Question

So i have described my problem, later on. Is there any solutions for my problems ?

My filter function

 $scope.filtered = function(){
    for(var i = 0; i < $scope.inventories.length; i++) {
        $scope.filtered_inventories = new Inventory().query($scope.filtrs).$object;
    }
 };

In my template

<select ng-model="filtrs.model" ng-change="filtered()">
<option value="">Model</option>
<option ng-repeat="inventory in inventories" value="{{inventory.model}}">{{inventory.model}}</div>
</select>
<!--filter inventory Manufacturer-->
<select ng-model="filtrs.manufacturer" ng-change="filtered()">
    <option value="">Manufacturer</option>
    <option ng-repeat="inventory in inventories" value="{{inventory.manufacturer}}">{{inventory.manufacturer}}</div>
</select>

When i filter using one select, it filtrs

{
  "manufacturer": "Samsung"
}

When i diselect value and choose Manufacturer it shows

{
  "manufacturer": ""
}

So when i choose Model and deselected manufacturer it does n`t show anything because manufacturer still appears as nothing PROBLEM NR1

{
  "manufacturer": "",
  "model": "M2112"
}

Second problem is that i choose two different values in those selects it display again everything, but logically it should show anything. PROBLEM NR2

Was it helpful?

Solution

That's not how you define options inside a select in angularjs

Please have a look at: http://docs.angularjs.org/api/ng/directive/select.

That's how your code should look like:

<select ng-model="filtrs.model" 
        ng-change="filtered()"
        ng-options="inventory.model as inventory.model for inventory in inventories">
  <option value="">Model</option>
</select>
<!--filter inventory Manufacturer-->
<select ng-model="filtrs.manufacturer" 
        ng-change="filtered()" 
        ng-options="inventory.manufacturer as inventory.manufacturer for inventory in inventories">
    <option value="">Manufacturer</option>
</select>

Once, you've changed your code, let me know if you still have issues.

Licensed under: CC-BY-SA with attribution
Not affiliated with StackOverflow
scroll top