I have the following languages or language locale codes in a URL and i am trying to identify through REGEX. I was partially successful in identifying them but it is failing for some scenarios
Languages that i am testing with
en-us -- Passes
us -- Fails
Here is the REGEX that i have
([a-zA-Z]{2}|[a-zA-Z]{2}-[a-zA-Z]{2}\/)c\/(deals-and-tips\/)?
For instance:
https://forum.leasehackr.com/en-us/c/deals-and-tips (passes)
https://forum.leasehackr.com/us/c/deals-and-tips (fails)
What am I missing in the above REGEX?
The regex you wanted is:
([a-zA-Z]{2}|[a-zA-Z]{2}-[a-zA-Z]{2})\/c\/(deals-and-tips\/)?
The difference from your regex is that I moved the first \/ from inside the parenthesis to outside (to sit with c\/).
Test here.
The last / fails the match in any case since your urls doesn't have it, in any way I would rewrite your regex as this: ([a-zA-Z]{2})(-[a-zA-Z]{2})?\/c\/(deals-and-tips)?.
This way it always looks for the first part (en) and consider the second (-us) as optional.
Alternatively use (\w{2})(-\w{2})?\/c\/(deals-and-tips)?, if you don't mind risking to match underscores and similar simbols
The reason your pattern does not match us is because the alternation ([a-zA-Z]{2}|[a-zA-Z]{2}-[a-zA-Z]{2}\/) only matches the \/ in the second part of the alternation.
Also it does not match the last group with deals-and-tips because there is no trailing \/ in the example data.
Your updated pattern might look like
([a-zA-Z]{2}|[a-zA-Z]{2}-[a-zA-Z]{2})\/c\/(deals-and-tips)?
Regex demo
You could shorten the pattern a bit by using an optional non capturing group (?:-[a-zA-Z]{2})? inside the first capturing group to optionally match the part starting with a hyphen.
As in the example data you could match the leading \/ in front of the capturing group to get a more efficient match.
\/([a-zA-Z]{2}(?:-[a-zA-Z]{2})?)\/c\/(deals-and-tips)?
In parts
\/ To be a bit more precise, match the leading /
( Capture group 1
[a-zA-Z]{2} Match 2 chars a-z
(?:-[a-zA-Z]{2})? Optionally match - and 2 chars a-z
) Close group
\/c\/ Match /c/deals-and-tips`
(deals-and-tips)? Optional capture group 2 match deals-and-tips
Regex demo
Note that if you use another delimiter than / you don't have to escape the forward slash.
Related
This is the regex that I'm currently testing
[\w\. ]+(?=[\.])
My ultimate goal is to include a regex expression to extract using regexp_extract in Impala/Hive query.
regexp_extract(col, '[\w\. ]+(?=[\.])', 1)
This doesn't work in Impala however.
Examples of path to extract from:
D:\mypath\Temp\abs\device\Program1.lua
D:\mypath\Temp\abs\device\SE1_Test-program.lua
D:\mypath\Temp\abs\device\Test_program.lua
D:\mypath\Temp\abs\device\Device_Test_Case-general.lua
The regex I've tested extracts the term I'm looking for but it's not good enough, for the second and third, fourth cases I would need to extract only the part after the last underscore.
My expections are:
Program1
Test-program
program
Case-general
Any suggestions? I'm also open to using something other than regexp_extract.
Note that Impala regex does not support lookarounds, and thus you need a capturing group to get a submatch out of the overall match. Also, if you use escaping \ in the pattern, make sure it is doubled.
You can use
regexp_extract(col, '([^-_\\\\]+)\\.\\w+$', 1)
See the regex demo.
The regex means
([^-_\\]+) - Group 1: one or more chars other than -, _ and \
\. - a dot
\w+ - one or more word chars
$ - end of string.
Using \w also matches an underscore, instead you can use [a-zA-Z0-9] instead.
Add matching a dot and hyphen in the character class, capture that in group 1 and match the expected trailing dot.
Note that you don't have to escape dots in a character class.
([a-zA-Z0-9.-]+)[.]
See a regex101 demo
Example using regexp_extract where the , 1 gets the group 1 value:
regexp_extract(col, '([a-zA-Z0-9.-]+)[.]', 1)
If it should be at the end of the string only, matching the last dot without matching any backslashes in between:
regexp_extract(col, '([a-zA-Z0-9.-]+)[.][^\\\\.]+$', 1)
The goal of my regular expression adventure is to create a matcher for a mechanism that could add a trailing slash to URLs, even in the presence of parameters denoted by # or ? at the end of the URL.
For any of the following URLs, I'm looking for a match for segment as follows:
https://example.com/what-not/segment matches segment
https://example.com/what-not/segment?a=b matches segment
https://example.com/what-not/segment#a matches segment
In case there is a match for segment, I'm going to replace it with segment/.
For any of the following URLs, there should be no match:
https://example.com/what-not/segment/ no match
https://example.com/what-not/segment/?a=b no match
https://example.com/what-not/segment/#a no match
because here, there is already a trailing slash.
I've tried:
This primitive regex and their variants: .*\/([^?#\/]+). However, with this approach, I could not make it not match when there is already a trailing slash.
I experimented with negative lookaheads as follows: ([^\/\#\?]+)(?!(.*[\#\?].*))$. In this case, I could not get rid of any ? or # parts properly.
Thank you for your kind help!
Lookahead and lookbehind conditionals are so powerful!
(?<=\/)[\w]+(?(?=[\?\#])|$)
P.s: I just added [\w]+ that means [a-zA-Z0-9_]+.
Of course URLs can contain many other character like - or ~ but for the examples provided it works nicely.
If you want to match urls, you might use
\b(https?://\S+/)[^\s?#/]+(?![^\s?#])
Explanation
\b A word boundary to prevent a partial word match
( Capture group 1
https?://\S+/ Match the protocol, 1+ non whitespace chars and then the last occurrence of /
) Close group 1
[^\s?#/]+ Match 1+ chars other than a whitespace char ? # /
(?![^\s?#]) Negative lookahead, assert that directly to the right is not a non whitespace char other than ? or #
See a regex demo.
In the replacement use group 1 followed by segment/
For a match only instead of a capture group:
(?<=\bhttps?://\S+/)[^\s?#/]+(?![^\s?#])
See another regex demo.
I have a regex
[a-zA-Z][a-z]
I have to change this regex such that the regex should not accept string that starts with "de","DE","dE" and "De" .I cannot use look behind or look ahead because my system does not support it?
There's a solution without a lookahead or lookbehind, but you need to be able to use groups.
The idea there is to create a sort of "honeypot" that will match your negative results and keep only the results that do interest you.
In your case, that would write:
[dD][eE].*|(<your-regex>)
If the proposition is de<anything> (case insensitive here), it will match, but group(1) will be null.
On the other hand, matching diZ for instance would match not match what is before the or and would therefore fall into the group(1).
Finally, if the proposition doesn't start with de and doesn't match your regex, well, there will be no groups to get at all.
If you need to be sure that your proposition will match the whole provided string, you can update the regex thus:
^(?:[dD][eE].*|(<your-regex>))$
Note that ?: is not a lookahead of any kind, it serves to mark the group as non-capturing, so that <your-regex> will still be captured by group(1) (would become group(2) otherwise and the capture of a group is not always a transparent operation, performance-wise).
Simply ignore those characters:
[a-ce-z][a-df-z][a-gi-kwxyzWZXZ]
Make sure the flag is set to case insensitive. Also, [a-gi-kwxyzWZXZ] can then be modified to [a-gi-kwxyz].
EDIT:
As pointed out in this comment, the regex here won't support other words that start with d but are not followed by e. In this case, negative lookahead is a possible solution:
^(?!de)[a-z]+
This matches anything not starting with "DE" (case insensitive, without look arounds, allowing leading whitespace):
^ *+(?:[^Dd].|.[^Ee])<your regex for rest of input>
See live demo.
The possessive quantifier *+ used for whitespace prevents [^Dd] from being allowed to match a space via backtracking, making this regex hardened against leading spaces.
You can use an alternation excluding matching the d and D from the first character, or exclude matching the e as the second character.
Note that the pattern [a-zA-Z][a-z] matches at least 2 characters, so will the following pattern:
^(?:[abce-zABCE-Z][a-z]|[a-zA-Z][a-df-z]).*
^ Start of string
(?: Non capture group
[abce-zABCE-Z][a-z] Match a char a-zA-Z without d and D followed by a lowercase char a-z
| or
[a-zA-Z][a-df-z] Match a char a-zA-Z followed by a lowercase chars a-z without e
) Close non capture grou
.* Match 0+ times any char except a newline
Regex demo
Another option is to use word boundaries \b instead of an anchor ^
\b(?:[abce-zABCE-Z][a-z]|[a-zA-Z][a-df-z])[a-zA-Z]*\b
Regex demo
I have strings which look like this:
/xxxxx/xxxxx-xxxx-xxxx-338200.html
With my regex:
(?<=-)(\d+)(?=\.html)
It matches just the numbers before .html.
Is it possible to write a regex that matches everything that surrounds the numbers (matches the .html part and the part before the numbers)?
In your current pattern you already use a capturing group. In that case you might also match what comes before and after instead of using the lookarounds
-(\d+)\.html
To get what comes before and after the digits, you could use 2 capturing groups:
^(.*-)\d+(\.html)$
Regex demo
In the replacement use the 2 groups.
This should do the job:
.*-\d+\.html
Explanation: .* will match anything until -\d+ say it should match a - followed by a sequence of digits before a \.html (where \. represents the character .).
To capture groups, just do (.*-)(\d+)(\.html). This will put everything before the number in a group, the number in another group and everything after the number in another group.
I have this regex pattern which I made myself (I'm a noob though, and made it through following tutorials):
^([a-z0-9\p{Greek}].*)\s(Ε[0-9\p{Greek}]+|Θ)\s[\(]([a-z1-9\p{Greek}]+.*)[\)]\s-\s([a-z0-9\p{Greek}]+$)
And I'm trying to match the following sentences:
ΠΡΟΓΡΑΜΜΑΤΙΣΤΙΚΕΣ ΕΦΑΡΜ ΣΤΟ ΔΙΑΔΙΚΤΥΟ Ε2 (Ε.Β.Δ.) - ΔΗΜΗΤΡΙΟΥ
ΠΡΟΓΡΑΜΜΑΤΙΣΜΟΣ 1 Θ (ΑΜΦ) - ΜΑΣΤΟΡΟΚΩΣΤΑΣ
ΕΙΣΑΓΩΓΗ ΣΤΗΝ ΠΛΗΡΟΦΟΡΙΚΗ Θ (ΑΜΦ) - ΒΟΛΟΓΙΑΝΝΙΔΗΣ
And so on.
This pattern splits the string into 4 parts.
For example, for the string:
ΠΡΟΓΡΑΜΜΑΤΙΣΤΙΚΕΣ ΕΦΑΡΜ ΣΤΟ ΔΙΑΔΙΚΤΥΟ Ε2 (Ε.Β.Δ.) - ΔΗΜΗΤΡΙΟΥ
The first match is: ΠΡΟΓΡΑΜΜΑΤΙΣΤΙΚΕΣ ΕΦΑΡΜ ΣΤΟ ΔΙΑΔΙΚΤΥΟ (Subject's Name)
Second match is: Ε2 (Class)
Third match is: Ε.Β.Δ. (Room)
And the forth match is: ΔΗΜΗΤΡΙΟΥ (Teacher)
Now in some entries E*/Θ is not defined, and I want to get the 3 matches without the E*/Θ. How should I modify my pattern so that (Ε[0-9\p{Greek}]+|Θ) is an optional match?
I tried ? so far, but because in my previous matches i'm defining \s and \s it requires 2 whitespaces to get 3 matches and i only have one in my string.
I think you need to do two things:
Make .* lazy (i.e. .*?)
Enclose (?:\s(Ε[0-9\p{Greek}]+|Θ))? with a non-capturing optional group.
The regex will look like
^([a-z0-9\p{Greek}].*?)(?:\s(Ε[0-9\p{Greek}]+|Θ))?\s[\(]([a-z1-9\p{Greek}]+.*)[\)]\s-\s([a-z0-9\p{Greek}]+)$
^^ ^^ ^
See demo
If you do not make the first .* lazy, it will eat up the second group that is optional. Making it lazy will ensure that if there is some text that can be matched by the second capturing group, it will be "set".
Note you call capture groups matches, which is wrong. Matches are whole texts matched by the entire regular expression and captures are just substrings matched by parts of regexp enclosed in unescaped round brackets. See more on capture groups at regular-expressions.info.
You can use something like:
(E[0-9\p{Greek}]+|0)?
The whole group will be optional (?).