Question

这是PHP Laravel，SQL LeftJoin查询问题。我有2个表分别命名为“ stories”和“ favorite_story”。我想返回“故事”表中的所有值，并基于“ story_id”返回“ favorite_story”表中的匹配值，这是两个表中的通用字段。

以数组形式显示的结果可以看到“ story_id”，“ user_id”的结果为空，不应为空。 Fav_id为空，可以，因为“ favorite_story”表为空。

我希望从“故事”表中获取“ story_id”，并希望从“ favorite_story”表中获取“ user_id”。

这是我的控制器代码：

public function fanfiction(){
    $user_id = Session('userid');
    $data['stories'] =  DB::table('stories')
    ->leftJoin('favorite_story', 'favorite_story.story_id','=','stories.story_id')
    ->Select('*')
    ->get();

    return view('fanfiction', $data);
}

这是上面代码以数组形式的结果。

        [0] => stdClass Object
            (
                [story_id] => 
                [story_title] => Third Story
                [story_desc] => This is "Times" story summary.
                [img] => 2018-12-26-70pgFK87ZN.png
                [story_genre] => Fighting
                [story_character] => Aoba Y.
                [story_rating] => Adult Only
                [story_warning] => Drug Use
                [story_date] => December 26, 2018
                [series_id] => 9999999
                [user_id] => 
                [fav_id] => 
            )

        [1] => stdClass Object
            (
                [story_id] => 
                [story_title] => Second Story
                [story_desc] => This is summary of this new story.
                [img] => 2018-12-19-e6JnEpo3ky.jpg
                [story_genre] => Friendship
                [story_character] => Aijsai
                [story_rating] => G
                [story_warning] => Rape (Actual)
                [story_date] => December 19, 2018
                [series_id] => 9999999
                [user_id] => 
                [fav_id] => 
            )

任何帮助将不胜感激。

Answer 1

将查询更改为：

public function fanfiction()
{
    $user_id = session->get('userid');

    $data['stories'] =  DB::table('stories')
        ->leftJoin('favorite_story', 'favorite_story.story_id','=','stories.story_id')
        ->select('stories.*') // add other fields that you need from favorite_story table here
        ->get();

    return view('fanfiction', $data);
}

发生这种情况的原因是，它是从user_id表中获取story_id和favorite_story，在这种情况下为空。

如何从故事表中获取story_id和user_id。

1 个答案: