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 This topic created in 3201 days ago, the information mentioned may be changed or developed.
 | 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 |
 | 2 sxm Aug 7, 2017 via Android (a(\d+))+ |
 | 3 L2AKnG8GXx60bc6P Aug 7, 2017 via iPhone you need regex |
 | 4 L2AKnG8GXx60bc6P Aug 7, 2017 via iPhone i mean import regex as re |
 | 5 momocraft Aug 7, 2017 ruby 中操作叫 scan. 你可以看看 py 有有似的 |
 | 8 inflationaaron Aug 8, 2017 没办法，与其用 regex 不如` split('a')` |
 | 9 code42 Aug 20, 2017 re.findall(r'a(\d+)', 'a12a34a56a78') |
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