You could try:
r'?id=([a-zA-Z.]+)'
For your regex, like so:
def get_id(toParse)
regex = r'?id=([a-zA-Z.]+)'
x = re.findall(regex, toParse)[0]
return x
Regex -
By adding r
before the actual regex code, we specify that it is a raw string, so we don't have to add multiple backslashes before every command, which is better explained here.
?
holds special meaning for the regex system, so to match a question mark, we precede it by a backslash like ?
id=
matches the id=
part of the extraction
([a-zA-Z.]+)
is the group(0) of the regex, which matches the id of the URL. Hence, by saying [0]
, we are able to return the desired text.
Note - I have used re.findall
for this, because it returns an array []
whose element at index 0 is the extracted text.
I recommend you take a look at rexegg.com for a full list of regex syntax.
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