使用Laravel中控制器/视图中的模型中的函数

时间:2013-07-07 17:43:28

标签: sql controller laravel laravel-4 models

我正在尝试从我在模型中的函数中放置的查询中访问数据。我试图在控制器中调用该函数,然后将该数据发送到视图。到目前为止,我一直没有成功。这是代码:

型号:Fanartist.php

public function fan_likes() {
        $fan_likes = DB::table('fanartists')
                    ->join('artists', 'fanartists.fan_id', '=', 'artists.id')
                    ->where('fanartists.fan_id', '=', Auth::user()->id)
                    ->select('artists.id', 'artists.stage_name', 'artists.city', 'artists.state', 'artists.image_path', 'artists.description');

           }

控制器:FansController.php

public function getHome() {
            return View::make('fans.home')
            ->with('fans', Fan::all())
            ->with('fanartists', Fanartist::fan_likes());

        }

查看:

@foreach($fanartists as $fanartist)

{{$fanartist}}

@endforeach

当我运行时,我收到错误:

Non-static method Fanartist::fan_likes() should not be called statically, assuming $this from incompatible context

感谢您的帮助和建议。

更新:

新错误。我正在返回视图,但现在,尝试运行它:

@foreach($fanartists as $fanartist)
                {{$fanartist->artists.id}}

            @endforeach

我收到错误:

log.ERROR:异常'ErrorException',消息'试图获取非对象的属性'在/ Applications / MAMP / htdocs / crowdsets / laravel-master / app / storage / views / 2ed7fc8952dab08cf4cb4f4e3d40d1ab:100

更新2:

跑步:

@foreach($fanartists as $fanartist)
<?php var_dump($fanartist); ?>          
            @endforeach

我得到以下输出:

NULL array(6) { [0]=> string(10) "artists.id" [1]=> string(18) "artists.stage_name" [2]=> string(12) "artists.city" [3]=> string(13) "artists.state" [4]=> string(18) "artists.image_path" [5]=> string(19) "artists.description" } bool(false) string(10) "fanartists" array(1) { [0]=> object(Illuminate\Database\Query\JoinClause)#201 (3) { ["type"]=> string(5) "inner" ["table"]=> string(7) "artists" ["clauses"]=> array(1) { [0]=> array(4) { ["first"]=> string(17) "fanartists.fan_id" ["operator"]=> string(1) "=" ["second"]=> string(10) "artists.id" ["boolean"]=> string(3) "and" } } } } array(1) { [0]=> array(5) { ["type"]=> string(5) "Basic" ["column"]=> string(17) "fanartists.fan_id" ["operator"]=> string(1) "=" ["value"]=> string(1) "1" ["boolean"]=> string(3) "and" } } NULL NULL NULL NULL NULL NULL

2 个答案:

答案 0 :(得分:0)

Fanartist::fan_likes()不应该静态调用,但如果要静态调用它,请将fan_likes函数更改为static。 为你的例子

public static function fan_likes() {
    $fan_likes = DB::table('fanartists')
                ->join('artists', 'fanartists.fan_id', '=', 'artists.id')
                ->where('fanartists.fan_id', '=', Auth::user()->id)
                ->select('artists.id', 'artists.stage_name', 'artists.city', 'artists.state', 'artists.image_path', 'artists.description');
return $fan_likes;
       }

答案 1 :(得分:0)

如果你的MyClass控制器中有这样的非静态函数:

    public function my_func( ) {
            // do stuff
    return ( $stuff ) ;
}

然后在您的控制器中,您可以使用以下代码调用它:

$local_obj = MyClass::findOrFail( $myobj_id );
$rval = $local_obj->my_func();

如果你想从你的视图中调用它,我认为你需要将对象传递给类似于此的视图:

$data['my_obj'] = $local_obj;
return View::make('view_name.show')->with('data',$data);

然后在视图中使用这样的东西:

{{{$data['my_obj']->my_func() }}}

(上面是从工作代码中复制和编辑的 - 如果我犯了错字,就会开始编辑)