如何在函数中传递过滤操作的结果

Question

在这里:http://jsfiddle.net/CBgcP/742/

我试图通过onclick将自定义过滤器的结果传递到一个作用域函数:

 <span ng-click="open(friend.name | custom)")

这个语法不正确。筛选操作的结果可以通过on-click传递给函数吗?

SRC:

<div ng:app="myApp">
 <div ng-controller="HelloCntl">
 <ul>
    <li ng-repeat="friend in friends | weDontLike:'Adam'">
        <span ng-click="open(friend.name | custom)")>{{friend.name}}</span>
        <span>{{friend.phone}}</span>
    </li>
 </ul>
 </div>
</div>

span
{
 width: 200px;   
}

angular.module('myApp',[]).filter('weDontLike', function(){
    return function(items, name){
        var arrayToReturn = [];        
        for (var i=0; i<items.length; i++){
            if (items[i].name != name) {
                arrayToReturn.push(items[i]);
            }
        }
        return arrayToReturn;
    };
});
angular.module('myApp', []).filter('custom', function() {
  return function(input) {
    return input.substring(0 , 2)
  };
});
function HelloCntl($scope) {
    $scope.open = function(){
        console.log('in open');
    }

     $scope.friends = [
        {
        name: 'John',
        phone: '555-1276'},
    {
        name: 'Mary',
        phone: '800-BIG-MARY'},
    {
        name: 'Mike',
        phone: '555-4321'},
    {
        name: 'Adam',
        phone: '555-5678'},
    {
        name: 'Julie',
        phone: '555-8765'}
    ];
}

Answer 1

您可以使用$filter服务来完成此操作:

function HelloCntl($scope, $filter) {
    $scope.open = function(friend){
        console.log('in open with friend's name: ' + friend);
        console.log('friend's name with filter applied: ' + $filter('custom')(friend);
    }
    // ...
}