Question

我在like函数中尝试Django原始sql查询，但结果为空。我尝试mysql客户端工具这个查询并获得许多记录。如何解决这个问题？

我的疑问：

  SELECT s.* , s.id as pk 
    FROM d_status as s, 
         (select post_id 
            from p_useractions 
           where from_user_id in 
                 (select to_user_id 
                    from p_fallowers 
                   where from_user_id = 1)
         ) as a 
   WHERE s.from_user_id = 1 OR 
         s.text like '%@Mustafa Yontar2123%' OR 
         s.id = a.post_id 
GROUP BY s.id 
ORDER BY s.last_update 
   LIMIT 25

Answer 1

尝试用%替换每个%%。因此，示例中的相关部分如下所示：'%%@Mustafa Yontar2123%%'。

Answer 2

引用MySQL的文档：

With LIKE you can use the following two wildcard characters in the pattern.

    Character  Description
    %          Matches any number of characters, even zero characters
    _          Matches exactly one character

To test for literal instances of a wildcard character, precede it by the 
escape character. If you do not specify the ESCAPE character, \ is assumed.

所以你的模式必须是'\%@Mustafa Yontar2123\%'。查询的Python代码如下所示：

query = r"""SELECT s.* , s.id as pk FROM d_status as s, 
                   (SELECT post_id FROM p_useractions WHERE from_user_id in
                     (SELECT to_user_id FROM p_fallowers WHERE from_user_id = 1)) as a
            WHERE s.from_user_id = 1 or
                  s.text like '\%@Mustafa Yontar2123\%' or
                  s.id = a.post_id
            GROUP BY s.id
            ORDER BY s.last_update
            LIMIT 25"""

PostgreSQL和SQLite的文档提到了同样的事情。

django原始查询百分号问题

2 个答案: