javascript - 我想匹配 Angular foreach 中的特定 id

Question

我正在尝试匹配id，并且我需要获取特定的描述，我怎样才能得到这个？？

我该如何解决这个问题？

$scope.getAlbumList = '[{"album_desc":"Test1","id":1,"album_title":"Test","ImageName":"Desert.jpg","album_pics":[{"media_type":"image/jpeg","last_modified_date":1247549551000,"thumnail_pic_loc":"Desert.jpg","large_pic_loc":"Desert.jpg","filter_type":"image/jpeg","pic_id":"d5bd"}]},{"album_desc":"Test1","id":2,"album_title":"Test1","ImageName":"Koala.jpg","album_pics":[{"media_type":"image/jpeg","last_modified_date":1247549551000,"thumnail_pic_loc":"Koala.jpg","large_pic_loc":"Koala.jpg","filter_type":"image/jpeg","pic_id":"d5bd"}]},{"album_desc":"Album Title","id":3,"album_title":"Album Title","ImageName":"Tulips.jpg","album_pics":[{"media_type":"image/jpeg","last_modified_date":1247549551000,"thumnail_pic_loc":"Tulips.jpg","large_pic_loc":"Tulips.jpg","filter_type":"image/jpeg","pic_id":"d5bd"}]},{"album_desc":"fsdfdsfds fsdfds fds fdsfdsfdsfsdfdsfbdsfkdskjfh dskjfh skjdfhdskjhfk sjdhfdskjhfksjdhfkjdsfh sf dfsjfnkjdshf dsjhf ksksfhkjfhds fdshfhkjdsfhdskjhfkj dshf dskjfh skjdhfskjdhfkjdsfhdskjfhkjdsfhsd fkjdshfkjsdhfkj hjshdjfhdskjfhs fdsjhf dskjhfdskjfh dskjfhdskjfhsdkjfhdskjfsdk jdshfkjdsh kjfdshkjfshkjdfshkj fdshkjdskjsh kjds hfkjsd kdshfkjdshfkjsdhfkjdshfkjdshfkjsdkjf fkjdsh fkjshdfkjdshf dskjf","id":4,"album_title":"sdfds","ImageName":"Lighthouse.jpg","album_pics":[{"media_type":"image/jpeg","last_modified_date":1247549551000,"thumnail_pic_loc":"Lighthouse.jpg","large_pic_loc":"Lighthouse.jpg","filter_type":"image/jpeg","pic_id":"d5bd"}]}]';
$scope.getalbumarray = 'http://localhost/learn/comment/#/album_details/1';

        angular.forEach($scope.getAlbumList, function(item) {
        $scope.albumDescription = item.id;      
        //console.log(item.id);
        //console.log($scope.getalbumarray[$scope.getalbumarray.length - 1]);
        if(item.id == $scope.getalbumarray[$scope.getalbumarray.length - 1]){
            $scope.albumDescription = item.album_desc;
            //console.log('1');
        }else{
            //console.log('2');
            $scope.albumDescription = item.album_desc;
        }
         });

         console.log($scope.albumDescription);
        $scope.showDescrption = $scope.albumDescription;

Answer 1

使用它会对你有用

  $scope.getalbumarray = window.location.href.split('/');

        $scope.albumDescr = '';
        var albumId = $scope.getalbumarray[$scope.getalbumarray.length - 1];
        var albums = $scope.getAlbumList.filter(function(item) {

            if(item.id == albumId){             
                $scope.albumDescr += item.album_desc;
            }           
            //return $scope.albumDescription = item.album_desc;;
        })[0];

        $scope.albumDescription = $scope.albumDescr;
        console.log('$scope.albumDescription');