我有以下字符串。
Page load for http://xxxx?roxy=www.yahoo.com&sendto=https://mywebsite?ent took 4001 ms (Ne: 167 ms, Se: 2509 ms, Xe: 1325 ms)<br><br><br>Topic: Yahoo!! My website is a good website | Mywebsite<br>
OR
Page load for http://xxxx?roxy=www.yahoo.com&sendto=https://mywebsite?ent took too long (12343 ms Ne: 167 ms, Se: 2509 ms, Xe: 1325 ms)<br><br><br>Topic: Yahoo!! My website is a good website | Mywebsite<br>
我想从4001 OR中提取上面的12343或ent took 4001 ms
ent took too long (12343 ms并将其分配给变量
tt = int(re.search(r"\?ent\s*took\s*(\d+)",message).group(1))
这个正则表达式确实与第一部分匹配，并且确实返回4001.我如何逻辑或表达式r"\?ent\s*\took\s*too\s*long\s*\((\d+)"
从第二部分中提取12343？
答案 0 :(得分：3)
正则表达式开头的问号不会跟随任何可以选择的问题。如果您想在那里匹配文字问号，请写下\?：
x = int(re.findall(r"\?ent\s*took\s*([^m]*)",message)[0])
答案 1 :(得分：1)
首先，您需要在模式的前导处转义?，因为?标记是正则表达式字符，并且使字符串可选，并且必须以字符串开头！因此，如果您想要数学?，您还需要使用\?作为一种更有效的方式，您可以在模式中使用re.search和\d+，并拒绝额外的索引：
>>> int(re.search(r"\?ent\s*took\s*(\d+)",s).group(1))
4001
对于第二个例子，你可以这样做：
>>> re.search(r'\((\d+)',s).group(1)
'12343'
对于两种情况下的匹配，请使用以下模式：
(\d+)[\s\w]+\(|\((\d+)
>>> s1="Page load for http://xxxx?roxy=www.yahoo.com&sendto=https://mywebsite?ent took too long (12343 ms Ne: 167 ms, Se: 2509 ms, Xe: 1325 ms)<br><br><br>Topic: Yahoo!! My website is a good website | Mywebsite<br>"
>>> s2="Page load for http://xxxx?roxy=www.yahoo.com&sendto=https://mywebsite?ent took 4001 ms (Ne: 167 ms, Se: 2509 ms, Xe: 1325 ms)<br><br><br>Topic: Yahoo!! My website is a good website | Mywebsite<br>"
>>> re.search(r'(\d+)[\s\w]+\(|\((\d+)',s1).group(2)
'12343'
>>> re.search(r'(\d+)[\s\w]+\(|\((\d+)',s2).group(1)
'4001'
答案 2 :(得分：1)
这一次匹配两种模式并提取所需的数字：
tt = int(re.search(r"\?ent took (too long \()?(?P<num>\d+)",message).group('num'))