Deleted.

Here is the plunker

Thanks in advance.

回答1:

You have to use externalScopes.

In your markup define the gridholder like this.

<div ui-grid="gridOptions" external-scopes="states" class="grid"></div>

And in your controller use this code:

var statusTxt = ['Active', 'Non Active', 'Deleted'];

$scope.states = {
  showMe: function(val) {
    return statusTxt[val];
  }
};

var statusTemplate = '<div>{{getExternalScopes().showMe(row.entity.status)}}</div>';
$scope.gridOptions = {
  columnDefs: [{
    field: 'code'
  }, {
    field: 'name'
  }, {
    field: 'status',
    cellTemplate: statusTemplate
  }]
};

Or use an angular filter.

Note that this only renders text. The best approach would be to transform myData to have real text states before using it in ui-grid. Just in case you want to do some text based filtering later on.

Here is a Plunker

回答2:

I would suggest to use ng-if solve this problem.

$scope.gridOptions = {
    columnDefs: [{
        field: 'code'
    }, {
        field: 'name'
    }, {
        field: 'status',
        cellTemplate: '<div ng-if="row.entity.status == 0">Active</div><div ng-if="row.entity.status == 1">Non Active</div>'
    }]
};

回答3:

I have got another solution for you without using external scopes:

The Template looks like this:

var statusTemplate = '<div>{{COL_FIELD == 0 ? "Active" : (COL_FIELD == 1 ? "Non Active" : "Deleted")}}</div>';

Here is the plunker:

http://plnkr.co/edit/OZtU7GrOskdqwHW5FIVz?p=preview

回答4:

Use a cellFilter.

columnDefs: [{
    field: 'code'
}, {
    field: 'name'
}, {
    field: 'status',
    cellFilter: 'mapStatus'
}]


app.filter('mapStatus', function() {

    var statusMap = ['Active', 'Non Active', 'Deleted'];

    return function(code) {
        if (!angular.isDefined(code) || code < 0 || code > 2) {
            return '';
        } else {
            return statusMap[code];
        }
    };
});

plunker

回答5:

You must change your template. When you are referring to external scopes in angular-ui-grid you may use grid.appScope.

var statusTemplate = '<div>{{grid.appScope.status[row.entity.status]}}</div>';

回答6:

Try below script. It is working for me.

  app.controller('MainCtrl', ['$scope',
  function($scope) {

  var statusTxt = ['Active', 'Non Active', 'Deleted'];

$scope.showMe= function(val) {
    return statusTxt[val];
  };

var statusTemplate = '<div>{{grid.appScope.showMe(row.entity.status)}}</div>';
$scope.gridOptions = {
  columnDefs: [{
    field: 'code'
  }, {
    field: 'name'
  }, {
    field: 'status',
    cellTemplate: statusTemplate
  }]
};

$scope.gridOptions.data = [{
  "code": "Cox",
  "name": "Carney",
  "status": 0
}, {
  "code": "Lorraine",
  "name": "Wise",
  "status": 1
}, {
  "code": "Nancy",
  "name": "Waters",
  "status": 2
  }];
  }
]);

来源：https://stackoverflow.com/questions/26688746/conditional-cell-template-in-ui-grid-angularjs

标签

angularjs

angular-ui-grid