使用AngularJS链接多个过滤器

Question

I am working on chaining multiple filters, but while thinking it through i realised that it creates an extreme amount of code (redudancy would be the best term in this use-case), which made me wonder if this could be done more dynamically/generalized. Basically i have 5-6 dropdown menu where a user can select one or more options to filter the data with. I have made this based on one dropdown menu but i want to extend this further. In this code sample i have one dropdown menu which sorts out based on one column.

JavaScript:

<script>
'use strict';
var App = angular.module('App',['ngResource','App.filters']);

App.controller('ExerciseCtrl', ['$scope','$http', function($scope, $http) {

    $scope.selectedMainMuscle = [];

    $http.get('/rest/exercises/')
       .then(function(res){
          $scope.exercises = res.data;                
    });

    $scope.orderProp = 'name';    

    $http.get('/rest/muscles/')
       .then(function(res){
          $scope.muscles = res.data;                
    });    

    $scope.isChecked = function () {
         if (_.contains($scope.selectedMainMuscle, this.muscle.id)) {
             return 'glyphicon glyphicon-ok pull-right';
         }
         return false;
     }; 

     $scope.setSelectedMainMuscle = function () {
         var id = this.muscle.id;
         if (_.contains($scope.selectedMainMuscle, id)) {
             $scope.selectedMainMuscle = _.without($scope.selectedMainMuscle, id);
         } else {
             $scope.selectedMainMuscle.push(id);
         }
         return false;
     };
}]);

angular.module('App.filters', []).filter('mainMuscleFilter', [function () {
    return function (exercises, selectedMainMuscle) {
         if (!angular.isUndefined(exercises) && !angular.isUndefined(selectedMainMuscle) && selectedMainMuscle.length > 0) {
             var tempClients = [];
             angular.forEach(selectedMainMuscle, function (id) {
                 angular.forEach(exercises, function (exercise) {
                     if (angular.equals(exercise.main_muscle.id, id)) {
                         tempClients.push(exercise);
                     }
                 });
             });
             return tempClients;
         } else {
             return exercises;
         }
     };
 }]);
 </script>

HTML:

<div class="btn-group" ng-class="{open: dd2}">
    <button type="button" class="btn btn-default dropdown-toggle" ng-click="dd2=!dd2">Main muscle <span class="caret"></span></button>
    <ul class="dropdown-menu">
        <li ng-repeat="muscle in muscles">
            <a href ng-click="setSelectedMainMuscle()">{%verbatim%}{{muscle.name}}{%endverbatim%} <span data-ng-class="isChecked()"></span></a>
        </li>
    </ul>
</div>    
<table class="table table-hover" >
    <tr><td><strong>Name</strong></td><td><strong>Main muscle</strong></td><td><strong>Equipment</strong></td></tr>
    <tr ng-repeat="exercise in filtered = (exercises | mainMuscleFilter:selectedMainMuscle)">
        {%verbatim%}<td>{{exercise.name}}</td><td>{{exercise.main_muscle.name}}</td><td>{{exercise.equipment.name}}</td>{%endverbatim%}
    </tr>
</table>

Answer 1

I would do all the checks in one filter, I don't see how a mainMuscleFilter would be reusable enough to warrant separation, let alone inclusion in a separate module. You can generalize your methods to toggle the inclusion of an id in an array and to check for the inclusion though:

$scope.selected = {
    muscle: [],
    equipment: []
};

$scope.isChecked = function (arr, id) {
     if (_.contains(arr, id)) {
         return 'glyphicon glyphicon-ok pull-right';
     }
     return false;
 }; 

 $scope.toggleInclusion = function (arr, id) {
     var index = arr.indexOf(id);
     if (index >= 0) {
         arr.splice(index, 1);
     } else {
         arr.push(id);
     }
     return false;
 };

HTML:

<li ng-repeat="muscle in muscles">
    <a href ng-click="toggleInclusion(selected.muscle, muscle.id)">
        {{muscle.name}}
        <span data-ng-class="isChecked(selected.muscle, muscle.id)"></span>
    </a>
</li>