一基础单表查询
1.1查询表结构
desc 表名
SQL> desc emp
Name Null? Type
EMPNO NOT NULL NUMBER(4)
ENAME VARCHAR2(10)
JOB VARCHAR2(9)
MGR NUMBER(4)
HIREDATE DATE
SAL NUMBER(7,2)
COMM NUMBER(7,2)
DEPTNO NUMBER(2)
1.2查找空值
使用 is null
SQL> select empno from emp where comm is null;
8 rows selected.
1.3 将空值转换成实际值，推荐使用coalesce
SQL> select empno,nvl(comm,0) from emp where comm is null;
SQL> select empno,nvl2(comm,comm,0) from emp where comm is null;
SQL> select empno,nullif(0,comm) from emp where comm is null;
SQL> select empno,coalesce(comm,0) from emp where comm is null;
NVL(expr1,expr2)
如果expr1和expr2的数据类型一致，则：
如果expr1为空(null),那么显示expr2，
如果expr1的值不为空，则显示expr1。
NVL2(expr1,expr2, expr3)
如果expr1不为NULL，返回expr2； expr1为NULL，返回expr3。
expr2和expr3类型不同的话，expr3会转换为expr2的类型，转换不了，则报错。
NULLIF(expr1,expr2)
如果expr1和expr2相等则返回空(NULL)，否则返回expr1。
coalesce(expr1, expr2, expr3….. exprn)
返回表达式中第一个非空表达式，如果都为空则返回空值。
所有表达式必须是相同类型，或者可以隐式转换为相同的类型，否则报错。
Coalese函数和NVL函数功能类似，只不过选项更多。
1.4 在SELECT语句中使用条件逻辑
SQL> select empno,
2 ename,
3 sal,
4 case
5 when sal<=2000 then '过低'
6 when sal>=4000 then '过高'
7 else 'OK'
8 end as status
9 from emp
10 where deptno=10;
1.5限制返回行数
SQL> select empno from emp where rownum<=2;
1.6从表中随机返回n条记录
SQL> select empno,ename from (select empno,ename from emp order by dbms_random.value()) where rownum<=3;
1.7 TRANSLATE替换
2.SQL> select TRANSLATE('ab 你好 abcdef','abcdef','123456') as newstring from dual;
12 你好 123456
SQL> select TRANSLATE('ab 你好 abcdef','abcdef','1234') as newstring from dual;
12 你好 1234
SQL> select TRANSLATE('ab 你好 abcdef','acdef','1234') as newstring from dual;
1b 你好 1b234
SQL> select TRANSLATE('ab 你好 abcdef','acdef','') as newstring from dual;
替换值为空，返回空
SQL> select TRANSLATE('ab 你好 abcdef','1abcdef','1') as newstring from dual;
你好
替换wei位置没有字符则删除
1.8 混合字符串按字母排序
SQL> set line 100
SQL> col TRANSLATE(EMPNO||''||ENAME,'-1234567890','-') format A40
SQL> select empno||' '||ename as data,translate(empno||' '||ename,'- 1234567890','-') from emp e order by 2 ;
7499 ALLEN ALLEN
7698 BLAKE BLAKE
7782 CLARK CLARK
7902 FORD FORD
7900 JAMES JAMES
7566 JONES JONES
7839 KING KING
7654 MARTIN MARTIN
7934 MILLER MILLER
7369 SMITH SMITH
7844 TURNER TURNER
7521 WARD WARD
12 rows selected.
SQL> select empno||' '||ename as data from emp e order by translate(empno||' '||ename,'- 1234567890','-') ;
7499 ALLEN
7698 BLAKE
7782 CLARK
7902 FORD
7900 JAMES
7566 JONES
7839 KING
7654 MARTIN
7934 MILLER
7369 SMITH
7844 TURNER
7521 WARD
1.9 NULL排序使用NULLS FIRST/LAST
1.10按条件区不同列中值来排序
3 sal
4 from emp
5 where deptno=30
6 order by Case
7 when sal>=1000 and sal <2000 then
8 empno
9 else ename
10 end,
11 sal;
ename,
*
ERROR at line 2:
ORA-00932: inconsistent datatypes: expected NUMBER got CHAR
7 when sal>=1000 and sal <2000 then 1
8 else 2
9 end, 3;
6 rows selected.
二 多表操作
2.1 union all与空字符串
SQL> select 'a' as c1 from dual
2 union all
3 select '' as c1 from dual;
a
2.2 union与or
SQL> select empno,ename from emp where empno=7782 or ename='WARD';
SQL> select empno,ename from emp where empno=7782
2 union
3 select empno,ename from emp where ename='WARD';
SQL> alter session set"_b_tree_bitmap_plans"=false;
Session altered.
SQL> explain plan for select empno,ename from emp where empno=7782 or ename='WARD';
Explained.
SQL> select * from table(dbms_xplan.display);
Plan hash value: 3956160932
1 - filter("EMPNO"=7782 OR "ENAME"='WARD')
13 rows selected.
SQL> explain plan for select empno,ename from emp where empno=7782
Plan hash value: 1027572458
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Ti
me |
| 0 | SELECT STATEMENT | | 2 | 20 | 6 (34)| 00
:00:01 |
| 1 | SORT UNIQUE | | 2 | 20 | 6 (34)| 00
| 3 | TABLE ACCESS BY INDEX ROWID| EMP | 1 | 10 | 1 (0)| 00
|* 4 | INDEX UNIQUE SCAN | PK_EMP | 1 | | 0 (0)| 00
|* 5 | TABLE ACCESS FULL | EMP | 1 | 10 | 3 (0)| 00
4 - access("EMPNO"=7782)
5 - filter("ENAME"='WARD')
18 rows selected.
实际上ENAME也可以建索引那样更快
需要注意的
SQL> select deptno from emp where EMPNO=7698 or job='SALESMAN' ORDER BY 1;
SQL> select deptno,empno from emp where EMPNO=7698 or job='SALESMAN' ORDER BY 1;
3 select deptno from emp where job='SALESMAN';
避免这样问题出现可以用唯一列，主键列或rowid
SQL> select deptno,empno from emp where EMPNO=7698
3 select deptno,empno from emp where job='SALESMAN';
SQL> with
2 e as (select rownum as sn,deptno,empno,job from emp)
3 select deptno
4 from
5 (
6 select sn,deptno from e where EMPNO=7698
7 union
8 select sn,deptno from e where job='SALESMAN'
9 )
10 order by 1;
2.3 组合相关的行
SQL> select e.empno,e.ename,d.dname,d.loc
2 from emp e
3 inner join dept d on (e.deptno=d.deptno)
4 where e.deptno =10;
3 inner join dept d using(deptno)
4 where deptno =10;
2.4 IN,EXISTS和INNER JOIN
alter session set"_b_tree_bitmap_plans"=false
ERROR at line 1:
ORA-12571: TNS:packet writer failure
SQL> conn scott/tiger@clonepdb_plug
Connected.
SQL> explain plan for select empno,ename,job,deptno,sal
2 from emp
3 where (empno,ename,sal) in (select empno,ename,sal from emp )
4 ;
SQL> select * from table(dbms_xplan.display());
1 - filter("ENAME" IS NOT NULL AND "SAL" IS NOT NULL)
2 from emp a
3 where exists (select null
4 from emp b
5 where b.ename=a.ename
6 and b.job=a.job
7 and b.sal=a.sal);
Plan hash value: 977554918
1 - access("B"."ENAME"="A"."ENAME" AND "B"."JOB"="A"."JOB" AND
"B"."SAL"="A"."SAL")
16 rows selected.
Plan hash value: 3638257876
1 - access("B"."JOB"="A"."JOB" AND "B"."ENAME"="A"."ENAME" AND
SQL> explain plan for select a.empno,ename,job,sal,a.deptno
2 from emp a inner join emp b using(job,ename,sal)
3 ;
1 - access("A"."SAL"="B"."SAL" AND "A"."ENAME"="B"."ENAME" AND
"A"."JOB"="B"."JOB")
2.5 INNER JOIN、LEFT JOIN、RIGHT JOIN、FULL JOIN区别
INNER JOIN 返回必配数据
LEFT JOIN 左表为主，右表只返回左表匹配数据，右表没有显示的为空 等同于右(+)
RIGHT JOIN与上面相反等同于左(+)
FULL JOIN 左右表均返回索引数据，匹配的显示一行
2.6 自关联
SQL> run/
1 select a.empno as "员工编号",
2 a.ename as "员工姓名",
3 a.job as "职位",
4 b.empno as "主管编号",
5 b.ename as "主管姓名"
6 from emp a
7 left join emp b on(a.mgr=b.empno)
8* order by 1
员工编号 员工姓名 职位 主管编号 主管姓名
2.7 NOT IN、NOT EXISTS和 LEFT JOIN
SQL> select count(*) from emp where deptno =40;
SQL> select * from dept where deptno not in (select deptno from emp where deptno is null);
no rows selected
SQL> select dept.* from dept left join emp on dept.deptno=emp.deptno where emp.deptno is null;
2.8 外连接的条件不能乱放
SQL> select dept.* from dept left join emp on(dept.deptno=emp.deptno and emp.deptno is null);
SQL> explain plan for select dept.* from dept left join emp on(dept.deptno=emp.deptno and emp.deptno is null);
Plan hash value: 2251696546
| 0 | SELECT STATEMENT | | 3 | 69 | 6 (17)| 00
| 1 | MERGE JOIN OUTER | | 3 | 69 | 6 (17)| 00
| 2 | TABLE ACCESS BY INDEX ROWID| DEPT | 3 | 60 | 2 (0)| 00
| 3 | INDEX FULL SCAN | PK_DEPT | 3 | | 1 (0)| 00
|* 4 | SORT JOIN | | 1 | 3 | 4 (25)| 00
|* 5 | TABLE ACCESS FULL | EMP | 1 | 3 | 3 (0)| 00
4 - access("DEPT"."DEPTNO"="EMP"."DEPTNO"(+))
filter("DEPT"."DEPTNO"="EMP"."DEPTNO"(+))
5 - filter("EMP"."DEPTNO"(+) IS NULL)
19 rows selected.
Plan hash value: 1353548327
| 0 | SELECT STATEMENT | | 1 | 23 | 6 (17)| 00
| 1 | MERGE JOIN ANTI | | 1 | 23 | 6 (17)| 00
|* 4 | SORT UNIQUE | | 12 | 36 | 4 (25)| 00
| 5 | TABLE ACCESS FULL | EMP | 12 | 36 | 3 (0)| 00
4 - access("DEPT"."DEPTNO"="EMP"."DEPTNO")
filter("DEPT"."DEPTNO"="EMP"."DEPTNO")
2.9 检查两个表中数据及对应数据条数是否相等
SQL> run
1 select a.empno,a.ename,b.empno,b.ename
3 full join emp b on(b.empno=a.empno)
4* where b.empno is null or b.empno is null
SQL> 4
4 where b.empno is null or b.empno is null
SQL> del
3 full join emp b on(b.empno=a.empno)
2.10多表查询的空值处理
比ALLEN提成低的
SQL> select a.ename,a.comm
3 where coalesce(a.comm,0)<(select b.comm from emp b where b.ename='ALLEN');
ENAME COMM
SMITH
JONES
BLAKE
CLARK
KING
TURNER 0
JAMES
FORD
MILLER
9 rows selected.
第三插入、更新与删除
SQL> create table test(
2 c1 varchar2(10) default '默认1',
3 c2 varchar2(10) default '默认2',
4 c3 varchar2(10) default '默认3',
5 c4 date default sysdate
6 );
Table created.
SQL> insert into test(c1,c2,c3) values(default,null,'test');
1 row created.
SQL> select * from test
2 ;
C1 C2 C3 C4
默认1 test 2017-12-26 09:46:20
3.1阻止对某几列插入
SQL> create or replace view v_test as select c1,c2,c3 from test;
View created.
SQL> insert into V_TEST values ('手输1',null,'不改4');
SQL> select * from test;
手输1 不改4 2017-12-26 09:57:36
SQL> insert into V_TEST values (default,null,'不改4');
insert into V_TEST values (default,null,'不改4')
ORA-32575: Explicit column default is not supported for modifying views
3.2复制表定义与结构
SQL> create table test1 as select * from test where 1=2;
SQL> select * from test1;
SQL> create table test2 as select * from test;
SQL> select * from test2;
3.3利用with check option限制数据输入
SQL> alter table test modify c3 not null;
Table altered.
SQL> create or replace view v_test1 as select c1,c2,c3 from test with check option;
SQL> insert into V_TEST1 values ('test',null,null);
insert into V_TEST1 values ('test',null,null)
ORA-01400: cannot insert NULL into ("SCOTT"."TEST"."C3")
3.4多表插入语句
无条件insert
SQL> insert all
2 into test1(c1,c2,c3) values ('1','2','3')
3 into test2(c1,c2,c3) values ('1','2','3')
4 into test(c1,c2,c3) values ('1','2','3')
5 select from test1 ;
插入次数取决于select 行数
要一行的话建议用select from dual;
有条件insert
1 insert all
2 when job in ('CLERK','SALESMAN') then
3 into test (c1,c2,c3) values (ENAME,JOB,mgr)
4 when job='MANAGER' then
5 into test1 (c1,c2,c3) values (ENAME,JOB,mgr)
6 else
7 into test2 (c1,c2,c3) values (ENAME,JOB,mgr)
8 select from emp
12 rows created.
1 2 3 2017-12-26 10:29:31
SMITH CLERK 7902 2017-12-26 10:39:54
ALLEN SALESMAN 7698 2017-12-26 10:39:54
WARD SALESMAN 7698 2017-12-26 10:39:54
MARTIN SALESMAN 7698 2017-12-26 10:39:54
TURNER SALESMAN 7698 2017-12-26 10:39:54
JAMES CLERK 7698 2017-12-26 10:39:54
MILLER CLERK 7782 2017-12-26 10:39:54
10 rows selected.
11 12 13
JONES MANAGER 7839
BLAKE MANAGER 7839
CLARK MANAGER 7839
21 22 23
KING PRESIDENT
FORD ANALYST 7566
SQL> insert first
4 when empno in (7900,7934,7566) then
8 select job,ename,mgr,empno from emp;
SMITH CLERK 7902 2017-12-26 10:53:18
ALLEN SALESMAN 7698 2017-12-26 10:53:18
WARD SALESMAN 7698 2017-12-26 10:53:18
MARTIN SALESMAN 7698 2017-12-26 10:53:18
TURNER SALESMAN 7698 2017-12-26 10:53:18
JAMES CLERK 7698 2017-12-26 10:53:18
MILLER CLERK 7782 2017-12-26 10:53:18
7 rows selected.
3.5Merge into用法总结
MERGE INTO table_name alias1
USING (table|view|sub_query) alias2
ON (join condition)
WHEN MATCHED THEN
UPDATE table_name
SET col1 = col_val1,
col2 = col_val2
WHEN NOT MATCHED THEN
INSERT (column_list) VALUES (column_values);
严格意义上讲，”在一个同时存在Insert和Update语法的Merge语句中，总共Insert/Update的记录数，就是Using语句中alias2的记录数”。
3.6删除重复记录
SQL> insert into test values (1,2,3,default)
SQL> insert into test values (1,2,3,default);
1 2 3 2017-12-26 11:08:14
1 2 3 2017-12-26 11:08:18
SQL> select rowid as rid,
2 c1,
3 row_number() over(partition by c1 order by c4) as seq
4 from test
5 order by 2,3;
RID C1 SEQ
AAASXpAALAAAACuAAA 1 1
AAASXpAALAAAACuAAB 1 2
SQL> delete
2 from test
3 where rowid in (select rid
4 from (select rowid as rid,
5 row_number() over(partition by c1 order by c4) as seq
6 from test)
7 where seq>1);
1 row deleted.
2 from test a
3 where exists(select /+hash_sj/ null from test b where b.c1=a.c1 and b.rowid>a.rowid);
保留最新的<保留老的
1 2 3 2017-12-26 13:32:18
第四字符串
4.1 遍历字符串
SQL> select level from dual connect by level<=4;
SQL> select "拼音",level,substr("拼音",level,1) from (select 'TTXS' as "拼音" FROM DUAL) connect by level <=4;
拼音 LEVEL SUB
TTXS 1 T
TTXS 2 T
TTXS 3 X
TTXS 4 S
4.2 字符串'
SQL> select 'g''day mate' qmarks from dual;
g'day mate
下面是10g
SQL> select q'[g'day mate]' qmarks from dual;
SQL> select q'{g'day mate}' qmarks from dual;
SQL> select q'<g'day mate>' qmarks from dual;
SQL> select q'(g'day mate)' qmarks from dual;
4.3 统计字符串出现次数
11g
SQL> select regexp_count('wo shi wo','o') from dual;
SQL> select length(translate('wo shi wo','wo shi wo','o')) from dual;
4.4 从字符里面删除不需要的
SQL> select ename,translate(ename,'1AEIOU','1') from emp;
ENAME TRANSLATE(ENAME,'1AEIOU','1')
SMITH SMTH
ALLEN LLN
WARD WRD
JONES JNS
MARTIN MRTN
BLAKE BLK
CLARK CLRK
KING KNG
TURNER TRNR
JAMES JMS
FORD FRD
MILLER MLLR
SQL> select ename,regexp_replace(ename,'[AEIOU]') from emp;
ENAME REGEXP_REPLACE(ENAME,'[AEIOU]'
4.5 将字母与数字分开
SQL> select dname||deptno,translate(dname||deptno,'a0123456789','a') as data,translate(DNAME||DEPTNO,'0123456789'||dname||deptno,'0123456789') as data1 from dept;
ACCOUNTING10 ACCOUNTING 10
RESEARCH20 RESEARCH 20
SALES30 SALES 30
SQL> select dname||deptno,regexp_replace(dname||deptno,'[0-9]','') as data,regexp_replace(DNAME||DEPTNO,'[^0-9]','') as data1 from dept;
4.6 ^,$.,+意义
表示自少匹配6零次
4.7 姓名字母首字大写
SQL> select regexp_replace('Michael Hartstein','([[:upper:]])(.)([[:upper:]])(.)','\1.\3') from dual;
M.H
4.8 按字符串中数字排序
SQL> select dname||deptno||loc from dept order by translate(dname||deptno||loc,'0123456789'||dname||deptno||loc,'0123456789') desc;
SALES30CHICAGO
RESEARCH20DALLAS
ACCOUNTING10NEW YORK
SQL> select dname||deptno||loc from dept order by regexp_replace(dname||deptno||loc,'[^0-9]') desc;
4.9 创建分割列表
SQL> select deptno,sal,ename from emp;
SQL> col TOTAL_SAL format 999999
SQL> col TOTAL_name format A100
SQL> select deptno,
2 sum(sal) as total_sal,
3 listagg(ename,',') within group(order by ename) as total_name
5 group by deptno;
4.10 提取第n个分割子串
1 with
2 a as
3 (
4 select listagg(ename,',') within group(order by ename) as name from emp where deptno in(10,20) group by deptno
5 )
6* select regexp_substr(a.name,'[^,]+',1,2) as "子串" from a
4.11 分解ip地址
1 select regexp_substr(v.ip,'[^.]+',1,1 ) a
2 ,regexp_substr(v.ip,'[^.]+',1,2 ) b
3 ,regexp_substr(v.ip,'[^.]+',1,3) c
4 ,regexp_substr(v.ip,'[^.]+',1,4 ) d
5* from (select '192.168.0.1' as ip from dual) v
A B C D
192 168 0 1
4.12 将分个数据转换成多值IN
SQL> var v_emps varchar2(30);
SQL> exec :v_emps :='CLARK,KING,MILLER';
PL/SQL procedure successfully completed.
SQL> SET LINE 1000
1 SELECT FROM EMP WHERE ENAME IN
2 (
3 SELECT REGEXP_SUBSTR(:v_emps,'[^,]+',1,level) as ename from dual
4 connect by level <=(length(translate(:v_emps,','||:v_emps,','))+1)
5 )
第五 使用数字
5.1 累计和
4 sum(sal) over (order by empno)
5 from emp
6 where deptno=30
7 order by empno;
5.2 返回各部门排行前三的员工
1 select deptno,
2 empno,
4 row_number() over (partition by deptno order by sal desc) as row_num,
5 rank() over (partition by deptno order by sal desc) as rank,
6 dense_rank() over (partition by deptno order by sal desc) as dense_rank
7 from emp
8 where deptno in (20,30)
9* order by 1,3 desc
5.3 返回最大值所在行数据
3 max(ename) keep(dense_rank first order by sal) over (partition by deptno),
4 max(ename) keep(dense_rank last order by sal) over (partition by deptno),
5 ename,
6 sal
8 where deptno=10
9* order by 1,6 desc
3 first_value(ename) over (partition by deptno),
4 ename,
5 sal
6 from emp
7 where deptno=10
8 order by 1,5 desc;
5.4 求和百分比
3 ename,
4 sal,
5 round(ratio_to_report(sal) over(partition by deptno)100,2)
7 order by 1,2
第六 日期
6.1 年月日加减
SQL> select hiredate,
2 hiredate -5,
3 hiredate +5,
4 add_months(hiredate,-5),
5 add_months(hiredate,5),
6 add_months(hiredate,-512),
7 add_months(hiredate,512)
8 from emp
9 where rownum<=1;
HIREDATE HIREDATE-5 HIREDATE+5 ADD_MONTHS(HIREDATE ADD_MONTHS(HIREDATE ADD_MONTHS(HIREDATE ADD_MONTHS(HIREDATE
1980-12-17 00:00:00 1980-12-12 00:00:00 1980-12-22 00:00:00 1980-07-17 00:00:00 1981-05-17 00:00:00 1975-12-17 00:00:00 1985-12-17 00:00:00
6.2 时分秒加减
1 select hiredate,
2 hiredate -5/24/60/60,
3 hiredate +5/24/60/60,
4 hiredate -5/24/60,
5 hiredate +5/24/60,
6 hiredate -5/24,
7 hiredate +5/24
9* where rownum<=1
HIREDATE HIREDATE-5/24/60/60 HIREDATE+5/24/60/60 HIREDATE-5/24/60 HIREDATE+5/24/60 HIREDATE-5/24
HIREDATE+5/24
1980-12-17 00:00:00 1980-12-16 23:59:55 1980-12-17 00:00:05 1980-12-16 23:55:00 1980-12-17 00:05:00 1980-12-16 19:00:00 1980-12-17 05:00:00
6.3 时间间隔
SQL> select max(hiredate)-min(hiredate),
2 (max(hiredate)-min(hiredate))24,
3 (max(hiredate)-min(hiredate))2460,
4 (max(hiredate)-min(hiredate))246060
6 where ename in('WARD','ALLEN')
7 ;
MAX(HIREDATE)-MIN(HIREDATE) (MAX(HIREDATE)-MIN(HIREDATE))24 (MAX(HIREDATE)-MIN(HIREDATE))2460 (MAX(HIREDATE)-MIN(HIREDATE))246060
172800
6.4 日期间隔
1 select max(hiredate)-min(hiredate),
2 months_between(max(hiredate),min(hiredate)),
3 months_between(max(hiredate),min(hiredate))/12
4* from emp
MAX(HIREDATE)-MIN(HIREDATE) MONTHS_BETWEEN(MAX(HIREDATE),MIN(HIREDATE)) MONTHS_BETWEEN(MAX(HIREDATE),MIN(HIREDATE))/12
6.5 当前记录和下一条记录差
3 hiredate,
4 lead(hiredate) over(order by hiredate)
6* where deptno=10
4 lag(hiredate) over(order by hiredate)
+04-02
SQL> select interval '99' day as day from dual;
+99 00:00:00
SQL> select interval '80' hour as hour from dual;
+03 08:00:00
SQL> select interval '5' year as year from dual;
+05-00
6.8 extract
SQL> select extract(year from systimestamp) as year from dual;
SQL> select extract(month from systimestamp) as month from dual;
SQL> select extract(day from systimestamp) as day from dual;
SQL> select extract(hour from systimestamp) as hour from dual;
第七 报表和数据仓库
7.1 行转列
SQL> select job,
2 case deptno when 10 then sal end as deptno10,
3 case deptno when 20 then sal end as deptno20,
4 case deptno when 30 then sal end as deptno30,
7 order by 1;
JOB DEPTNO10 DEPTNO20 DEPTNO30 SAL
ANALYST 3000 3000
CLERK 1300 1300
CLERK 950 950
CLERK 800 800
MANAGER 2975 2975
MANAGER 2850 2850
MANAGER 2450 2450
PRESIDENT 5000 5000
SALESMAN 1500 1500
SALESMAN 1250 1250
SALESMAN 1600 1600
2 sum(case deptno when 10 then sal end) as deptno10,
3 sum(case deptno when 20 then sal end) as deptno20,
4 sum(case deptno when 30 then sal end) as deptno30,
5 sum(sal) as sal
7 group by job
8 order by 1;
CLERK 1300 800 950 3050
MANAGER 2450 2975 2850 8275
SALESMAN 5600 5600
SQL> select *
2 from (select job,
4 ,deptno
5 from emp)
6 pivot(sum(sal) as s
7 for deptno in (10 as d10,
8 20 ,
9 30 as d30)
10 )
11 order by 1;
JOB D10_S 20_S D30_S
ANALYST 3000
CLERK 1300 800 950
MANAGER 2450 2975 2850
PRESIDENT 5000
SALESMAN 5600
7.2 控制结果集重复值
SQL> select job ,ename from emp where deptno=30 order by emp.job,ename;
JOB ENAME
CLERK JAMES
MANAGER BLAKE
SALESMAN ALLEN
SALESMAN MARTIN
SALESMAN TURNER
SALESMAN WARD
SQL> select case
2 when lag(job) over(order by job,ename)=job then
3 null
4 else
5 job
6 end as job,
7 ename
9 where deptno=30
10 order by emp.job,ename;
MARTIN
TURNER
WARD
7.3 简单小计
SQL> select deptno,sum(sal) as s_sal from emp group by rollup(deptno)
7.4 分组函数详解
SQL> select DEPTNO,sum(sal) from emp group by deptno;
grouping值为0时说明这个值是数据库中本来的值，为1说明是统计的结果
SQL> select DEPTNO,sum(sal) from emp group by cube(deptno) order by 1;
仔细观察一下，CUBE与ROLLUP之间的细微差别
rollup(a,b) 统计列包含：(a,b)、(a)、()
rollup(a,b,c) 统计列包含：(a,b,c)、(a,b)、(a)、()
……以此类推ing……
cube(a,b) 统计列包含：(a,b)、(a)、(b)、()
cube(a,b,c) 统计列包含：(a,b,c)、(a,b)、(a,c)、(b,c)、(a)、(b)、(c)、()
CUBE在ROLLUP的基础上进一步从各种维度上给出细化的统计汇总结果。
SQL> select DEPTNO,job,sum(sal) from emp group by grouping sets(deptno,job) order by 1;
Grouping_id()的返回值事实上就是參数中的每列的grouping()值的二进制向量。假设grouping(a)=1，grouping(b)=1，则grouping_id(A,B)的返回值就是二进制的11。转成10进制就是3。
參数能够是多个，但必须为group by中出现的列。
7.5 不同组进行统计
SQL> select ename,deptno,count() over(partition by deptno),job,count()over(partition by job),count(*) over()
2 from emp;
ENAME DEPTNO COUNT()OVER(PARTITIONBYDEPTNO) JOB COUNT()OVER(PARTITIONBYJOB) COUNT(*)OVER()
KING 10 3 PRESIDENT 1 12
CLARK 10 3 MANAGER 3 12
MILLER 10 3 CLERK 3 12
JONES 20 3 MANAGER 3 12
SMITH 20 3 CLERK 3 12
FORD 20 3 ANALYST 1 12
ALLEN 30 6 SALESMAN 4 12
WARD 30 6 SALESMAN 4 12
TURNER 30 6 SALESMAN 4 12
MARTIN 30 6 SALESMAN 4 12
JAMES 30 6 CLERK 3 12
BLAKE 30 6 MANAGER 3 12
7.6 移动范围内值计算
2 sal,
3 sum(sal) over(order by hiredate range between interval '3' month preceding and current row)
6 order by 1;
HIREDATE SAL SUM(SAL)OVER(ORDERBYHIREDATERANGEBETWEENINTERVAL'3'MONTHPRECEDINGANDCURRENTROW)
1981-02-20 00:00:00 1600 1600
1981-02-22 00:00:00 1250 2850
1981-05-01 00:00:00 2850 5700
1981-09-08 00:00:00 1500 1500
1981-09-28 00:00:00 1250 2750
1981-12-03 00:00:00 950 3700
第八 分层查询
8.1 简单树形结构
1 select empno,
3 mgr,
4 prior ename
6 start with empno=7566
7 connect by(prior empno)=mgr
8*
8.2 根节点，分支节点，叶子节点
1 select lpad('-',(level-1)2,'-')||empno as empno,
4 level,
5 decode(level,1,1) as root,
6 decode(connect_by_isleaf,1,1) as leaf,
7 case
8 when(connect_by_isleaf=0 and level>1) then
9 1
10 end as fenzi
11 from emp
12 start with empno=7566
13 connect by (prior empno)=mgr
EMPNO ENAME MGR LEVEL ROOT LEAF FENZI
7566 JONES 7839 1 1
--7902 FORD 7566 2 1
----7369 SMITH 7902 3 1
8.3 sys_connect_by_path ==listagg
4 sys_connect_by_path(ename,',') as enames
7* connect by (prior empno)=mgr
EMPNO ENAME MGR ENAMES
7566 JONES 7839 ,JONES
7902 FORD 7566 ,JONES,FORD
7369 SMITH 7902 ,JONES,FORD,SMITH
8.4 树形查询排序
SQL> select lpad('-',(level-1)*2,'-')||empno as empno,
3 mgr
5 start with empno=7839
6 connect by (prior empno)=mgr
7 order siblings by emp.empno desc;
EMPNO ENAME MGR
--7782 CLARK 7839
----7934 MILLER 7782
--7698 BLAKE 7839
----7900 JAMES 7698
----7844 TURNER 7698
----7654 MARTIN 7698
----7521 WARD 7698
----7499 ALLEN 7698
--7566 JONES 7839
----7902 FORD 7566
------7369 SMITH 7902
8.5 树型查询使用where
2 mgr,
4 deptno
5 from(select * from emp where deptno=20) emp
6 start with mgr is null
7 connect by(prior empno)=mgr;
8.6 查询树型的一个分支
4 level
6 start with empno=7698
EMPNO MGR ENAME LEVEL
7698 7839 BLAKE 1
7499 7698 ALLEN 2
7521 7698 WARD 2
7654 7698 MARTIN 2
7844 7698 TURNER 2
7900 7698 JAMES 2
8.7 减去一个分支
6 start with mgr is NULL
7839 KING 1
7566 7839 JONES 2
7902 7566 FORD 3
7369 7902 SMITH 4
7698 7839 BLAKE 2
7499 7698 ALLEN 3
7521 7698 WARD 3
7654 7698 MARTIN 3
7844 7698 TURNER 3
7900 7698 JAMES 3
7782 7839 CLARK 2
7934 7782 MILLER 3
SQL> RUN
7 connect by (prior empno)=mgr
8* and empno !=7698
第九 调优案例分享
9.1 不建议使用标量子查询，使用left join优化标量子查询
4 deptno,
5 (select dname from dept where dept.deptno=emp.deptno)
6 from emp;
EMPNO ENAME SAL DEPTNO (SELECTDNAMEFR
7369 SMITH 800 20 RESEARCH
7499 ALLEN 1600 30 SALES
7521 WARD 1250 30 SALES
7566 JONES 2975 20 RESEARCH
7654 MARTIN 1250 30 SALES
7698 BLAKE 2850 30 SALES
7782 CLARK 2450 10 ACCOUNTING
7839 KING 5000 10 ACCOUNTING
7844 TURNER 1500 30 SALES
7900 JAMES 950 30 SALES
7902 FORD 3000 20 RESEARCH
7934 MILLER 1300 10 ACCOUNTING
SQL> select e.empno,
2 e.ename,
3 e.sal,
4 e.deptno,
5 d.dname
6 from emp e
7 left join dept d on(e.deptno=d.deptno);
EMPNO ENAME SAL DEPTNO DNAME
1 select /+use_nl(e,d)/
7* left join dept d on(e.deptno=d.deptno)
ENAME SAL DEPTNO DNAME
SMITH 800 20 RESEARCH
ALLEN 1600 30 SALES
WARD 1250 30 SALES
JONES 2975 20 RESEARCH
MARTIN 1250 30 SALES
BLAKE 2850 30 SALES
CLARK 2450 10 ACCOUNTING
KING 5000 10 ACCOUNTING
TURNER 1500 30 SALES
JAMES 950 30 SALES
FORD 3000 20 RESEARCH
MILLER 1300 10 ACCOUNTING
9.2 使用left jion 优化标量子查聚合
SQL> select d.department_id,
2 d.department_name,
3 d.location_id,
4 nvl((select sum(e.salary)
5 from employees e
6 where e.department_id=d.department_id),
7 0) as sum_sal
8 from departments d;
DEPARTMENT_ID DEPARTMENT_NAME LOCATION_ID SUM_SAL
27 rows selected.
4 COALESCE(e.sum_sal,0) as sum_sal
5 from departments d
6 left join (select e.department_id,sum(e.salary) as sum_sal
7 from employees e
8 group by e.department_id) e on ( e.department_id=
9 d.department_id);
本文转自whshurk 51CTO博客，原文链接：http://blog.51cto.com/shurk/2054155，如需转载请自行联系原作者