日期:2014-05-16  浏览次数:20756 次

请教一个sql,用一个sql写出来
工种 人数 金额
工种1 1 6
工种2 1 6
工种3 1 6
工种4 1 6
工种5 1 56
工种6 1 46455
工种7 1 456
工种8 1 0
工种9 1 0
程汝梅 1 0



希望变成
工种 工种1 工种2 工种3 工种4 工种5
人数 1 1 1 1 1
金额 6 6 6 6 55


假设表为a



------解决方案--------------------
SQL code

SQL> WITH t AS (
  2       SELECT 'wt1' work_type,1 num,100 amount FROM DUAL UNION ALL
  3       SELECT 'wt2' work_type,2 num,200 amount FROM DUAL UNION ALL
  4       SELECT 'wt3' work_type,3 num,300 amount FROM DUAL UNION ALL
  5       SELECT 'wt4' work_type,4 num,400 amount FROM DUAL UNION ALL
  6       SELECT 'wt5' work_type,5 num,500 amount FROM DUAL UNION ALL
  7       SELECT 'wt6' work_type,6 num,600 amount FROM DUAL
  8  )
  9  SELECT 'work_type' work_type,
 10         'wt1' wt1,
 11         'wt2' wt2,
 12         'wt3' wt3
 13    FROM DUAL
 14  UNION ALL
 15  SELECT 'num' num,
 16         TO_CHAR(MAX(DECODE(t.work_type, 'wt1', t.num))) wt1,
 17         TO_CHAR(MAX(DECODE(t.work_type, 'wt2', t.num))) wt2,
 18         TO_CHAR(MAX(DECODE(t.work_type, 'wt3', t.num))) wt3
 19    FROM t
 20  UNION ALL
 21  SELECT 'amount' amount,
 22         TO_CHAR(MAX(DECODE(t.work_type, 'wt1', t.amount))) wt1,
 23         TO_CHAR(MAX(DECODE(t.work_type, 'wt2', t.amount))) wt2,
 24         TO_CHAR(MAX(DECODE(t.work_type, 'wt3', t.amount))) wt3
 25    FROM t
 26  ;

WORK_TYPE WT1                                      WT2                                      WT3
--------- ---------------------------------------- ---------------------------------------- ----------------------------------------
work_type wt1                                      wt2                                      wt3
num       1                                        2                                        3
amount    100                                      200                                      300

------解决方案--------------------
SQL code

select 
sum(decode(工种,'工种1',1,0)) "工种1",
sum(decode(工种,'工种2',1,0)) "工种2",
sum(decode(工种,'工种3',1,0)) "工种3",
sum(decode(工种,'工种4',1,0)) "工种4",
sum(decode(工种,'工种5',1,0)) "工种5"
from a
union all
select 
sum(decode(工种,'工种1',金额,0)) "工种1",
sum(decode(工种,'工种2',金额,0)) "工种2",
sum(decode(工种,'工种3',金额,0)) "工种3",
sum(decode(工种,'工种4',金额,0)) "工种4",
sum(decode(工种,'工种5',金额,0)) "工种5"
from a

------解决方案--------------------
SQL code

建这么一张表:
create table A(
       work_type varchar2(40),
       person_num number,
       money number
)
统计sql如下:
select 
'工种' type,
sum(decode(work_type,'工种1',person_num,0)) "工种1",
sum(decode(work_type,'工种2',person_num,0)) "工种2",
sum(decode(work_type,'工种3',person_num,0)) "工种3",
sum(decode(work_type,'工种4',person_num,0)) "工种4",
sum(decode(work_type,'工种5',person_num,0)) "工种5"
from A
union 
select 
'金额' type,
sum(decode(work_type,'工种1',money,0)) "工种1",
sum(decode(work_type,'工种2',money,0)) "工种2",
sum(decode(work_type,'工种3',money,0)) "工种3",
sum(decode(work_type,'工种4',money,0)) "工种4",
sum(decode(work_type,'工种5',money,0)) "工种5"
from A
order by type

1    工种    1    1    1    1    1
2    金额    6    6    6    6    56