Select (floor(([序号]-1)/10) + 1) as '组',[学校],[姓名],Row_Number() OVER(PARTITION by floor(([序号]-1)/10) Order by [序号],[学校]) as '序号' From( SELECT [学校],[姓名],Row_Number() OVER(PARTITION BY [学校] Order by newid()) AS '序号' FROM test ) as tb2
列顺序保持一致就可以了, insert into table1 Select (floor(([序号]-1)/10) + 1) as '组',[学校],[姓名],Row_Number() OVER(PARTITION by floor(([序号]-1)/10) Order by [序号],[学校]) as '序号' From( SELECT [学校],[姓名],Row_Number() OVER(PARTITION BY [学校] Order by newid()) AS '序号' FROM test ) as tb2不一致的话就 insert into table1(col,col2) Select (floor(([序号]-1)/10) + 1) as '组',[学校],[姓名],Row_Number() OVER(PARTITION by floor(([序号]-1)/10) Order by [序号],[学校]) as '序号' From( SELECT [学校],[姓名],Row_Number() OVER(PARTITION BY [学校] Order by newid()) AS '序号' FROM test ) as tb2
From(
SELECT [学校],[姓名],Row_Number() OVER(PARTITION BY [学校] Order by newid()) AS '序号'
FROM test
) as tb2
列顺序保持一致就可以了,
insert into table1
Select (floor(([序号]-1)/10) + 1) as '组',[学校],[姓名],Row_Number() OVER(PARTITION by floor(([序号]-1)/10) Order by [序号],[学校]) as '序号'
From(
SELECT [学校],[姓名],Row_Number() OVER(PARTITION BY [学校] Order by newid()) AS '序号'
FROM test
) as tb2不一致的话就
insert into table1(col,col2)
Select (floor(([序号]-1)/10) + 1) as '组',[学校],[姓名],Row_Number() OVER(PARTITION by floor(([序号]-1)/10) Order by [序号],[学校]) as '序号'
From(
SELECT [学校],[姓名],Row_Number() OVER(PARTITION BY [学校] Order by newid()) AS '序号'
FROM test
) as tb2
1 001 四分卫 1
1 002 数 2
1 003 3
1 004 色弱 4
1 005 收到公司打工 5
1 006 TTT 6
1 008 rrr 7
1 001 是否 8
1 002 ③ 9
1 003 2342 10
1 004 水电费 11
1 005 i比你们 12
1 006 QQQQ 13
1 008 rrrr 14
1 001 阿 15
1 002 是 16
1 003 456 17
1 004 水电费 18
1 005 啊沙发沙发 19
1 008 rrrrrr 20
1 001 啊 21
1 003 是否 22
1 004 fff 23
1 005 受到广泛受到该 24
1 001 22 25
1 003 发 26
1 004 水电费 27
1 005 是大概是个 28
1 001 二我 29
1 003 30
1 004 hhhh 31
1 005 收到公司打工是 32
1 001 是否 33
1 003 飞 34
1 004 uu 35
1 005 同意同意 36
1 001 女不女 37
1 003 38
1 005 是公司是否 39
1 001 沃尔沃 40
1 003 234 41
1 001 儿童 42
1 003 水电费 43
2 001 问 1
2 003 玩儿 2
2 001 问问 3
2 003 收发室 4
2 003 656 5