MYSQL |在 COUNT() 中使用子查询时出现语法错误答案

【问题标题】：MYSQL | Getting syntax error for using subquery in COUNT()MYSQL |在 COUNT() 中使用子查询时出现语法错误
【发布时间】：2015-02-14 09:13:48
【问题描述】：

SELECT 
E.`employee_id`,
E.`full_name`,
LE.`no_of_leaves` AS AllocatedLeaves,
MLLT.`leave_type` AS LeaveTypeName,
COUNT(SELECT * FROM leave_approval WHERE employee_id = 1) AS TotalLeavesTaken
FROM employee E
INNER JOIN leave_entitlement LE
ON E.`employee_id` = LE.`employee_id`
INNER JOIN `ml_leave_type` MLLT
ON MLLT.`ml_leave_type_id` = LE.`ml_leave_type_id`
LEFT JOIN leave_approval LA
ON E.`employee_id` = LA.`employee_id`
LEFT JOIN leave_application LAPP
ON LAPP.`application_id` = LA.`leave_application_id`
LEFT JOIN ml_leave_type MLLTLA
ON MLLTLA.`ml_leave_type_id` = LAPP.`ml_leave_type_id`

我在计数附近遇到语法错误，但我试图找出语法错误但我找不到任何？

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 'select * from leave_approval where employee_id = 1) AS TotalLeavesTaken
from emp' at line 6

真的是语法错误吗。还是我在这里遗漏了什么？？

【问题讨论】：

标签： mysql sql select count group-by

【解决方案1】：

改变

COUNT(SELECT * FROM leave_approval WHERE employee_id = 1) AS TotalLeavesTaken

到

 (SELECT count(*) FROM leave_approval WHERE employee_id = 1) AS TotalLeavesTaken

【讨论】：

【解决方案2】：

这一行：

COUNT(SELECT * FROM leave_approval WHERE employee_id = 1) AS TotalLeavesTaken

不正确。如果不将select * 子查询放在括号中，您将无法对其进行计数，但即便如此，您也需要group by 和其他逻辑。更好的方法是：

(Select count(*) from leave_approval where employee_id = 1) AS TotalLeavesTaken

【讨论】：

感谢您的快速回答。完美的答案..问题解决了.. :)

【解决方案3】：

在您的情况下，无需使用子查询进行计数，请检查以下查询

试试这个：

SELECT E.employee_id, E.full_name, LE.no_of_leaves AS AllocatedLeaves,
       MLLT.leave_type AS LeaveTypeName,
       SUM(CASE WHEN LA.employee_id = 1 THEN 1 ELSE 0 END) AS TotalLeavesTakenByEmplyeeNo1
FROM employee E
INNER JOIN leave_entitlement LE ON E.employee_id = LE.employee_id
INNER JOIN `ml_leave_type` MLLT ON MLLT.ml_leave_type_id = LE.ml_leave_type_id
LEFT JOIN leave_approval LA ON E.employee_id = LA.employee_id
LEFT JOIN leave_application LAPP ON LAPP.application_id = LA.leave_application_id
LEFT JOIN ml_leave_type MLLTLA ON MLLTLA.ml_leave_type_id = LAPP.ml_leave_type_id
GROUP BY E.employee_id;

【讨论】：

【解决方案4】：

试试这个查询

SELECT E.`employee_id`,E.`full_name`,LE.`no_of_leaves` AS AllocatedLeaves,MLLT.`leave_type` AS LeaveTypeName,
SUM( CASE WHEN LA.employee_id = '1' THEN 1 ELSE 0 END ) as TotalLeavesTaken
FROM employee E
INNER JOIN leave_entitlement LE
ON E.`employee_id` = LE.`employee_id`
INNER JOIN `ml_leave_type` MLLT
ON MLLT.`ml_leave_type_id` = LE.`ml_leave_type_id`
LEFT JOIN leave_approval LA
ON E.`employee_id` = LA.`employee_id`
LEFT JOIN leave_application LAPP
ON LAPP.`application_id` = LA.`leave_application_id`
LEFT JOIN ml_leave_type MLLTLA
ON MLLTLA.`ml_leave_type_id` = LAPP.`ml_leave_type_id`

只有你必须将 COUNT(SELECT * FROM leave_approval WHERE employee_id = 1) 替换为 SUM(CASE WHEN LA.employee_id = '1' THEN 1 ELSE 0 END)

谢谢，试试这个，因为我已经尝试过这个查询，它会给你完美的结果

【讨论】：