我需要在sql中回答这个棘手的问题,我尝试了很多方法,最终得到了这个查询,它只检索max sum或min sum,而不检索相应的雇员。
此代码仅检索最大费用总和值,但我需要它的最大总和值(按职务代码分组)与员工id或姓名作为问题中的请求。
select MAX(t.SUM_CHARGE)
from
(select DISTINCT e.JOB_CODE2019,a.EMP_NUM2019,a.PROJ_NUM2019,SUM(a.ASSIGN_CHARGE) as SUM_CHARGE
from ASSIGNMENT2019 a inner join EMPLOYEE2019 e
on a.EMP_NUM2019 = e.EMP_NUM2019
inner join JOB2019 j
on e.JOB_CODE2019 = j.JOB_CODE2019
group by a.EMP_NUM2019)t
group by t.JOB_CODE2019 ;
下面是表create和insert语句。
create table JOB2019(
JOB_CODE2019 int,
JOB_DESCRIPTION2019 varchar(30),
JOB_CHG_HOUR2019 decimal(4,2),
PRIMARY KEY(JOB_CODE2019 )
);
create table EMPLOYEE2019(
EMP_NUM2019 int,
EMP_LNAME2019 varchar(20),
EMP_FNAME2019 varchar(20),
EMP_INITIAL2019 varchar(1),
EMP_HIREDATE2019 varchar(50),
JOB_CODE2019 int,
primary key (EMP_NUM2019),
foreign key (JOB_CODE2019) references JOB2019(JOB_CODE2019)
);
create table PROJECT2019(
PROJ_NUM2019 int,
PROJ_NAME2019 varchar(20),
EMP_NUM2019 int,
primary key (PROJ_NUM2019),
foreign key (EMP_NUM2019) references employee2019(EMP_NUM2019)
);
create table ASSIGNMENT2019(
ASSIGN_NUM2019 int,
ASSIGN_DATE2019 varchar(50),
PROJ_NUM2019 int,
EMP_NUM2019 int,
ASSIGN_HOURS2019 decimal(5,2),
ASSIGN_CHG_HOUR2019 decimal(5,2),
ASSIGN_CHARGE2019 decimal(7,2) ,
primary key (ASSIGN_NUM2019),
foreign key (PROJ_NUM2019) references project2019(PROJ_NUM2019),
foreign key (EMP_NUM2019) references employee2019(EMP_NUM2019)
);
insert into JOB2019
values (500,'Programmer',35.75)
,(501,'Systems Analyst',96.75)
,(502,'Database Designer',105.00)
,(503,'Electrical Engineer',84.50)
,(504,'Mechanical Engineer',67.90)
,(505,'Civil Engineer',55.78)
,(506,'Clerical Support',26.87)
,(507,'DSS Analyst',45.95)
,(508,'Applications Designer',48.10)
,(509,'Bio Technician',34.55)
,(510,'General Support',18.36);
insert into EMPLOYEE2019
values (101,'News','John','G','08Nov2000'd,502)
,(102,'Senior','David','H','12Jun1989'd,501)
,(103,'Arbough','June','E','01Dec1997'd,503)
,(104,'Ramoras','Anne','K','15Nov1988'd,501)
,(105,'Johnson','Alice','K','01Feb1994'd,502)
,(106,'Smithfield','William',' ','22Jun2005'd,500)
,(107,'Alonzo','Maria','D','10Oct1994'd,500)
,(108,'Washington','Ralph','B','22Aug1889'd,501)
,(109,'Smith','Larry','W','18Jul1999'd,501)
,(110,'Olenko','Gerald','A','11Dec1996'd,505)
,(111,'Wabash','Geoff','B','04Apr1989'd,506)
,(112,'Smithson','Darlene','M','23Oct1995'd,507)
,(113,'Joenbrood','Delbert','K','15Nov1994'd,508)
,(114,'Jones','Annelise',' ','20Aug1991'd,508)
,(115,'Bawangi','Travis','B','25Jan1990'd,501)
,(116,'Pratt','Gerald','L','05Mar1995'd,510)
,(117,'Williamson','Angie','H','19Jun1994'd,509)
,(118,'Frommer','James','J','04Jan2006'd,510);
insert INTO PROJECT2019
values (15,'Evergreen',105)
,(18,'Amber Wave',104)
,(22,'Rolling Tide',113)
,(25,'Starflight',101);
insert into ASSIGNMENT2019(ASSIGN_NUM2019,ASSIGN_DATE2019,PROJ_NUM2019,EMP_NUM2019,ASSIGN_HOURS2019,ASSIGN_CHG_HOUR2019)
values(1001,'04Mar2012'd,15,103,2.6,84.50)
,(1002,'04Mar2012'd,18,118,1.4,18.36)
,(1003,'05Mar2012'd,15,101,3.6,105.00)
,(1004,'05Mar2012'd,22,113,2.5,48.10)
,(1005,'05Mar2012'd,15,103,1.9,84.50)
,(1006,'05Mar2012'd,25,115,4.2,96.75)
,(1007,'05Mar2012'd,22,105,5.2,105.00)
,(1008,'05Mar2012'd,25,101,1.7,105.00)
,(1009,'05Mar2012'd,15,105,2.0,105.00)
,(1010,'06Mar2012'd,15,102,3.8,96.75)
,(1011,'06Mar2012'd,22,104,2.6,96.75)
,(1012,'06Mar2012'd,15,101,2.3,105.00)
,(1013,'06Mar2012'd,25,114,1.8,48.10)
,(1014,'06Mar2012'd,22,111,4.0,26.87)
,(1015,'06Mar2012'd,25,114,3.4,48.10)
,(1016,'06Mar2012'd,18,112,1.2,45.95)
,(1017,'06Mar2012'd,18,118,2.0,18.36)
,(1018,'06Mar2012'd,18,104,2.6,96.75)
,(1019,'06Mar2012'd,15,103,3.0,84.50)
,(1020,'07Mar2012'd,22,105,2.7,105.00)
,(1021,'08Mar2012'd,25,108,4.2,96.75)
,(1022,'07Mar2012'd,25,114,5.8,48.10)
,(1023,'07Mar2012'd,22,106,2.4,35.75);
Update assignment2019
set ASSIGN_CHARGE2019 = ASSIGN_HOURS2019 * ASSIGN_CHG_HOUR2019;
我非常感谢你对解决上述问题的帮助。谢谢您。
1条答案
按热度按时间daolsyd01#
你需要帮助,所以你也必须努力呈现一个mre,为什么我要为一个非常简单的sql查询提供一个最小的可复制的例子?
这就是你想要的,给出了一个非常难听的问题
db<>在这里摆弄