I have path strings like these two:
tree/bee.horse_2021/moose/loo.se
bee.horse_2021/moose/loo.se
bee.horse_2021/mo.ose/loo.se
The path can be arbitrarily long after moose. Sometimes the first part of the path such as tree/ is missing, sometimes not. I want to capture tree in the first group if it exists and bee.horse in the second.
I came up with this regex, but it doesn't work:
path_regex = r'^(?:(.*)/)?([a-zA-Z]+\.[a-zA-Z]+).+$'
What am I missing here?
You can restrict the characters to be matched in the first capture group.
For example, you could match any character except / or . using a negated character class [^/\n.]+
^(?:([^/\n.]+)/)?([a-zA-Z]+\.[a-zA-Z]+).*$
Regex demo
Or you can restrict the characters to match word characters \w+ only
^(?:(\w+)/)?([a-zA-Z]+\.[a-zA-Z]+).*$
Regex demo
Note that in your pattern, the .+ at the end matches as least a single character. If you want to make that part optional, you can change it to .*
Related
Given a string such as this:
upstream-status=502; upstream-scheme=http; upstream-host=dfsdf-dsfsd88.dsfsdf99.sdfsdf.dfdf.in.sdfsf; upstream-url=%2FWebObjects%2Fdsdf.woa;
The regex that I wrote for matching and extracting the upstream-host is:
upstream-host=(?P<hostname>\S+(?=;))*
The ?P<hostname> allows me to create a named group.
The \S+ matches the actual hostname.
The ?=; says don't include the ; in the named group.
The last * says I don't care what comes after.
I have a nagging feeling that there is a better way to write this regex.
You can omit the lookahead and match the ; outside of the group, as the \S+ first captures all non whitespace chars and then you also match the last ; instead of asserting it.
Also, you can omit the quantifier * from the group, as repeating it zero or more times it can also match an empty string.
upstream-host=(?P<hostname>\S+);
Regex demo
I have two files:
/c/desktop/test.txt#edit
/c/desktop/test.txt
I am using regex: (.*desktop.*)(?:#.*)?
it should match everything before and after desktop but leave anything which is from #, which may or may not exists in that line.
But it's either matching everything or nothing.
One way of achieving what you want is by using the non-greedy operator *? in conjunction with the end of line operator: (.*desktop.*?)(?:$|#.*)
.*? says match as few characters as possible
$|#.* says match either the end of line or a # followed by characters. This way, the .* from the first group does not match past the # because it is possible to match the pattern with fewer characters if the second group takes it.
Tested here: https://regex101.com/r/7l1CQi/1
I am trying to parse UIDs from URLs. However regex is not something I am good at so seeking for some help.
Example Input:
https://example.com/d/iazs9fEil/somethingelse?foo=bar
Example Output:
iazs9fEil
What I've tried so far is
([/d/]+[\d\x])\w+
Which somehow works, but returns in with the /d/ prefix, so the output is /d/iazs9fEil.
How to change the regex to not contain the /d/ prefix?
EDIT:
I've tried this regex ([^/d/]+[\d\x])\w+ which outputs the correct string which is iazs9fEil, but also returns the rest of the url, so here it is somethingelse?foo=bar
In short, you may use
match = re.search(r'/d/(\w+)', your_string) # Look for a match
if match: # Check if there is a match first
print(match.group(1)) # Now, get Group 1 value
See this regex demo and a regex graph:
NOTE
/ is not any special metacharacter, do not escape it in Python string patterns
([/d/]+[\d\x])\w+ matches and captures into Group 1 any one or more slashes or digits (see [/d/]+, a positive character class) and then a digit or (here, Python shows an error: sre_contants.error incomplete escape \x, probably it could parse it as x, but it is not the case), and then matches 1+ word chars. You put the /d/ into a character class and it stopped matching a char sequence, [/d/]+ matches slashes and digits in any order and amount, and certainly places this string into Group 1.
Try (?<=/d/)[^/]+
Explanation:
(?<=/d/) - positive lookbehind, assure that what's preceeding is /d/
[^/]+ - match one or more characters other than /, so it matches everything until /
Demo
You could use a capturing group:
https?://.*?/d/([^/\s]+)
Regex demo
I am trying to make regex that can match all of them:
word
word-hyphen
word-hyphen-again
that is -\w+could be many depends on words in a term. How can I make it optional
Thing I made so far is given here:- https://regex101.com/r/Atpwze/1
Try using
\w+(-\w+)* for matching 0 or more hyphenated words after first word
\w+(-\w+){0,} same as first case
based on your exact requirement.
In order to eliminate some extreme cases like a-+-+---, you could use \w+(-\w+)*[^\W]
\W matches all non-word characters and ^\W negates the matching of non-word characters
To catch all of your examples, I think you could use:
^\w+(?:\w+\-?|\-\w+)+$
Beginning of the string ^
Match a word character one or more times \w+
Start a non capturing group (?:
Match a word character one or more times with an optional hyphen \w+\-?
Or |
A hyphen with one or more word characters \-\w+
Close the non capturing group )
End of the string $
I have string in a text file containing some text as follows:
txt = "java.awt.GridBagLayout.layoutContainer"
I am looking to get everything before the Class Name, "GridBagLayout".
I have tried something the following , but I can't figure out how to get rid of the "."
txt = re.findall(r'java\S?[^A-Z]*', txt)
and I get the following: "java.awt."
instead of what I want: "java.awt"
Any pointers as to how I could fix this?
Without using capture groups, you can use lookahead (the (?= ... ) business).
java\s?[^A-Z]*(?=\.[A-Z]) should capture everything you're after. Here it is broken down:
java //Literal word "java"
\s? //Match for an optional space character. (can change to \s* if there can be multiple)
[^A-Z]* //Any number of non-capital-letter characters
(?=\.[A-Z]) //Look ahead for (but don't add to selection) a literal period and a capital letter.
Make your pattern match a period followed by a capital letter:
'(java\S?[^A-Z]*?)\.[A-Z]'
Everything in capture group one will be what you want.
This seems to do what you want with re.findall(): (java\S?[^A-Z]*)\.[A-Z]