I'm trying to add a #param to url's I want to add it to all urls that doesn't already have the # to avoid double param. My urls doesn't look like urls they are made up of handlebar parameters.
they can look like following:
{{app.url}}
{{root.app.url}}
{{app.url}}#param
{{root.app.url}}#param
So I came up with a regex that matches the handlebar tag ({{(root.)?app.url}})
only problem is that when I later uses regexp_replace(url, '({{(root\.)?app\.url}})', '\1#param')
my result looks like this:
{{app.url}}#param
{{root.app.url}}#param
{{app.url}}#param#param
{{root.app.url}}#param#param
One solution I can think of is doing it in two steps, and the 2nd step should look for duplicate #param#param and replace that with single #param.
But it had me wondering if there was a way using regex to exclude the handlebar tags that are followed by # and completely cancel that match?
Here are some examples:
https://regex101.com/r/d3Zyvo/6
Note: this is for use in postgressql update queries. The regex is POSIX/PCRE. I must use regex_replace with back reference since there might be content before and after the hanbdlebar tags, I simply cannot just concatenate the param. (see the link).
You may use a negative lookahead (?!#):
({{(root\.)?app\.url}})(?!#)
^^^^^
See the regex demo.
Details
({{(root\.)?app\.url}}) - Group 1 (later referred to with \1 from the replacement pattern):
{{ - {{ substring
(root\.)? - an optional Group 2 matching 1 or 0 occurrences of root.
app\.url}} - a literal app.url}} substring
(?!#) - a negative lookahead that fails the match if, immediately to the right of the current location, there is a # char.
See Table 9-17. Regular Expression Constraints:
(?!re) negative lookahead matches at any point where no substring matching re begins (AREs only)
PostgreSQL demo:
select regexp_replace('{{app.url}}
{{root.app.url}}
{{app.url}}#param
{{root.app.url}}#param',
'({{(root\.)?app\.url}})(?!#)',
'\1#param',
'g');
Related
Working on trying to figure out some regex to pull out the last 2 segments of an FQDN.
^.*\shostname=[\w-]+\.(?P<myfield>[^\t]+)
This RegEx works and takes out the first segment of an FQDN.
www.aaa.bbb.someurl.net --> aaa.bbb.someurl.net
But… I only want to keep the last 2 segments of any FQDN.
I need it to be --> someurl.net
Other restrictions:
The hostname field will always be at least 3 segments - don't know the max.
This is for Splunk so I can't use a script. I need it to be PCRE compatible regex.
Here is an example of data:
2021-07-20 18:19:14 reason=Not allowed to browse this category event_id=12345 protocol=HTTP action=Blocked transactionsize=16051 responsesize=789 requestsize=456 urlcategory=Blocked serverip=1.2.4.5 clienttranstime=0 requestmethod=GET refererURL=None useragent=Microsoft-Delivery location=Internal ClientIP=5.6.7.8 status=403 user=John url=dl.delivery.mp.microsoft.com/filestreamingservice/files/abcd-efgh-ijkl/pieceshash vendor=Zscaler hostname=dl.delivery.mp.microsoft.com
From this I data I need the field “myfield” to be: microsoft.com.
The original answer with a much simpler regex ((?:\s|^)hostname=(?:[^\s.]+\.)*(?P<myfield>[^\s.]+\.[^\s.]+)) that worked for OP is in the question history.
You can use
(?:\s|^)hostname=(?:[^\s.]+\.)*?(?P<myfield>[^\s.]+\.(?:(?:ac|co)\.uk|govt?\.uk|judiciary\.uk|l(?:ea|td)\.uk|m(?:e|il|od)\.uk|n(?:et|hs|ic)\.uk|orgn?\.uk|p(?:arliament|lc|olice)\.uk|(?:royal|sch)\.uk|[^\s.]+)(?!\S))
Or, to match the last hostname=... value on a line:
^.*\shostname=(?:[^\s.]+\.)*?(?P<myfield>[^\s.]+\.(?:(?:ac|co)\.uk|govt?\.uk|judiciary\.uk|l(?:ea|td)\.uk|m(?:e|il|od)\.uk|n(?:et|hs|ic)\.uk|orgn?\.uk|p(?:arliament|lc|olice)\.uk|(?:royal|sch)\.uk|[^\s.]+)(?!\S))
See the regex #1 demo and regex #2 demo. Details:
(?:\s|^) - either a whitespace or start of string
hostname= - a literal substring
(?:[^\s.]+\.)*? - zero or more (but as few as possible) occurrences of one or more chars other than whitespace and dot and then a dot
(?P<myfield>[^\s.]+\.(?:(?:ac|co)\.uk|govt?\.uk|judiciary\.uk|l(?:ea|td)\.uk|m(?:e|il|od)\.uk|n(?:et|hs|ic)\.uk|orgn?\.uk|p(?:arliament|lc|olice)\.uk|(?:royal|sch)\.uk|[^\s.]+)(?!\S)) - Group "myfield": one or more chars other than whitespace and dot, then a dot, then any second-level domain or any one or more chars other than whitespace and dot and then either a whitespace or end of string.
Note: the \.(?:(?:ac|co)\.uk|govt?\.uk|judiciary\.uk|l(?:ea|td)\.uk|m(?:e|il|od)\.uk|n(?:et|hs|ic)\.uk|orgn?\.uk|p(?:arliament|lc|olice)\.uk|(?:royal|sch)\.uk pattern part (built from a regex trie) matches this list:
.ac.uk
.co.uk
.gov.uk
.judiciary.uk
.ltd.uk
.me.uk
.mod.uk
.net.uk
.nhs.uk
.nic.uk
.org.uk
.parliament.uk
.plc.uk
.police.uk
.royal.uk
.sch.uk
.co.uk
.ltd.uk
.me.uk
.net.uk
.nic.uk
.org.uk
.plc.uk
.sch.uk
.govt.uk
.orgn.uk
.lea.uk
.mil.uk
If you want to add more second-level domain names, add more to the list and use https://www.myregextester.com or suchlike services to built the word list regex.
You could match all following non whitspace chars after hostname= and then use a capture group to capture the last part with a single dot.
^.*\shostname=(?:\S+\.)?([^\s.]+\.[^\s.]+)
^.*\shostname=
(?:\S+\.)? Optionally match a possible dot before
( Capture group 1
[^\s.]+\.[^\s.]+ Match 2 non dot parts with a . in between
) Close group
Regex demo
If you would like to account for country codes, I've previously answered this at: Get Domain Extension From Hostname
The regular expression would look something like (shortened version): \w+((\.[a-z]{2,3})(\.(uk|au))?)$
The full expression with all country codes: \w+((\.[a-z]{2,3})(\.(ad|ae|af|ag|ai|al|am|ao|aq|ar|as|at|au|aw|ax|az|ba|bb|bd|be|bf|bg|bh|bi|bj|bl|bm|bn|bo|bq|br|bs|bt|bv|bw|by|bz|ca|cc|cd|cf|cg|ch|ci|ck|cl|cm|cn|co|cr|cu|cv|cw|cx|cy|cz|de|dj|dk|dm|do|dz|ec|ee|eg|er|es|et|fi|fj|fk|fm|fo|fr|ga|gb|gd|ge|gf|gg|gh|gi|gl|gm|gn|gp|gq|gr|gs|gt|gu|gw|gy|hk|hm|hn|hr|ht|hu|id|ie|il|im|in|io|iq|ir|is|it|je|jm|jo|jp|ke|kg|kh|ki|km|kn|kp|kr|kw|ky|kz|la|lb|lc|li|lk|lr|ls|lt|lu|lv|ly|ma|mc|md|me|mf|mg|mh|mk|ml|mm|mn|mo|mp|mq|mr|ms|mt|mu|mv|mw|mx|my|mz|na|nc|ne|nf|ng|ni|nl|no|np|nr|nu|nz|om|pa|pe|pf|pg|ph|pk|pl|pm|pn|pr|ps|pt|pw|py|qa|re|ro|rs|ru|rw|sa|sb|sc|sd|se|sg|sh|si|sj|sk|sl|sm|sn|so|sr|ss|st|sv|sx|sy|sz|tc|td|tf|tg|th|tj|tk|tl|tm|tn|to|tr|tt|tv|tw|tz|ua|ug|uk|us|uy|uz|va|vc|ve|vg|vi|vn|vu|wf|ws|ye|yt|za|zm|zw))?)$
I'm having an issue with Regex.
I'm trying to match T0000001 (2, 3 and so on).
However, some of the lines it searches has what I can describe as positioners. These are shown as a question mark, followed by 2 digits, such as ?21.
These positioners describe a new position if the document were to be printed off the website.
Example:
T123?214567
T?211234567
I need to disregard ?21 and match T1234567.
From what I can see, this is not possible.
I have looked everywhere and tried numerous attempts.
All we have to work off is the linked image. The creators cant even confirm the flavour of Regex it is - they believe its Python but I'm unsure.
Regex Image
Update
Unfortunately none of the codes below have worked so far. I thought to test each code in live (Rather than via regex thinking may work different but unfortunately still didn't work)
There is no replace feature, and as mentioned before I'm not sure if it is Python. Appreciate your help.
Do two regex operations
First do the regex replace to replace the positioners with an empty string.
(\?[0-9]{2})
Then do the regex match
T[0-9]{7}
If there's only one occurrence of the 'positioners' in each match, something like this should work: (T.*?)\?\d{2}(.*)
This can be tested here: https://regex101.com/r/XhQXkh/2
Basically, match two capture groups before and after the '?21' sequence. You'll need to concatenate these two matches.
At first, match the ?21 and repace it with a distinctive character, #, etc
\?21
Demo
and you may try this regex to find what you want
(T(?:\d{7}|[\#\d]{8}))\s
Demo,,, in which target string is captured to group 1 (or \1).
Finally, replace # with ?21 or something you like.
Python script may be like this
ss="""T123?214567
T?211234567
T1234567
T1234434?21
T5435433"""
rexpre= re.compile(r'\?21')
regx= re.compile(r'(T(?:\d{7}|[\#\d]{8}))\s')
for m in regx.findall(rexpre.sub('#',ss)):
print(m)
print()
for m in regx.findall(rexpre.sub('#',ss)):
print(re.sub('#',r'?21', m))
Output is
T123#4567
T#1234567
T1234567
T1234434#
T123?214567
T?211234567
T1234567
T1234434?21
If using a replace functionality is an option for you then this might be an approach to match T0000001 or T123?214567:
Capture a T followed by zero or more digits before the optional part in group 1 (T\d*)
Make the question mark followed by 2 digits part optional (?:\?\d{2})?
Capture one or more digits after in group 2 (\d+).
Then in the replacement you could use group1group2 \1\2.
Using word boundaries \b (Or use assertions for the start and the end of the line ^ $) this could look like:
\b(T\d*)(?:\?\d{2})?(\d+)\b
Example Python
Is the below what you want?
Use RegExReplace with multiline tag (m) and enable replace all occurrences!
Pattern = (T\d*)\?\d{2}(\d*)
replace = $1$2
Usage Example:
I am using this regex
(rs)\w+/
to select strings that begin with the string 'rs', i.e.
..the biomarker rs4343 but not rs4342. However rs4343 ..
this returns: rs4343, rs4242, re4343
Is it possible to use regex to select only the first instance of a matched string to avoid duplication, i.e. to return: rs4343, rs4242
I can use JS or PHP regex.
Try this:
(rs\w+)(?!.*\1)
Regex101
Details:
(rs\w+) - Group the required match
(?!.*\1) - Use negative lookahead to assert that there is no same match after this
i am using match expression as https://([^/]*)/(.*) and replace expression as constantprefix/$2 and trying to rewrite incoming URL by adding '/constantprefix' to all URLs
for Below URLs it is working as expected:
https://hostname/incomingURI is converting to
/constantprefix/incomingURI
https://hostname/ is converting to /constantprefix/
https://hostname/login/index.aspx is converting to
/constantprefix/login/index.aspx
i am having problem for the URLs which already starting with /constantprefix, i am seeing two /constantprefix/constantprefix in the output URL which I am not looking for, is there any way we can avoid that ?
if incoming URL is https://hostname/constantprefix/login/index.aspx then output URL is becoming https://hostname/constantprefix/constantprefix/login/index.aspx
may i know how i can avoid /constantprefix/constantprefix from match expression ?
You can do it with:
https://[^/]*/(?!constantprefix(?:/|$))(.*)
using the replacement string:
constantprefix/$1
(?!...) is a negative lookahead and means not followed by. It's only a test and doesn't consume characters (this kind of elements in a pattern are also called "zero-width assertions" as a lookbehind or anchors ^ and $).
The first capture group in your pattern was useless, I removed it.
Consider the following input string (part of css file):
url('data:image/png;base64,iVBORw0KGgoAAAAN...');
url(example.png);
The objective is to take the url part using regex and do something with it. So the first part is easy:
url\(['"]?(.+?)['"]?\)
Basically, it takes contents from inside url(...) with optional quotes symbols. Using this regexp I get the following matches:
data:image/png;base64,iVBORw0KGgoAAAAN...
example.png
So far so good. Now I want to exclude the urls which include 'data:image' in their text. I think negative lookahead is the proper tool for that but using it like this:
url\(['"]?(?!data:image)(.+?)['"]?\)
gives me the following result for the first url:
'data:image/png;base64,iVBORw0KGgoAAAAN...
Not only it doesn't exclude this match, but the matched string itself now includes quote character at the beginning. If I use + instead of first ? like this:
url\(['"]+(?!data:image)(.+?)['"]?\)
it works as expected, url is not matched. But this doesn't allow the optional quote in url (since + is 1 or more). How should I change the regex to exclude given url?
You can use negative lookahead like this:
url\((['"]?)((?:(?!data:image).)+?)\1?\)
RegEx Demo