oracle实现如下效果
表 a
id name
1 a
2 b
3 c
4 d
需要结果集为
id name
1 b
2 c
3 d
4 a
有好的写法不
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这个结果是怎么出来的，大姐，解释一下行不!!
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测试数据：
- SQL code
CREATE TABLE T190( ID NUMBER(4), F1 VARCHAR2(20));INSERT INTO T190 VALUES(1, 'a');INSERT INTO T190 VALUES(2, 'b');INSERT INTO T190 VALUES(3, 'c');INSERT INTO T190 VALUES(4, 'd');
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估计楼主是想要分析函数的效果
lag() over 和lead() over()
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- SQL code
with t as(select 1 id, 'a' name from dualunion allselect 2,'b' from dualunion allselect 3,'c' from dualunion allselect 4,'d' from dual)select id,name from t order by mod(id,4) ID NAME---------- ---- 4 d 1 a 2 b 3 c
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- SQL code
with t as(select 1 id, 'a' name from dualunion allselect 2,'b' from dualunion allselect 3,'c' from dualunion allselect 4,'d' from dual)select id,name from t order by mod(id+2,4) ID NAME---------- ---- 2 b 3 c 4 d 1 a
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- SQL code
with t as(select 1 id, 'a' name from dualunion allselect 2,'b' from dualunion allselect 3,'c' from dualunion allselect 4,'d' from dual)select mod(id+2,4)+1 id,name from t order by id ID NAME---------- ---- 1 b 2 c 3 d 4 a