求助怎么才能查詢出與01的b2值完全相同的b1呢？
uj5u.com熱心網友回復：
啥子意思 你說下想得到的結果uj5u.com熱心網友回復：
查詢b1 條件是 b2完全等于01的b2
uj5u.com熱心網友回復：
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
*/
uj5u.com熱心網友回復：
呃，抱歉，沒怎么看懂，能解釋一蛤嗎？
uj5u.com熱心網友回復：
有沒大佬幫助一蛤的uj5u.com熱心網友回復：
如果你的B1是varchar型的數字
B2是number型的數字
那就直接to_number B1然后和B2去做比較
uj5u.com熱心網友回復：
select b1 from dual where dual in (select b2 from b where b1=01) 這樣寫呢uj5u.com熱心網友回復：
select b1 from dual where b2 in (select b2 from dual where b1=01)uj5u.com熱心網友回復：
已知：a = [(4,2,3), (5, 9, 1), (7,8,9)]希望將二維串列轉換成一維串列：["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']
笨辦法，提供一種思路
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']
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