qiu'zhu

[idontsay33]

求助怎么才能查询出与01的b2值完全相同的b1呢？

[学海无涯-回头是岸]

啥子意思 你说下想得到的结果

[idontsay33]

引用 1 楼 baidu_36457652 的回复:

啥子意思 你说下想得到的结果

查询b1 条件是 b2完全等于01的b2

[学海无涯-回头是岸]

with t1(b1,b2)
 as (select  '01',1 from dual union 
 select '01',2 from dual  union
 select '02', 1 from dual union 
 select '02' ,2 from dual )
 select * from t1 where regexp_substr(b1,'[^0]+',1,1)=b2
/*    	B1	B2
1	01	1
2	02	2
*/

[idontsay33]

引用 3 楼 baidu_36457652 的回复:

with t1(b1,b2)
 as (select  '01',1 from dual union 
 select '01',2 from dual  union
 select '02', 1 from dual union 
 select '02' ,2 from dual )
 select * from t1 where regexp_substr(b1,'[^0]+',1,1)=b2
/*    	B1	B2
1	01	1
2	02	2
*/

呃，抱歉，没怎么看懂，能解释一蛤吗？

[idontsay33]

有没大佬帮助一蛤的

[mj845573397]

select b1 from dual where dual in (select b2 from b where b1=01) 这样写呢

[mj845573397]

select b1 from dual where b2 in (select b2 from dual where b1=01)

[碧水幽幽泉]

已知：a = [(4,2,3), (5, 9, 1), (7,8,9)]
希望将二维列表转换成一维列表：["4,2,3", "5, 9, 1", "7,8,9"]

>>> a = [(4,2,3), (5, 9, 1), (7,8,9)]
>>> from itertools import chain
>>> list(chain.from_iterable(a))
[4, 2, 3, 5, 9, 1, 7, 8, 9]
>>> from tkinter import _flatten # python2.7也可以from compiler.ast import flatten
>>> _flatten(a)
(4, 2, 3, 5, 9, 1, 7, 8, 9)

>>> [','.join(map(str,t)) for t in a]
['4,2,3', '5,9,1', '7,8,9']
>>> from itertools import starmap
>>> list(starmap('{},{},{}'.format,a))
['4,2,3', '5,9,1', '7,8,9']

笨办法，提供一种思路
a = [(4, 2, 3), (5,9,1), (7,8,9)]
i=0
while i<3:
a[i]=str(a[i])[1:3*3-1]
i=i+1
print (a[0:3])

>>>
['4, 2, 3', '5, 9, 1', '7, 8, 9']

[alex259]

引用 5 楼 idontsay33 的回复:

有没大佬帮助一蛤的

如果你的B1是varchar型的数字 B2是number型的数字 那就直接to_number B1然后和B2去做比较