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Join 如何在laravel中获取外键表的值_Join_Laravel 4_Foreign Keys - Fatal编程技术网

Join 如何在laravel中获取外键表的值

Join 如何在laravel中获取外键表的值,join,laravel-4,foreign-keys,Join,Laravel 4,Foreign Keys,嗨,我正在尝试获取一个引用了两个表的表的值。我的桌子看起来像这样 pages_content table page_id | content_id ------------------------ 1 | 4 6 | 10 pages table id | title ----------------- 1 | home 6 | contact content table id | content -

嗨,我正在尝试获取一个引用了两个表的表的值。我的桌子看起来像这样

pages_content table

page_id   |  content_id
------------------------
1     |       4
6     |       10

pages table

id   |   title 
-----------------
1    |   home
6    |   contact


content table


id   |  content
-----------------
4    |  home page
10   |  contact us    
我需要引用pages\u content表并获取其他表中的值。 我试过这个

$content = DB::select('select * from pages_content pxc, content c where page_id = '.$page_id.' and content_id = c.id');
我有一个语法错误

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '<p>home page</p> and content_id = c.id' at line 1 (SQL: select * from pages_content pxc, content c where page_id = <p>home page</p> and content_id = c.id) (View: /Applications/MAMP/htdocs/test/app/views/public/content.blade.php)
SQLSTATE[42000]:语法错误或访问冲突:1064您的SQL语法有错误;检查与MySQL服务器版本对应的手册,以了解在第1行的“主页

和content_id=c.id”附近使用的正确语法(SQL:select*from pages_content pxc,content c其中page_id=主页

和content_id=c.id)(视图:/Applications/MAMP/htdocs/test/app/views/public/content.blade.php)

我也尝试了一些方法。如果您需要我提供我尝试过的其他信息,或者如果您需要更多信息,请让我知道

嗯,错误是不言自明的,您在查询中有语法错误,这是因为var$page_id等于主页

(出于某种原因)

right syntax to use near '<p>home page</p> and content_id = c.id' at line 1
$contents = Page::find($page_id)->content;

foreach($contents as $c){
    $c->pivot->created_at //or whatever you want to access
}