关于正则表达的问题
问题1：是哪里打错了呢，既然报错了
import re
re.search(r'(({0,1}\d{0,1}\d|2\d|25\.){3}({0,1}\d{0,1}\d|2\d|25)', 'other192.168.1.1other')
Traceback (most recent call last):
File "<pyshell#4>", line 1, in <module>
re.search(r'(({0,1}\d{0,1}\d|2\d|25\.){3}({0,1}\d{0,1}\d|2\d|25)', 'other192.168.1.1other')
File "E:\pyhthon\lib\re.py", line 201, in search
return _compile(pattern, flags).search(string)
File "E:\pyhthon\lib\re.py", line 304, in _compile
p = sre_compile.compile(pattern, flags)
File "E:\pyhthon\lib\sre_compile.py", line 764, in compile
p = sre_parse.parse(p, flags)
File "E:\pyhthon\lib\sre_parse.py", line 948, in parse
p = _parse_sub(source, state, flags & SRE_FLAG_VERBOSE, 0)
File "E:\pyhthon\lib\sre_parse.py", line 443, in _parse_sub
itemsappend(_parse(source, state, verbose, nested + 1,
File "E:\pyhthon\lib\sre_parse.py", line 836, in _parse
raise source.error("missing ), unterminated subpattern",
re.error: missing ), unterminated subpattern at position 0
问题2：红色是一个小组，那绿色的哪个小括号是干嘛的
re.search(r'(({0,1}\d{0,1}\d|2\d|25\.){3}({0,1}\d{0,1}\d|2\d|25)', 'other192.168.1.1other') # 这样可以匹配 192.168.1.1
import re
patterm = '{1,3}\.{1,3}\.{1,3}\.{1,3}'
string = 'other192.168.1.1other'
print(re.search(patterm, string)) 缺括号吧 r"(??+.){3}??+" wp231957 发表于 2021-11-27 11:20
缺括号吧
没有呀，和小甲鱼的一模一样 提示很清楚了，差一个右括号。
re.search(r'(({0,1}\d{0,1}\d|2\d|25)\.){3}({0,1}\d{0,1}\d|2\d|25)', 'other192.168.1.1other') txxcat 发表于 2021-11-27 16:49
提示很清楚了，差一个右括号。
re.search(r'(({0,1}\d{0,1}\d|2\d|25)\.){3}({0,1}\d{0 ...
re.search(r'({0,1}\d{0,1}\d|2\d|25\.){3}({0,1}\d{0,1}\d|2\d|25)', 'other192.168.1.1other')
这样子不是也可以代表红色小组重复三次嘛?
为什么要多一组括号呢，谢谢回答！ W1ND123 发表于 2021-11-28 10:03
re.search(r'({0,1}\d{0,1}\d|2\d|25\.){3}({0,1}\d{0,1}\d|2\d|25)', 'oth ...
abc{3}# 只能匹配3个重复的c，即abccc
(abc){3}# abc成为一组，可以匹配3组abc，即abcabcabc
基本功先练好，事半功倍。https://www.runoob.com/python3/python3-reg-expressions.html
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