Question

此SQL命令中的错误

select account_id,fname,lname ,
    (select email from account where id=account_id) as email 
from user_account 
where account_id in (
       select destination_id 
       from friendship 
       where source_id=$user_id) 
    and (concat(fname,lname) like '%a%' or email like '%a%')

email是account表

中的列

错误消息为：Unknown column email

我想收到电子邮件并将其与fname + lname

一样进行测试

Answer 1

听起来您的表user_account没有在WHERE子句中引用的名为email的列。请尝试使用JOIN代替：

   select ua.account_id,
          ua.fname,
          ua.lname, 
          a.email 
from user_account ua
inner join  account a where ua.id = a.account_id
and a.account_id  in ( select destination_id from friendship where source_id=$user_id) 
and (concat(fname,lname) like '%a%' or email like '%a%');

Answer 2

重新格式化查询的易读性：

SELECT account_id, fname, lname,
       (SELECT email FROM account WHERE id = account_id) AS email
  FROM user_account 
 WHERE account_id IN (SELECT destination_id FROM friendship WHERE source_id = $user_id)
   AND (CONCAT(fname, lname) LIKE '%a%' OR email LIKE '%a%')

最可能的原因是列email在主WHERE子句中可以引用的唯一表是user_account，但email列只能在account表。

最简单的解决方法是删除选择列表中的子查询并改为使用常规连接：

SELECT u.account_id, u.fname, u.lname, a.email
  FROM user_account AS u
  JOIN account      AS a ON a.id = u.account_id
 WHERE account_id IN (SELECT destination_id
                        FROM friendship AS f
                       WHERE f.source_id = $user_id)
   AND (CONCAT(u.fname, u.lname) LIKE '%a%' OR a.email LIKE '%a%');

我假设$user_id是一个输入值，并且符号在您的系统中有意义;它通常在SQL中无效。

目前尚不清楚IN条款有多大好处;也可以与其他表格建立联接。

Answer 3

我假设您在Accout表中的ID与User_Account中的Account_id以及友谊表中的目标ID相同。

select ua.account_id,
       ua.fname,
       ua.lname, 
       a.email 
from user_account ua
inner join  account a where ua.id = a.account_id
and a.account_id  
in ( select destination_id from friendship where source_id=$user_id) and (concat(fname,lname) like '%a%' or email like '%a%')

Answer 4

此SQL命令中的错误

select account_id,fname,lname ,
(select email from account where id=account_id) as email from user_account 
where account_id 
  in ( select destination_id from friendship where source_id=$user_id) and          (concat(fname,lname) like '%a%' or email like '%a%')

问题在于，在以下语法中，电子邮件是表别名，而不是列名

(select email from account where id=account_id) as email

看起来你已经从p.campell获得了更好的方法

这个SQL命令有什么错误？

4 个答案: