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This topic created in 3193 days ago, the information mentioned may be changed or developed.
import re
s = 'a12a34a56a78 ‘
reg1 = r'a(\d+)a(\d+)a(\d+)a(\d+)'
reg2 = r'(?:a(\d+)){4}' #我的错误示范
假如 reg1 重复 n 次，如何将 reg1 改成 reg2 的形式还能正常的捕获？我这样写 reg2 只能捕获 78 求大神不吝赐教，刚入正则坑
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1
shoaly Aug 7, 2017
真要是这种有规律的东西 , 用循环加字符串切割吧
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sxm Aug 7, 2017 via Android
(a(\d+))+
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L2AKnG8GXx60bc6P Aug 7, 2017 via iPhone
you need regex
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4
L2AKnG8GXx60bc6P Aug 7, 2017 via iPhone
i mean
import regex as re |
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5
momocraft Aug 7, 2017
ruby 中這個操作叫 scan. 你可以看看 py 有沒有類似的
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8
inflationaaron Aug 8, 2017
没办法，与其用 regex 不如` split('a')`
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code42 Aug 20, 2017
re.findall(r'a(\d+)', 'a12a34a56a78')
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