Question

我正在尝试在mySQL的IF子句中使用LIKE并收到错误。

SELECT nc.web_name AS company_name,
    IF((b.booth_number LIKE '%Cafe%', b.booth_number, substring(b.booth_number,4)) AS booth_number, 
    nc.site_url AS website, IF (nc.ismi_status = 1, 'Member','') AS member_status 
    FROM booth_sale bs JOIN {ismi.booth} b ON bs.booth_id = b.booth_id 
    JOIN ismi.new_people np ON bs.contact_id = np.id 
    JOIN ismi.new_company nc ON nc.id = np.company_id 
    WHERE b.show_event_id = 298 AND status IN(44,45) ORDER BY 3

如果我这样做，查询工作正常：

IF((b.booth_number = 'Cafe', b.booth_number, substring(b.booth_number,4)) AS booth_number

但是我需要检查多种可能性 - 可能有Cafe，Cafe1，Cafe2等，我想返回b.booth_number以获取该字段值包含Cafe的任何记录。

这是我得到的错误：

PDOException：SQLSTATE [42000]：语法错误或访问冲突：1064 您的SQL语法有错误;检查手册对应于您的MySQL服务器版本，以便使用正确的语法靠近'AS booth_number，nc.site_url AS网站，IF（nc.ismi_status = 1， 'Member'，''）'在第2行：SELECT nc.web_name AS company_name， IF（（b.booth_number LIKE'％Cafe％'，b.booth_number， substring（b.booth_number，4））AS booth_number，nc.site_url AS网站， IF（nc.ismi_status = 1，'Member'，''）AS member_status FROM {ismi.booth_sale} bs JOIN {ismi.booth} b ON bs.booth_id = b.booth_id JOIN {ismi.new_people} np ON bs.contact_id = np.id JOIN {ismi.new_company} nc ON nc.id = np.company_id WHERE b.show_event_id = 298 AND状态IN（44,45）ORDER BY 3

任何人都可以告诉我我做错了什么，或者是否还有其他方法可以做到这一点？

Answer 1

您的IF语句有一个太多的空心括号。删除第一个，你应该是好的（当然，等待任何其他SQL错误）：

...
IF(b.booth_number LIKE '%Cafe%', b.booth_number, substring(b.booth_number,4)) AS booth_number, 
...

在mysql的IF中使用LIKE

1 个答案: