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.