Tips for displaying only the items that are currently visible in an AngularJS Dropdown

I am currently working with an AngularJs object that looks like this:

$scope.data = {'options': [{ "id": 1, "text": "Option 1", "isHidden": 0 }, { "id": 2, "text": "Option 2", "isHidden": 1 }, { "id": 3, "text": "Option 3", "isHidden": 0 }]};

Using the code snippet below, I am able to render a dropdown menu containing all the items in the object:

<select ng-model="data.selectedValue" ng-options="option.text for option in data.options" class="form-control"></select>

Is there a way to only display elements where "isHidden = 0"? For example, I would like to show only "Option 1" and "Option 3" in the dropdown menu.

javascriptangularjs

Answer №1

Implement a customized filtration on control.options. It is possible to avoid creating this filter by using an expression, but it's generally considered best practice to minimize logic in the view.

Take a look at this scenario:

Example

View Section

<select ng-model="control.uiSelValue"
        ng-options="option.text for option in control.options | filter:myFilter"
        class="form-control">
</select>

Controller Details

$scope.control = {
    options: [{
        "id": 1,
        "text": "Option 1",
        "isHidden": 0
    }, {
        "id": 2,
        "text": "Option 2",
        "isHidden": 1
    }, {
        "id": 3,
        "text": "Option 3",
        "isHidden": 0
    }]
};

$scope.myFilter = function (value) {
    return value.isHidden === 0;
};

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