Question

我有函数返回员工编号整数

CREATE  FUNCTION `getUser`(s CHAR(20)) RETURNS char(50) CHARSET latin1
RETURN  @user;

当我用（0.00秒）

执行返回时

但它执行此

select e.emp_number from hs_hr_employee e where 

hie_code_1 in ( select hie_code_1 from hs_hr_emp_level L,hs_hr_users U   where L.emp_number=U.emp_number and L.emp_number=getUser2() and  ( U.def_level=1 or U.def_level=4 )   )
or
 hie_code_3 in ( select hie_code_3 from hs_hr_emp_level L,hs_hr_users U   where L.emp_number=U.emp_number and L.emp_number=getUser2()  and  U.def_level=2   )
or
  hie_code_4 in ( select hie_code_4 from hs_hr_emp_level L,hs_hr_users U   where L.emp_number=U.emp_number and L.emp_number=getUser2()   and  U.def_level=3 )

变得更慢;

但当我硬编码为

时

select e.emp_number from hs_hr_employee e where 

hie_code_1 in ( select hie_code_1 from hs_hr_emp_level L,hs_hr_users U   where L.emp_number=U.emp_number and L.emp_number=5 and  ( U.def_level=1 or U.def_level=4 )   )
or
 hie_code_3 in ( select hie_code_3 from hs_hr_emp_level L,hs_hr_users U   where L.emp_number=U.emp_number and L.emp_number=5  and  U.def_level=2   )
or
  hie_code_4 in ( select hie_code_4 from hs_hr_emp_level L,hs_hr_users U   where L.emp_number=U.emp_number and L.emp_number=5   and  U.def_level=3 )

这将以（0.00秒）时间返回。请给出optimiste函数的解决方案;

Answer 1

你的样品会让我的眼睛受伤，但这是尝试一种可能的选择

SELECT DISTINCT e.emp_number 

FROM hs_hr_employee e 

INNER JOIN hs_hr_emp_level L
ON L.emp_number = e.emp_number
AND (
    e.hie_code_1 = U.hie_code_1
    OR e.hie_code_3 = U.hie_code_3
    OR e.hie_code_4 = U.hie_code_4
)

INNER JOIN hs_hr_users U 
ON U.emp_number = e.emp_number
AND U.def_level IN (1,2,3,4)

-- You stated getUser2() returns paramter @user, so just accessing directly
WHERE e.emp_number = @user

假设如下

hs_hr_employee.emp_number == hs_hr_emp_level.emp_number == hs_hr_emp_level.emp_number
hs_hr_employee.hie_code_*n* == hs_hr_users.hie_code_ 名词的
getUser2()返回@user，这是一个员工编号

上面的语法完成了复杂查询的功能。由于上述情况不正确，因此很难提供建议 - 如果您不提供完整的表格结构/适当的细节，那么我无法进一步帮助。

mysql函数类型转换错误

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