如何仅匹配特定字符的正则表达式模式

Question

line =abcsdfs?a=name&ab=fsdfsd&c=sfssf
pattern = '\?a=(.+?)&'
match_pattern = re.search(pattern, line)

I get result as ?a=name& and and match_pattern.group(1) as name

But if I remove word name I don't get NONE return?

line =abcsdfs?a=&ab=fsdfsd&c=sfssf
pattern = '\?a=(.+?)&'
match_pattern = re.search(pattern, line)

I get result as ?a=&ab=fsdfsd& and match_pattern.group(1) as &ab=fsdfsd

How can I stop it to go further. I mean how to get result None/ error if do match_pattern.group(1)

Answer 1

Try this instead:

pattern = '\?a=(.*?)&'

Note that .+? matches 1 or more characters; .*? matches 0 or more characters.

Of course, if what you are trying to do is to match URL query strings, regex is the wrong tool. Try the urlparse module:

import urlparse
url = 'abcsdfs?a=name&b=&ab=fsdfsd&c=sfssf'
url = urlparse.urlsplit(url)
qs = url.query
qs = urlparse.parse_qs(qs, keep_blank_values=True)
print [qs['a'][0], qs['b'][0], qs['c'][0]]

Answer 2

您应该匹配除“&”之外的所有符号：

pattern = '\?a=([^&]*)'