re.search() 逻辑上或 re.search() 中的两个模式

Question

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>

或

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ent took 4001 ms OR 中提取上面的4001 OR 12343 OR 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匹配第一部分并返回 4001。我如何在逻辑上或表达式 r"\?ent\s*\took\s*too\s*long\s*\((\d+)" 从第二部分提取12343？

Answer 1

正则表达式开头的问号不跟随任何可以设为可选的内容。如果你想在那里匹配一个文字问号，写\?：

x = int(re.findall(r"\?ent\s*took\s*([^m]*)",message)[0])

Answer 2

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'))

Answer 3

首先，您需要在模式的开头转义?，因为? 标记是一个正则表达式字符，并且使字符串可选并且必须以字符串开头！所以如果你想计算?，你需要使用\?作为一种更有效的方式，你可以在你的模式中使用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'

对于这两种情况的匹配，请使用以下模式：

(\d+)[\s\w]+\(|\((\d+)

Demo

>>> 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'