Regex Matching in Python 2.7 for HTTP GET Headers

Question

I am trying to form a regular expression that will match as follows:

One or more of any character except a colon or space, that then ends with a colon
Followed by a space
Followed by one or more number of any character except a colon or space
Followed directly by /r/n

As specified above, this is for HTTP GET requests so any of the following would work:

Host: www.stackoverflow.com/r/n
a-b-sads&^*@hgsdafAS&FTD: sjal;dfh9S^&D^F&(SDfsdgafs/r/n

and the following would not:

Host : www.stackoverflow.com
H:o:s:t: www.stackoverflow.com
Host: www.:::stackoverflow.com
Host: www.stackoverflow.com/n

I am currently using re.compile(r"^.{1,}: .{1,}[/r/n]$") but am not sure how to exclude colons from certain subsets of the string.

EDIT: I believe what I want to start with is ^ to signify the beginning of a string. Then, I want one or more number of any character except a colon so .{1,}, but I am not sure how I would exclude colon from this list. Then I want a colon and a space, so just :, and then any character except a colon .{1,} with the same problem as before of excluding colons. Finally, I want it to end with [\r\n]$. This still does not seem to work, even if I exclude the no colon character requirement. So something like ^.{1,}: .{1,}\r\n$, but I still need to figure out how to exclude colons.

why does the first "Host : www.stackoverflow.com" not match your requirements? — Veltzer Doron
– Veltzer Doron, Commented Jan 20, 2018 at 23:10
@VeltzerDoron Because it goes Host space : space where the first space should not be present. — ThoseKind
– ThoseKind, Commented Jan 20, 2018 at 23:12
So, you want to exclude spaces from the first string as well, or just before the colon? — Veltzer Doron
– Veltzer Doron, Commented Jan 20, 2018 at 23:14
@Veltzer Doron Ah yes, the first string up to the colon should exclude spaces and colons. Then a colon followed by a single space, and then a second string that excludes spaces and colons that ends with /r/n. — ThoseKind
– ThoseKind, Commented Jan 20, 2018 at 23:17

Veltzer Doron · Accepted Answer · 2018-01-20 23:27:31Z

1

{1,} is simply +
excluding colons is done by [^:]*
If you want to exclude spaces and colons, use [^ :]
catching end of string with $ following \r\n seems strange to me, it means a single string ending with an eoln and nothing after it (also I hope you know about the difference between unix and windows regarding this)
Also: eoln is \r\n, putting something in square brackets means either of the characters contained will match which is not what you need

In total, the following should work

^([^ :]+): ([^ :]+)$

giving Host in group 1 and the url in group 2

Test it here

edited Jan 20, 2018 at 23:27

answered Jan 20, 2018 at 23:22

Veltzer Doron

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1 Comment

ThoseKind Over a year ago

Ok, it makes sense now what exactly I was doing wrong. Thank you for your help.

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