Question

我有一个原始的SQL查询代码，我想使用laravel中的laravel查询构建器运行：

 $q = "SELECT * FROM wp_posts WHERE post_parent = ?  ORDER BY ? ? LIMIT ?,   ?";    

  $values = ['40','post_status', "ASC" ,'1','10'];
   $q = $db->connection()->select($q,$values);

此查询将抛出错误消息

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 '? LIMIT ?, ?' at line 1 (SQL:    
SELECT * FROM wp_posts WHERE post_parent = `40` ORDER BY `post_status` ASC LIMIT 1, 10)

现在，当我从错误输出中复制查询并将其粘贴到phpmyadmin sql runner中时，它将运行时没有任何错误。

任何想法？

Answer 1

可能是post_parent不是字符串而且不需要背景

 $q = "SELECT * FROM wp_posts WHERE post_parent = ?  ORDER BY ? ? LIMIT ?,   ?";    

$values = [ 40,'post_status', "ASC" ,'1','10'];

Answer 2

你可以写

  $q = "SELECT * FROM wp_posts WHERE post_parent = ?  ORDER BY ? ? LIMIT ?,   ?";

  $values = [ 40,'post_status', "ASC" , 1, 10];

希望这将解决您的问题

Answer 3

如果您想使用查询构建器执行此操作，这应该可以。

$posts = DB::table('wp_posts')->where('post_parent', 40)->orderBy('post_status', 'ASC')->take(10)->get();

Laravel会自动将值绑定到PDO。

Answer 4

似乎查询构建器不允许将SQL命令放入绑定变量中，当我从其工作的绑定变量中删除“ASC”时：

$q = "SELECT * FROM wp_posts WHERE post_parent = ?  ORDER BY ? ASC LIMIT ?,   ?"; 
$values = [40,'post_status', 1,10];
$q = $db->connection()->select($q,$values);

eloquent - 使用绑定运行原始sql的查询构建器

4 个答案: