I am trying to write a regex which will strip away the rest of the path after a particular folder name.
If Input is:
/Repository/Framework/PITA/branches/ChangePack-6a7B6/core/src/Pita.x86.Interfaces/IDemoReader.cs
Output should be:
/Repository/Framework/PITA/branches/ChangePack-6a7B6
Some constrains:
ChangePack- will be followed change pack id which is a mix of numbers or alphabets a-z or A-Z only in any order. And there is no limit on length of change pack id.
ChangePack- is a constant. It will always be there.
And the text before the ChangePack can also change. Like it can also be:
/Repository/Demo1/Demo2/4.3//PITA/branches/ChangePack-6a7B6/core/src/Pita.x86.Interfaces
My regex-fu is bad. What I have come up with till now is:
^(.*?)\-6a7B6
I need to make this generic.
Any help will be much appreciated.
Below regex can do the trick.
^(.*?ChangePack-[\w]+)
Input:
/Repository/Framework/PITA/branches/ChangePack-6a7B6/core/src/Pita.x86.Interfaces/IDemoReader.cs
/Repository/Demo1/Demo2/4.3//PITA/branches/ChangePack-6a7B6/core/src/Pita.x86.Interfaces
Output:
/Repository/Framework/PITA/branches/ChangePack-6a7B6
/Repository/Demo1/Demo2/4.3//PITA/branches/ChangePack-6a7B6
Check out the live regex demo here.
^(.*?ChangePack-[a-zA-Z0-9]+)
Try this.Instead of replace grab the match $1 or \1.See demo.
https://regex101.com/r/iY3eK8/17
Will you always have '/Repository/Framework/PITA/branches/' at the beginning? If so, this will do the trick:
/Repository/Framework/PITA/branches/\w+-\w*
Instead of regex you could can use split and join functions. Example python:
path = "/a/b/c/d/e"
folders = path.split("/")
newpath = "/".join(folders[:3]) #trims off everything from the third folder over
print(newpath) #prints "/a/b"
If you really want regex, try something like ^.*\/folder\/ where folder is the name of the directory you want to match.
Related
I was trying to find solution for my problem.
Input: prd-abcd-efgh-i-0dflnk55f5d45df
Output: prd-abcd-efgh
Tried Splunk Query : index=aws-* (host=prd-abcd-efgh*) | rex field=host "^(?<host>[^.]+)"| dedup host | stats count by host,methodPath
I want to remove everything comes after "-i-" using simple regex.I tried with regex "^(?[^.]+)" listed here
https://answers.splunk.com/answers/77101/extracting-selected-hosts-with-regex-regex-hosts-with-exceptions.html
Please help me to solve it.
replace(host, "(?<=-i-).*", "")
Example here: https://regex101.com/r/blcCcQ/2
This (?<=-i-) is a lookbehind
I have no knowledge of Splunk. but the normal way to do that would be to match the part you don't want and replace it with an empty string.
The regex for doing that could be:
-i-.*
Then replace the match with an empty string.
Something simple like this should work:
([a-z-]+)-i-.+
The first capture group will return only the part preceding -i-.
i have some corrupt rtf files with lines like this:
{\s39\li0\fi0\ri0\sb0\sa0\ql\vertalt\fs22 Fußzeile Zchn;}
^----------------------------^
i want to replace all [^a-zA-Z0-9_\{}; ]
but only in lines beginning with "{\s" and ending with "};" from the first "space" to "};"
the first "space" and "};" should not be replaced.
You didn't specify language, here is Regex101 example:
({\\s.+?\s)(.*)(})
So, I'm unsure what language/technology you'd like to use here, but if using C# is an option, you can check out this previous question. The answer gets you almost the way there.
For your example:
var text = #"{\s39\li0\fi0\ri0\sb0\sa0\ql\vertalt\fs22 Fußzeile Zchn;}";
var pattern = #"^({\\s\S*\s[a-zA-Z0-9_\{}; ]*)([^a-zA-Z0-9_\{}; ]*)([^}]*})";
var replaced = System.Text.RegularExpressions.Regex.Replace(text, pattern, "$1$3");
This will get you to replace one contiguous blob of bad characters, which addresses your example, but unfortunately, not your question. There is probably a more elegant solution, but I think you'll have to iteratively run that expression until the input and output of Regex.Replace() are equal.
If you can use sed in a terminal, you could do something like this.
sed -i 's/^\({\\s[^ ]*\s\).*\(\;}\)\(}\)\?$/\1\2/' filename
Turned my file containing:
{\s39\li0\fi0\ri0\sb0\sa0\ql\vertalt\fs22 Fußzeile Zchn;}
To:
{\s39\li0\fi0\ri0\sb0\sa0\ql\vertalt\fs22 ;}
I'm using the following regex to find URLs in a text file:
/http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+/
It outputs the following:
http://rda.ucar.edu/datasets/ds117.0/.
http://rda.ucar.edu/datasets/ds111.1/.
http://www.discover-earth.org/index.html).
http://community.eosdis.nasa.gov/measures/).
Ideally they would print out this:
http://rda.ucar.edu/datasets/ds117.0/
http://rda.ucar.edu/datasets/ds111.1/
http://www.discover-earth.org/index.html
http://community.eosdis.nasa.gov/measures/
Any ideas on how I should tweak my regex?
Thank you in advance!
UPDATE - Example of the text would be:
this is a test http://rda.ucar.edu/datasets/ds117.0/. and I want this to be copied over http://rda.ucar.edu/datasets/ds111.1/. http://www.discover-earth.org/index.html). http://community.eosdis.nasa.gov/measures/).
This will trim your output containing trail characters, ) .
import re
regx= re.compile(r'(?m)[\.\)]+$')
print(regx.sub('', your_output))
And this regex seems workable to extract URL from your original sample text.
https?:[\S]*\/(?:\w+(?:\.\w+)?)?
Demo,,, ( edited from https?:[\S]*\/)
Python script may be something like this
ss=""" this is a test http://rda.ucar.edu/datasets/ds117.0/. and I want this to be copied over http://rda.ucar.edu/datasets/ds111.1/. http://www.discover-earth.org/index.html). http://community.eosdis.nasa.gov/measures/). """
regx= re.compile(r'https?:[\S]*\/(?:\w+(?:\.\w+)?)?')
for m in regx.findall(ss):
print(m)
So for the urls you have here:
https://regex101.com/r/uSlkcQ/4
Pattern explanation:
Protocols (e.g. https://)
^[A-Za-z]{3,9}:(?://)
Look for recurring .[-;:&=+\$,\w]+-class (www.sub.domain.com)
(?:[\-;:&=\+\$,\w]+\.?)+`
Look for recurring /[\-;:&=\+\$,\w\.]+ (/some.path/to/somewhere)
(?:\/[\-;:&=\+\$,\w\.]+)+
Now, for your special case: ensure that the last character is not a dot or a parenthesis, using negative lookahead
(?!\.|\)).
The full pattern is then
^[A-Za-z]{3,9}:(?://)(?:[\-;:&=\+\$,\w]+\.?)+(?:\/[\-;:&=\+\$,\w\.]+)+(?!\.|\)).
There are a few things to improve or change in your existing regex to allow this to work:
http[s]? can be changed to https?. They're identical. No use putting s in its own character class
[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),] You can shorten this entire thing and combine character classes instead of using | between them. This not only improves performance, but also allows you to combine certain ranges into existing character class tokens. Simplifying this, we get [a-zA-Z0-9$-_#.&+!*\(\),]
We can go one step further: a-zA-Z0-9_ is the same as \w. So we can replace those in the character class to get [\w$-#.&+!*\(\),]
In the original regex we have $-_. This creates a range so it actually inclues everything between $ and _ on the ASCII table. This will cause unwanted characters to be matched: $%&'()*+,-./0123456789:;<=>?#ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_. There are a few options to fix this:
[-\w$#.&+!*\(\),] Place - at the start of the character class
[\w$#.&+!*\(\),-] Place - at the end of the character class
[\w$\-#.&+!*\(\),] Escape - such that you have \- instead
You don't need to escape ( and ) in the character class: [\w$#.&+!*(),-]
[0-9a-fA-F][0-9a-fA-F] You don't need to specify [0-9a-fA-F] twice. Just use a quantifier like so: [0-9a-fA-F]{2}
(?:%[0-9a-fA-F][0-9a-fA-F]) The non-capture group isn't actually needed here, so we can drop it (it adds another step that the regex engine needs to perform, which is unnecessary)
So the result of just simplifying your existing regex is the following:
https?://(?:[$\w#.&+!*(),-]|%[0-9a-fA-F]{2})+
Now you'll notice it doesn't match / so we need to add that to the character class. Your regex was matching this originally because it has an improper range $-_.
https?://(?:[$\w#.&+!*(),/-]|%[0-9a-fA-F]{2})+
Unfortunately, even with this change, it'll still match ). at the end. That's because your regex isn't told to stop matching after /. Even implementing this will now cause it to not match file names like index.html. So a better solution is needed. If you give me a couple of days, I'm working on a fully functional RFC-compliant regex that matches URLs. I figured, in the meantime, I would at least explain why your regex isn't working as you'd expect it to.
Thanks all for the responses. A coworker ended up helping me with it. Here is the solution:
des_links = re.findall('http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+', des)
for i in des_links:
tmps = "/".join(i.split('/')[0:-1])
print(tmps)
I need to put together a regex that matches a patter only if string does not begin with 'N'.
Here is my pattern so far [A-E]+[-+]?.
Now I want to make sure that it does not match something like:
N\A
NA
NB+
NB-
NCAB
This is for REGEXP_SUBSTR command in Oracle SQL DB
UPDATE
It looks like I should have been more specific, sorry
I want to extract from a string [A-E]+[-+]? but if the string also matches ^(N|n) then I want my regex to return nothing.
See examples below:
String Returns
N/A
F1/AAA AAA
NABC
FABC ABC
To match a character between A and E not preceded by N, you can use:
([^N]|^)[A-E]+
If you want to avoid fields that contains N[A-E] use a negation in your query using the pattern N[A-E] (in other words, use two predicates, this one to exclude NA and the first to find A)
To be more clear:
WHERE NOT REGEXP_LIKE(coln, 'N[A-E]') AND REGEXP_LIKE(coln, '[A-E]')
Ok I figured it out, I broadened the scope of the problem a little, I realized that I can also play with other parameters of REGEXP_SUBSTR in this case that I can have returned only second substring.
REGEXP_SUBSTR(field1, '^([^NA-D][^A-D]*)?([A-D]+[-+]?)',1,1,'i',2)
I still have to give you guys the credit, lot of good ideas that led me to here.
Just throw a [^N]? in front. That should do it.
OOPS...
That actually needs to include an " OR ^ "...
It should look like this:
([^N]|^)[A-E]+[-+]?
Sorry about that...It looks like the right answer already got posted anyway.
I'm trying to solve a regex riddle. Let's say I have rows of hrefs looking like this:
anchor1.in
an3.php
setup.exe
What I want the regex (or any other solution) to do is to take the href title and copy it over to the actual url with a foward slash in front of it.
A successful result would become:
anchor1.in
an3.php
setup.exe
If you can solve this please explain how you did it.
You can use the following to match:
(<a\s+href=")(.*?)(">)(.*?)(<\/a>)
And replace with:
\1\2/\4\3\4\5
See DEMO and Explanation