列转行的技巧：求租户六个月中的最高支付金额

有一个表的数据是这样的：

SQL> select * from tb_halfyear where rownum<20;

    USERID       PAY1       PAY2       PAY3       PAY4       PAY5       PAY6
---------- ---------- ---------- ---------- ---------- ---------- ----------
     58972      28212       1861      10061      18060      23402       6482
     15400       3643      10652       7258      20062       4009      17374
     77468      25930      11279      15535      17379       4905      18634
     13111      20085      15247      19790      22442       8886      22813
     99274      17378      11625      11013       1468       9519       7945
     54844      14820      25413       2777      24721      22930      14547
      3724      26339      15024      28935      25647       8447       4697
     30829      10373      17209      26303      12700      12108      12177
     32982       4617      25208      24179      18292       4676      26188
     68572      26515      26073      23818      10484       7451      14882
     92319      17247       4809       4059      14611       2759      11173     25825      29283      18142      18066      12155       4519      23635
     26640      11650      18506       4590      13568       6048      27161
     54964      26773      28542      13488      11044      17087       7363
     51879       6300       7184       1100      29654       1832       8114
     93611      20111      19418      24163       2242      10397       6280
     20660      24278       6772      25644      18662       1077      26924
     89533      25974       6244      28506      19444      16181      22237
     70409      16488      20378      11479       1262      25172       2684

userid是租户的id号，pay1~pay6则是半年来租户缴纳的金额。

明显这是一个中式表格，我们需要将其转化为真正的表数据。

先创建一个表格：

create table tb_halfyearflow(
   id number(8,0) primary key,
   userid number(8,0) not null,
   pay number(5,0) not null,
   month number(3,0) not null
)

然后插值：

insert into tb_halfyearflow(id,userid,pay,month) select rownum,userid,pay1,1 from tb_halfyear;
insert into tb_halfyearflow(id,userid,pay,month) select 100000+rownum,userid,pay2,2 from tb_halfyear;
insert into tb_halfyearflow(id,userid,pay,month) select 200000+rownum,userid,pay3,3 from tb_halfyear;
insert into tb_halfyearflow(id,userid,pay,month) select 300000+rownum,userid,pay4,4 from tb_halfyear;
insert into tb_halfyearflow(id,userid,pay,month) select 400000+rownum,userid,pay5,5 from tb_halfyear;
insert into tb_halfyearflow(id,userid,pay,month) select 500000+rownum,userid,pay6,6 from tb_halfyear;

插值后，tb_halfyearflow就是纯粹的流水表了，求每个租户单月最高缴纳金额就方便了：

select userid,max(pay) from tb_halfyearflow group by userid order by userid

让我们看看id从1~9的用户最高缴纳了多少：

SQL> select * from (
  2  select userid,max(pay) from tb_halfyearflow group by userid order by userid
  3  )
  4  where rownum<10;

    USERID   MAX(PAY)
---------- ----------
         1      24931
         2      27306
         3      25578
         4      27853
         5      29227
         6      26800
         7      23628
         8      22326
         9      26010

已选择9行。

再看看userid=2的租户其它消费情况：

SQL> select * from tb_halfyearflow where userid=2  order by month;

        ID     USERID        PAY      MONTH
---------- ---------- ---------- ----------
     50656          2      27050          1
    150656          2      19124          2
    250656          2       6690          3
    350656          2      27306          4
    450656          2       5520          5
    550656          2      20530          6

已选择6行。

果然是27306最高。

而下面这个代码能直接输出用户2的最高消费金额和消费月：

select a.userid,a.pay,a.month from tb_halfyearflow a
          right join (
          select * from (
select userid,max(pay) as maxpay from tb_halfyearflow group by userid order by userid
) tb
where tb.userid=2 ) b
          on a.userid=b.userid and a.pay=b.maxpay
order by a.userid desc

SQL> select a.userid,a.pay,a.month from tb_halfyearflow a
  2            right join (
  3    select * from (
  4  select userid,max(pay) as maxpay from tb_halfyearflow group by userid order by userid
  5  ) tb
  6  where tb.userid=2 ) b
  7    on a.userid=b.userid and a.pay=b.maxpay
  8  order by a.userid desc;

    USERID        PAY      MONTH
---------- ---------- ----------
         2      27306          4

也可以换一种写法：

SQL> select a.userid,a.pay,a.month from (
  2    select * from (
  3  select userid,max(pay) as maxpay from tb_halfyearflow group by userid order by userid
  4  ) tb
  5  where tb.userid=2 ) b left join tb_halfyearflow a
  6  on a.userid=b.userid and a.pay=b.maxpay;

    USERID        PAY      MONTH
---------- ---------- ----------
         2      27306          4

好了，到此完成需求。

--2020年2月16日--

给tb_halfyear建表：

create table tb_halfyear(
    userid number(8,0) primary key,
    pay1 number(5,0),
    pay2 number(5,0),
    pay3 number(5,0),
    pay4 number(5,0),
    pay5 number(5,0),
    pay6 number(5,0)
)

给tb_halfyear充值：

insert into tb_halfyear
select rownum,
        dbms_random.value(1000,30000),
        dbms_random.value(1000,30000),
        dbms_random.value(1000,30000),
        dbms_random.value(1000,30000),
        dbms_random.value(1000,30000),
        dbms_random.value(1000,30000)
from dual
connect by level<=100000
order by dbms_random.random

--END--