I have the following scenario sending Auth Headers to an application that can range from the following:
"APIAuth 5b6b7ed3b9708d1168455da4:hW1ZeYYLJFGBP8tEHAEGoiGD1xM="
"APIAuth-HMAC-SHA256 5b6b7ed3b9708d1168455da4:hW1ZeYYLJFGBP8tEHAEGoiGD1xM="
etc.
What I'd like to do is to be able to capture APIAuth and APIAuth-HMAC-SHA256 from the header leaving me the client_id:signature like so:
string = '5b6b7ed3b9708d1168455da4:hW1ZeYYLJFGBP8tEHAEGoiGD1xM='
I want to be able to grab this value from any APIAut-WHATEVER-ENCRYPTION
I've been playing around with regex's but the best I have was this /\ABearer\s+/i. I thought this would have worked to grab both because the \s+ is more than one of any single character so I don't know why its not working. Could someone please assist? Regexs are not my strong suit. Thank you.
For the example strings, you could match the parts that you want:
\bAPIAuth(?:-\S+)?\s+\K[^\s:"]+:[^\s:"]+
Explanation
\bAPIAuth A word boundary, followed by APIAuth
(?:-\S+)? Optionally match - and 1+ non whitespace chars
\s+\K Match 1+ whitespace chars and forget what is matched so far using \K
[^\s:"]+:[^\s:"]+ Match : surrounded by chars other than a whitespace char or : or " if those are also part of the string
See a rubular regex demo.
You could also match only the first part, and then replace with an empty string.
\bAPIAuth(?:-\S+)?\s+
See another regex demo
Related
I have a string and would like to match a part of it.
The string is Accept: multipart/mixedPrivacy: nonePAI: <sip:4168755400#1.1.1.238>From: <sip:4168755400#1.1.1.238>;tag=5430960946837208_c1b08.2.3.1602135087396.0_1237422_3895152To: <sip:4168755400#1.1.1.238>
I want to match PAI: <sip:4168755400#
the whitespace can be a word so i would like to use .* but if i used that it matches most of the string
The example on that link is showing what i'm matching if i use the whitespace instead of .*
(PAI: <sip:)((?:\([2-9]\d{2}\)\ ?|[2-9]\d{2}(?:\-?|\ ?))[2-9]\d{2}[- ]?\d{4})#
The example on that link is showing what i'm trying to achieve with .* but it should only match PAI: <sip:4168755400#
(PAI:.*<sip:)((?:\([2-9]\d{2}\)\ ?|[2-9]\d{2}(?:\-?|\ ?))[2-9]\d{2}[- ]?\d{4})#
I tried lookaround but failing.
Any idea?
thanks
Matching the single space can be updated by using a character class matching either a space or a word character and repeat that 1 or more times to match at least a single occurrence.
Note that you don't have to escape the spaces, and in both occasions you can use an optional character class matching either a space or hyphen [ -]?
If you want the match only, you can omit the 2 capturing groups if you want to.
(PAI:[ \w]+<sip:)((?:\([2-9]\d{2}\) ?|[2-9]\d{2}[ -]?)[2-9]\d{2}[- ]?\d{4})#
Regex demo
The regex should be like
PAI:.*?(<sip:.*?#)
Explanation:
PAI:.*? find the word PAI: and after the word it can be anything (.*) but ? is used to indicate that it should match as few as possible before it found the next expression.
(<sip:.*?#) capturing group that we want the result.
<sip:.*?# find <sip: and after the word it can be anything .*? before it found #.
Example
I am looking for regex to match following set:
/VIDEO_PRE_MINE
/VIDEO_PRE
/VIDEO_PRE/
/VIDEO_PRE/SOMETHING
And I want exclude expresions like this:
/VIDEO_PRESOMETHING
/VIDEO_PREsomething/something
In other words after expression '_PRE' cannot be any literal character, but it can be end of the string.
Here are regexes that i tried:
1. ^\/[^\/]*_PRE[^a-z|A-Z]
2. ^\/[^\/]*_PRE[^a-z|A-Z]?$
However I didn't manage to cover all use cases from sets with those regex.
I would really appreciate any help with this.
Thanks
For your example data, you could add an optional group (?:[_/].*)? to match either a _ or / followed by matching any char except a newline 0+ times until the end of the string $
^/[^/]*_PRE(?:[_/].*)?$
^ Start of string
/[^/]* Match /, then 0+ times any char except /
_PRE Match literally
(?: Non capturing group
[_/].* Match either _ or / followed by 0+ times any char except a newline
)? Close non capturing group and make it optional
$ End of string
Regex demo
Note that the forward slashes are not escaped. Depending on the language or delimiters you might have to escape them.
My guess is that we might want to have some right boundaries, such as
^\/VIDEO_PRE(?:\b\/?|\/[^\/\s]+\/?|_[^\/\s]+\/?)$
in specified form, and in general form:
^\/[^_]+_PRE(?:\b\/?|\/[^\/\s]+\/?|_[^\/\s]+\/?)$
which might work. You would likely want to test and modify the expression, which is explained on the top right panel of regex101.com, if you wish to explore/simplify it, and in this link, you can watch how it would match against some sample inputs, if you like.
DEMO
I'm trying to report on a set of URLs that catches all potential URL parameters and I'm having an issue defining the RegEx properly.
We have this RegEx to capture a few variations of our URLs to feed into our reporting but I need to be able to end the string with a $ but when I do, it doesn't show any results.
The RegEx:
/join/$|/join/\?product.*|/join/\.*
For another account, we only use one variation which is outlined below (which works):
^/join/$
I believe the issue is in that after \?product.*, I'm not ending the string (or even starting it).
So far I have tried: ^/join/$|(^[/join/\?product.*]$)|(^[/join/\.*]$) with no luck.
If you want to match the dollar sign literally you have to escape it \$ or else it would mean an anchor to assert the end of the string / line.
This pattern ^/join/$ would therefore only match /join/
In your pattern you use an alternation where the last part /join/\.* would match /join/ but also /join/..... because when you escape the dot you will match it literally and the * quantifier repeats 0+ times.
Perhaps you are looking for:
^/join/(?:\?product.*\$)?$
This will match /join/ followed by an optional part (?:\?product.*\$)? that will match ?product, followed by any char 0+ times and will end on $.
Regex demo
Please, make the pattern lazy and $ is a special character for regex so need to escape that. (Regarding escaping part, google analytics may follow something else.) [] is used to capture a character in a range, be careful with that as well, as you are trying to capture a group I think.
\?product.*?\$
I need to create regex to find last underscore in string like 012344_2.0224.71_3 or 012354_5.00123.AR_3.335_8
I have wanted find last part with expression [^.]+$ and then find underscore at found element but I can not handle it.
I hope you can help me :)
Just use a negative character class [^_] that will match everything except an underscore (this helps to ensure no other underscores are found afterwards) and end of string $
Pattern would look as such:
(_)[^_]*$
The final underscore _ is in a capturing group, so you are wanting to return the submatch. You would replace the group 1 (your underscore).
See it live: Regex101
Notice the green highlighted portion on Regex101, this is your submatch and is what would be replaced.
The simplest solution I can imagine is using .*\K_, however not all regex flavours support \K.
If not, another idea would be to use _(?=[^_]*$)
You have a demo of the first and second option.
Explanation:
.*\K_: Fetches any character until an underscore. Since the * quantifier is greedy, It will match until the last underscore. Then \K discards the previous match and then we match the underscore.
_(?=[^_]*$): Fetch an underscore preceeded by non-underscore characters until the end of the line
If you want nothing but the "net" (i.e., nothing matched except the last underscore), use positive lookahead to check that no more underscores are in the string:
/_(?=[^_]*$)/gm
Demo
The pattern [^.]+$ matches not a dot 1+ times and then asserts the end of the string. The will give you the matches 71_3 and 335_8
What you want to match is an underscore when there are no more underscores following.
One way to do that is using a negative lookahead (?!.*_) if that is supported which asserts what is at the right does not match any character followed by an underscore
_(?!.*_)
Pattern demo
I'm trying to match first occurrence of window.location.replace("http://stackoverflow.com") in some HTML string.
Especially I want to capture the URL of the first window.location.replace entry in whole HTML string.
So for capturing URL I formulated this 2 rules:
it should be after this string: window.location.redirect("
it should be before this string ")
To achieve it I think I need to use lookbehind (for 1st rule) and lookahead (for 2nd rule).
I end up with this Regex:
.+(?<=window\.location\.redirect\(\"?=\"\))
It doesn't work. I'm not even sure that it legal to mix both rules like I did.
Can you please help me with translating my rules to Regex? Other ways of doing this (without lookahead(behind)) also appreciated.
The pattern you wrote is really not the one you need as it matches something very different from what you expect: text window.location.redirect("=") in text window.location.redirect("=") something. And it will only work in PCRE/Python if you remove the ? from before \" (as lookbehinds should be fixed-width in PCRE). It will work with ? in .NET regex.
If it is JS, you just cannot use a lookbehind as its regex engine does not support them.
Instead, use a capturing group around the unknown part you want to get:
/window\.location\.redirect\("([^"]*)"\)/
or
/window\.location\.redirect\("(.*?)"\)/
See the regex demo
No /g modifier will allow matching just one, first occurrence. Access the value you need inside Group 1.
The ([^"]*) captures 0+ characters other than a double quote (URLs you need should not have it). If these URLs you have contain a ", you should use the second approach as (.*?) will match any 0+ characters other than a newline up to the first ").