好玩的SQL

<span style="color: #000000;"><span style="color: #ff0000;">SQL> select a||‘+‘||b||‘=‘||(a+b) from (select rownum a from all_objects where rownum<4), (select rownum b from all_objects where rownum<4);</span>
A||‘+‘||B||‘=‘||(A+B)
------------------------------------------------------------------------------------------------------------------------
1+1=2
1+2=3
1+3=4
2+1=3
2+2=4
2+3=5
3+1=4
3+2=5
3+3=6</span>
9 rows selected.

2. 做一个5*5的乘法表

<span style="color: #ff0000;">with multiplier as (select rownum n from dual connect by rownum<6)
select a.n||‘*‘||b.n||‘=‘||(a.n*b.n) from multiplier a, multiplier b</span>

3. 不用connect by，只用dual表，构造出1到128

<span style="color: #ff0000;">with a as (select 1 from dual union all select 1 from dual)
select rownum from a,a,a,a,a,a,a</span>

4. 池塘边上有牛和鹅若干，小华总共看到15个头42条腿，请问牛和鹅各有多少？

<span style="color: #ff0000;">with a as (select 1 from dual union all select 1 from dual),
b as (select rownum n from a,a,a,a)
select x.n num_of_bull, y.n num_of_goose from b x, b y where x.n*4+y.n*2=42 and x.n+y.n=15</span>

5. 百钱买鸡兔：老母鸡3块1只，小母鸡4块5只，大白兔2块1只，小白兔3块4只，要求买回来的动物总共100只，并且脚不少于240条不多于320条。花100块钱来买这些动物，要求每种动物都至少要购买一只且钱正好花完，输出所有的可能情况。

<span style="color: #ff0000;">with t as (select 1 from dual union all select 1 from dual),
t1 as (select rownum n from t,t,t,t,t)
select a.n lmj,5*b.n xmj,c.n dbt,4*d.n xbt from t1 a,t1 b,t1 c,t1 d where 3*a.n+b.n*4+c.n*2+d.n*3=100 and a.n+5*b.n+c.n+4*d.n=100 and (2*a.n+10*b.n+4*c.n+16*d.n between 240 and 320) and a.n<>0 and b.n<>0 and c.n<>0 and d.n<>0;</span>

6. 每个雇员的薪水（SAL）都对应到一个薪水级别（SALGRADE表中的GRADE字段），哪个薪水级别上的雇员数量最多？输出该薪水级别信息。本题需要用三种不同的写法作答。

第一种写法：

<span style="color: #ff0000;">select * from salgrade where grade=(select grade from (select s.grade,count(*) from emp e,salgrade s where e.sal between s.losal and s.hisal group by s.grade order by 2 desc) where rownum=1);</span>

第二种写法：

<span style="color: #ff0000;">with t as (select s.grade,count(*) num from emp e,salgrade s where e.sal between s.losal and s.hisal group by s.grade),
t1 as (select max(num) maxnum from t)
select s.* from salgrade s,t,t1 where s.grade=t.grade and t.num=t1.maxnum;</span>

第三种写法：

<span style="color: #ff0000;">select * from salgrade where exists (select 1 from (select grade from (select s.grade,count(*) from emp e,salgrade s where e.sal between s.losal and s.hisal group by s.grade order by 2 desc) where rownum=1) s where s.grade=salgrade.grade);</span>

好玩的SQL

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