数据量大，也是有助于查找问题的。

数据量大，不可否认管理和维护起来难度都会有所增加。但是我们应该看到另一个好的方面，数据量大了，也是有助于查找问题的。下面是论坛里一个朋友发帖子问的问题：

 

现有一张表 表名tabel1
        KHMC        PZM        GX      SL
          AA          A-01        S1      2
          AA          A-01        S3      1
          AA          A-01        S1      4
          AA          A-01        S3      1
          AA          A-01        S1      8
          AA          A-01        S3      5
          BB          B-01        S1      4
          BB          B-01        S2      2
          BB          B-01        S1      9   
          BB          B-01        S2      4 
          BB          B-01        S1      4
          BB          B-01        S2      2

  请帮忙写 汇总要求格式  按 KHMC PZM 汇总 GX要全部显示出来 没有的现实0
        KHMC        PZM        GX      SL 
          AA        A-01        S1      14 
          AA        A-01        S2      0
          AA        A-01        S3      7
          BB        B-01        S1      17
          BB        B-01        S2      8
          BB        B-01        S3      0 

 

我看到一个朋友的回帖，内容如下：

 

select a.khmc,a.pzm,a.gx,nvl(num,0) from
(select * from (select distinct khmc,pzm from tabel1) a,(select distinct gx from tabel1) b) a,
(select khmc,pzm,gx,sum(sl) num from tabel1 group by khmc,pzm,gx) b
where a.khmc = b.khmc(+) and a.pzm = b.pzm(+) and a.gx = b.gx(+)
order by 1,2,3;

 

我一开始没仔细看，不明白上面第二行红色的那句话的作用。就自己做了下实验：

SQL> create table tabel1(khmc varchar2(10),pzm varchar2(10),gx varchar2(10),sl n
umber);

Table created.

SQL> insert into tabel1 values('AA','A-01','S1',2);

1 row created.

SQL> insert into tabel1 values('AA','A-01','S1',4);

1 row created.

SQL> insert into tabel1 values('AA','A-01','S1',8);

1 row created.

SQL> insert into tabel1 values('AA','A-01','S3',1);

1 row created.

SQL> insert into tabel1 values('AA','A-01','S3',1);

1 row created.

SQL> insert into tabel1 values('AA','A-01','S3',5);

1 row created.

SQL> commit;

Commit complete.

SQL> insert into tabel1 values('AA','A-01','S2',0);

1 row created.

SQL> commit;

Commit complete.

SQL> select * from tabel1;

KHMC       PZM        GX                 SL
---------- ---------- ---------- ----------
AA         A-01       S1                  2
AA         A-01       S1                  4
AA         A-01       S1                  8
AA         A-01       S3                  1
AA         A-01       S3                  1
AA         A-01       S3                  5
AA         A-01       S2                  0

7 rows selected.

SQL> select a.khmc,a.pzm,a.gx,nvl(num,0) from
2  (select * from (select distinct khmc,pzm from tabel1) a,(select distinct gx
from tabel1) b) a,
3  (select khmc,pzm,gx,sum(sl) num from tabel1 group by khmc,pzm,gx) b
4  where a.khmc = b.khmc(+) and a.pzm = b.pzm(+) and a.gx = b.gx(+)
5  order by 1,2,3;

KHMC       PZM        GX         NVL(NUM,0)
---------- ---------- ---------- ----------
AA         A-01       S1                 14
AA         A-01       S2                  0
AA         A-01       S3                  7
--下面是我改过了的写法
SQL> select a.khmc,a.pzm,a.gx,nvl(num,0) from
2  (select distinct khmc,pzm,gx from tabel1) a,
3  (select khmc,pzm,gx,sum(sl) num from tabel1 group by khmc,pzm,gx) b
4  where a.khmc = b.khmc and a.pzm = b.pzm and a.gx = b.gx
5  order by 1,2,3;

KHMC       PZM        GX         NVL(NUM,0)
---------- ---------- ---------- ----------
AA         A-01       S1                 14
AA         A-01       S2                  0
AA         A-01       S3                  7

这样看起来好像结果没什么不同哦，可后来我又仔细看了下上面的SQL语句，又看了下楼主的要求里有这么一句话"GX要全部显示出来 没有的现实0 ”。就加了点数据，又试了试，果然就试出来问题了

SQL> insert into tabel1 values('BB','B-01','S1',4);

1 row created.

SQL> insert into tabel1 values('BB','B-01','S1',9);

1 row created.

SQL> insert into tabel1 values('BB','B-01','S2',2);

1 row created.

SQL> insert into tabel1 values('BB','B-01','S2',4);

1 row created.

SQL> commit;

Commit complete.

SQL> select a.khmc,a.pzm,a.gx,nvl(num,0) from
2  (select * from (select distinct khmc,pzm from tabel1) a,(select distinct gx
from tabel1) b) a,
3  (select khmc,pzm,gx,sum(sl) num from tabel1 group by khmc,pzm,gx) b
4  where a.khmc = b.khmc(+) and a.pzm = b.pzm(+) and a.gx = b.gx(+)
5  order by 1,2,3;

KHMC       PZM        GX         NVL(NUM,0)
---------- ---------- ---------- ----------
AA         A-01       S1                 14
AA         A-01       S2                  0
AA         A-01       S3                  7
BB         B-01       S1                 13
BB         B-01       S2                  6
BB         B-01       S3                  0

6 rows selected.

SQL> select a.khmc,a.pzm,a.gx,nvl(num,0) from
2  (select distinct khmc,pzm,gx from tabel1) a,
3  (select khmc,pzm,gx,sum(sl) num from tabel1 group by khmc,pzm,gx) b
4  where a.khmc = b.khmc and a.pzm = b.pzm and a.gx = b.gx
5  order by 1,2,3;

KHMC       PZM        GX         NVL(NUM,0)
---------- ---------- ---------- ----------
AA         A-01       S1                 14
AA         A-01       S2                  0
AA         A-01       S3                  7
BB         B-01       S1                 13
BB         B-01       S2                  6

看到了吧，BB中没有显示出S3相应的记录。

这也就是前面标注出红色那部分我一开始没看明白语句的作用

select a.khmc,a.pzm,a.gx,nvl(num,0) from
(select * from (select distinct khmc,pzm from tabel1) a,(select distinct gx from tabel1) b) a,
(select khmc,pzm,gx,sum(sl) num from tabel1 group by khmc,pzm,gx) b
where a.khmc = b.khmc(+) and a.pzm = b.pzm(+) and a.gx = b.gx(+)
order by 1,2,3;

通过第二句在两个select语句中选取出了khmc,pzm和gx，如果一起选的话，就不会有

BB         B-01       S3                  0

这条记录了。

而where条件中的(+)作用就显而易见了，外连接，即便BB中不存在相应S3的记录，也要拼接出一条来，通过NVL把SL字段置为0就可以了。