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I have two different forms of a string:
https://anexample.com/things/stuff
and
https:///things/stuff
I need a regular expression that will match everything in the string after the 3rd slash, no matter the syntax. Once it hits the 3rd slash, the rest of the string is matched. I have found a bunch of examples, but I can't seem to tweak the right way to get it to work. Thanks in advance.
You can use this
^[^/]*/[^/]*/[^/]*/(.*)$
You can use this regex:
^(?:[^\/]*\/){3}(.*)$
And use matched group #1
In javascript:
var s = 'https:///things/stuff';
m = s.match(/^(?:[^\/]*\/){3}(.*)$/);
// m[1] => things/stuff
Assuming PCRE, and that you won't have newlines in your string:
If the 3 slashes can be at any position (like your first example):
^[^/]/[^/]*/[^/]*/(.*)$
This could also be expressed as
^(:?[^/]*/){3}(.*)$
Using positive lookbehind, you could use the following, which should only match what you want instead of putting it into a capturing group:
(?<=^(:?[^/]*/){3}).*$
Any needed escaping due to used delimiters is left as an exercise to the reader of course ( if you use / as a delimiter, you have to escape all / in the expression, like \/)
And there's probably a million other alternatives, depending on what exact needs you have besides the ones you mentioned.
Something like this should work, however I'm writting it without any testing, but it should look for three sections of any character any number of times followed by slash and then catch last section which is everything until line end - you can of course change delimiter to whitespace or whatever.
^.*/{3}(.*$)
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I have a long output string in PowerShell with all complex characters
this is part of it:
{host-up|rp-web1|/images/logos/Generic_Host.gif|0|276|0 CRITICAL service-critical|rp-web1|ssl_expiration_bitwarden|26186|0|0|1|0 2023/02/06 ....
"service-critical" is a fixed string and appears several more times in the string
"rp-web1|ssl_expiration_bitwarden" - this is a dynamic string which comes right after "service-critical"
I was not able to write a regex that managed to extract all the dynamic strings in the text
Of course I tried to use 3 pipes between the dynamic string but without success
I expect to get all dynamic string after "service-critical" like:
rp-web1|ssl_expiration_bitwarden
Use the .NET [regex]::Matches() method to extract all desired matches of a given regex.
# Sample input string with 2 "service-critical" groups.
$str=#'
{host-up|...|0 CRITICAL service-critical|rp-web1|ssl_expiration_bitwarden1|26186|0|0|1|0 2023/02/06 ....
{host-up|...|0 CRITICAL service-critical|rp-web2|ssl_expiration_bitwarden2|26186|0|0|1|0 2023/02/06 ....
'#
# Define the regex.
$regex = '(?<=\bservice-critical\|)[^|]+\|[^|]+'
# Find and report all matches
# -> 'rp-web1|ssl_expiration_bitwarden1', 'rp-web2|ssl_expiration_bitwarden2'
[regex]::Matches($str, $regex).Value
For an explanation of the regex and the ability to experiment with it, see this regex101.com page.
Unfortunately, use of PowerShell's -matchoperator (with subsequent inspection of automatic $Matches variable) is not an option in this case, because it invariably looks only for one match.
GitHub issue #7867 is a green-lit proposal to introduce a -matchall operator in order to support finding all matches.
You can try this one:
(?:service-critical\|)\K(.*?\|.*?)(?=\|)
Here is a demo
(?:service-critical\|) - looks for service-critial| phrase but without grouping it
\K - omits matching of found service-critial|, as that is the part you do not want to see in results
NOTE: as #mklement0 pointed out it won't work in .NET regex engine (which is used by powershell). You can skip it in this case and get matching with service-critical| phrase or use positive lookbehind structure which is shown in #mklement0 answer
(.*?\|.*?) - captures a group of any character (or no characters at all, separated by single| in non-greedy way
(?=\|) - assures it is followed by | (it is called positive lookahead)
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I tried to find all comments beginning with // that don't have a space after the slashes.
I want to select only the slashes. No whitespace or text before that, no whitespace or text after that.
So far I've reached to [\s].(\/\/(?! )) but it catches the space before the slashes as well.
Basically I wanna make sure my line comments have a space after the slashes.
I'm trying to do this either in JavaScript or in any text editor.
Since javascript doesn't have the lookbehind feature, you can't.
The workaround (for instance, in a replacement context) is to use a capture group for the character before the two slashes and to start the replacement string with a reference to this group ('$1replacement'):
([^/\s]|^)//(?! )
You can use the following regex:
.*(\/\/(?= )) demo
The idea is to use positive lookahead and capture the // iff it is followed by a space.
EDIT: Just noticed that your question is contradictory. So if you want to capture if the // is not followed by a space, use this: .*(\/\/(?=\S)). Otherwise use the one above.
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I'm trying to extract a portion of a delimited string. The string is something like this:
272;#This is the text i want
I'd like to get everything after the "#". Anybody have any suggestions?
TL;DR
Language implementations matter. Not all languages support every regular expression operator or feature. There are some general approaches, though, such as zero-width assertions and capture groups.
Positive Lookbehind
Use a zero-width assertion to find the character preceding your string. For example, to capture just the text of interest using Ruby 2.0:
'272;#This is the text i want'.match /(?<=#).*/
pp $&
#=> "This is the text i want"
Capture Groups
Use capture and non-capture groups to match text, then extract the group you're interested in. For example, to capture your desired match in the first capturing group with Ruby 2.0:
'272;#This is the text i want'.match /(?:#)(.*)/
pp $1
#=> "This is the text i want"
You can use the regex #(.*) and extract the first capturing group - btw, what language are you trying to do this??
edit: if you can't access the capturing groups you can try lookbehind if it's supported by the engine:
(?<=#).*
Consider the following Regex...
(?<=#).*
Good Luck!
Try:
string text = "272;#";
int index = text.IndexOf("#");
string sub = text.Substring(index + 1);
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I am very confused how to replace a word before pattern ".ext".
example :
Before Replace : abcd.ext.com
After Replace : customer.ext.com
You can use something like [^.]+(?=\.) as the match and replace it by customer.
(?=\.) is a positive lookahead which will match when there is a dot following the part before, but it won't match any characters on its own.
E.g. in C# you can use
Regex.Replace(foo, #"[^.]+(?=\.)", "customer");
If you're doing this in C# then I would recommend you just doing something like this:
var newFileName = fileName.Replace(Path.GetFileName(fileName), "newFileNameValue");
If it's in VB.NET, it would look almost exactly the same:
Dim newFileName As String = fileName.Replace(Path.GetFileName(fileName), "newFileNameValue")
You can use a Regex, but it's probably a little overkill and less stable. See, when building a Regex you have to break it down to a really abstract level. You need to handle every extension that's in your domain and that list can grow pretty quickly. So then it's generally not feasible to include those extensions in the Regex itself.
To further add to the problem, a valid file name might be something like this, MyFile.v1.l1.ext1.txt. The extension of that file is .txt, but grabbing that with a Regex is tough.
On Unix you can use sed like this:
echo "$str"|sed 's/abcd\(\.ext\)/customer\1/'
i.e. look for abcd immediately followed by .ext (capture this in a group). Then replace it with customer and match group #1 (.ext)
If you're using any other platform/language approach should be same.
perl
$x =~ s/(.*)(\.ext\.com)/customer$2/;
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Is there any regex master that can help me?
I have a list of words like {first|second|third}, and I want just the last word, in this case {first|second|third}.
Can anyone help me please??
Edit:
After feedback I am adding more information.
I have a sentence such as "I am going to France {today|tomorrow|next week}" for example. But I want only "I am going to France next week".
I tried (?<=\{).*?(?=\|.*?\}) but this gives me |tomorrow|next week, I just want next week without the vertical lines.
PS it doesn't necessarily have to 3 words, I just want the last regardless.
The regex you need is obviously:
((\w|\s)+)\}
A python test of the regex:
>>> import re
>>> test = """ds like {first|second|third}, and I want just t"""
>>> re.findall('\|([^\|]+)\}', test)
['third']
To generally replace {a|bc|def} with def with no nesting or other complications, search for the regex
\{[}]+\|([^|}]+)\}
and substitute with the first (and only) parenthesized subgroup. In Perl, that would be something like
s/\{[}]+\|([^|}]+)\}/$1/g;
Many other languages have a similar syntax, though it might be less compact. In PHP, look for preg_replace.
The expression looks complex, but isn't really; we look for a literal { followed by the longest possible string of characters which are not }, as long as it's followed by a literal | and (subgroup) a string of characters which are neither | nor }, and finally, after the subgroup, a literal }.
Try this
/\{.+?([^|]+?)\}/
(you can easily test it by typing the following in your browser's JS console)
"{first|second|third}".replace( /\{.+?([^|]+?)\}/, "$1")
or even
"I am going on the {first|second|third}".replace(/\{.+?([^|]+?)\}/, "$1")
-> "I am going on the third"