I am a beginner at regex and have the following problem:
I want a regex where it will target only the "/new" string and not the "/news" string. Here are two examples of text I want to use it on:
/category/news/new
/category/news/new?t=week
/politics/new
Also, the /news will always precede /new in order. There will also never be another '/' after /new. (I hope that makes sense to you.)
(I use this to apply active classes to the navigation menu. But I get a problem in the news section where the page thinks the active sort type is 'new' when it is actually 'top' as I base this off the URL's path)
I attempted writing one but it didn't work:
/new[^s]
Any help would be greatly appreciated.
Thanks
You may use
/new(?:[?/]|$)
See the regex demo
Details:
/new - a literal /new substring
(?:[?/]|$) - either of the two alternatives:
[?/] - either a ? or /
| - or
$ - end of string
Related
I would like to ask for help from people that have more advanced regex understanding than me. I have spent many hours trying stuff and also gone thru the tutorials on youtube and I'm at my wits end because it kind of works but the regex pattern is kind of broad I'm unable to narrow it down
Basically I'm using this regex
apidocs\/static\/[^\/]+[.](?:png|jpg|jpeg|gif|pdf)
so it will consider this valid
./apidocs/static/mypicture.jpg
also this will be valid
apidocs/static/mypicture.jpg
regex demo:
https://regex101.com/r/p1EA9m/1
But I find that these are also valid which is not my intention
./traapidocs/static/mypicture.jpg
./whatever/apidocs/static/mypicture.jpg
How can i configure the regex so that only these 2 patterns are valid (root folders are ./apidocs or apidocs)
./apidocs/static/mypicture.jpg
apidocs/static/mypicture.jpg
I'm using this in a python script btw, and found that putting a caret infront in a group does not work. Maybe there is a simpler way to form the regex.
Thank you in advance to anyone that is able to help!
You can use
^(?:\.\/)?apidocs\/static\/[^\/]+\.(?:png|jpe?g|gif|pdf)$
See the regex demo.
Details:
^ - start of string
(?:\.\/)? - an optional ./ string
apidocs\/static\/ - a apidocs/static/ string
[^\/]+ - one or more chars other than /
\. - a dot
(?:png|jpe?g|gif|pdf) - png, jpg or jpeg, gif, pdf
$ - end of string
This ones seems to work for your inputs:
^(\.\/)?apidocs\/static\/[^\/]+[.](?:png|jpg|jpeg|gif|pdf)$
Demo: https://regex101.com/r/VA0bKz/1
I'm trying to create a "Azure front door" compliant regex that matches /users/* URL pathname patterns. But not /users/*/ or /users/*/profile or /users/*/<anything at all>
I've tried without escaping as it looks like front door escapes for you.
^/users/([^/]+?)(?:/)?$
and this
^/users/[^/]+?$
But this doesn't work, I'm assuming because of the "?" which would count as a back reference? Any ideas on how to create a compliant regex, happy to try anything for anyone who doesn't have front door to test.
docs:
https://learn.microsoft.com/en-us/azure/frontdoor/rules-match-conditions?pivots=front-door-standard-premium&tabs=portal#regular-expressions
=== EDIT ===
Iv'e accepted the below answer, but I think the culprit is RegEx's matching a preceeding "/", every other operator in front door rules seems to want a preceeding "/" except RegEx, without this it seems to work as expected in some basic testing.
With this in mind I've accepted the below answer but I have a feeling (without trying) that some of the original formats will work.
You need to use
^/users/[^/\s]+$
Details
^ - start of string
/users/ - a fixed string
[^/\s]+ - one or more chars other than / and whitespace
$ - end of string.
I am currently learning regex and I am trying to filter all links (eg: http://www.link.com/folder/file.html) from a document with notepad++. Actually I want to delete everything else so that in the end only the http links are listed.
So far I tried this : http\:\/\/www\.[a-zA-Z0-9\.\/\-]+
This gives me all links which is find, but how do I delete the remaining stuff so that in the end I have a neat list of all links?
If I try to replace it with nothing followed by \1, obviously the link will be deleted, but I want the exact opposite to have everything else deleted.
So it should be something like:
- find a string of numbers, letters and special signs until "http"
- delete what you found
- and keep searching for more numbers, letters ans special signs after "html"
- and delete that again
Any ideas? Thanks so much.
In Notepad++, in the Replace menu (CTRL+H) you can do the following:
Find: .*?(http\:\/\/www\.[a-zA-Z0-9\.\/\-]+)
Replace: $1\n
Options: check the Regular expression and the . matches newline
This will return you with a list of all your links. There are two issues though:
The regex you provided for matching URLs is far from being generic enough to match any URL. If it is working in your case, that's fine, else check this question.
It will leave the text after the last matched URL intact. You have to delete it manually.
The answer made previously by #psxls was a great help for me when I have wanted to perform a similar process.
However, this regex rule was written six years ago now: accordingly, I had to adjust / complete / update it in order it can properly work with the some recent links, because:
a lot of URL are now using HTTPS instead of HTTP protocol
many websites less use www as main subdomain
some links adds punctuation mark (which have to be preserved)
I finally reshuffle the search rule to .*?(https?\:\/\/[a-zA-Z0-9[:punct:]]+) and it worked correctly with the file I had.
Unfortunately, this seemingly simple task is going to be almost impossible to do in notepad++. The regex you would have to construct would be...horrible. It might not even be possible, but if it is, it's not worth it. I pretty much guarantee that.
However, all is not lost. There are other tools more suitable to this problem.
Really what you want is a tool that can search through an input file and print out a list of regex matches. The UNIX utility "grep" will do just that. Don't be scared off because it's a UNIX utility: you can get it for Windows:
http://gnuwin32.sourceforge.net/packages/grep.htm
The grep command line you'll want to use is this:
grep -o 'http:\/\/www.[a-zA-Z0-9./-]\+\?' <filename(s)>
(Where <filename(s)> are the name(s) of the files you want to search for URLs in.)
You might want to shake up your regex a little bit, too. The problems I see with that regex are that it doesn't handle URLs without the 'www' subdomain, and it won't handle secure links (which start with https). Maybe that's what you want, but if not, I would modify it thusly:
grep -o 'https\?:\/\/[a-zA-Z0-9./-]\+\?' <filename(s)>
Here are some things to note about these expressions:
Inside a character group, there's no need to quote metacharacters except for [ and (sometimes) -. I say sometimes because if you put the dash at the end, as I have above, it's no longer interpreted as a range operator.
The grep utility's syntax, annoyingly, is different than most regex implementations in that most of the metacharacters we're familiar with (?, +, etc.) must be escaped to be used, not the other way around. Which is why you see backslashes before the ? and + characters above.
Lastly, the repetition metacharacter in this expression (+) is greedy by default, which could cause problems. I made it lazy by appending a ? to it. The way you have your URL match formulated, it probably wouldn't have caused problems, but if you change your match to, say [^ ] instead of [a-zA-Z0-9./-], you would see URLs on the same line getting combined together.
I did this a different way.
Find everything up to the first/next (https or http) (then everything that comes next) up to (html or htm), then output just the '(https or http)(everything next) then (html or htm)' with a line feed/ carriage return after each.
So:
Find: .*?(https:|http:)(.*?)(html|htm)
Replace with: \1\2\3\r\n
Saves looking for all possible (incl non-generic) url matches.
You will need to manually remove any text after the last matched URL.
Can also be used to create url links:
Find: .*?(https:|http:)(.*?)(html|htm)
Replace: \1\2\3\r\n
or image links (jpg/jpeg/gif):
Find: .*?(https:|http:)(.*?)(jpeg|jpg|gif)
Replace: <img src="\1\2\3">\r\n
I know my answer won't be RegEx related, but here is another efficient way to get lines containing URLs.
This won't remove text around links like Toto mentioned in comments.
At least if there is nice pattern to all links, like https://.
CTRL+F => change tab to Mark
Insert https://
Tick Mark to bookmark.
Mark All.
Find => Bookmarks => Delete all lines without bookmark.
I hope someone who lands here in search of same problem will find my way more user-friendly.
You can still use RegEx to mark lines :)
i have this code in php that transforms URL inside a text to active html links.
For example in a string
Hey check this cool link http://www.example.com
this transforms to:
Hey check this cool link http://www.example.com
As you can see it just adds the correct < a > html tag
The code is this:
$active_links_text = ereg_replace("[[:alpha:]]+://[^<>[:space:]]+[[:alnum:]/]","\\0", $original_text);
My question is, how to do this to work EXCEPT if the URL is a youtube url.
So i want this result: In a string
Wow have you checked http://www.youtube.com/watch?v=dQw4w9WgXcQ its even better than http://www.example.com !!!
i want to be transformed to
Wow have you checked http://www.youtube.com/watch?v=dQw4w9WgXcQ its even better than http://www.example.com
As you can see the < a > html tag was added to the example.com's URL but NOT at the youtube's URL.
How can i make this happen???
I hope i described my problem good enough, i hope its easy to implement this! Last note: i am using this code in php 5.2.14
Thank you guys!
[EDIT : Wow, I had gotten your question completely wrong! Below's a better attempt at helping you.]
I gave it a go in js here, here is the original regex : /(http:\/\/(?!www.youtube)[^<>\s]+)\b/g, since i'm not a php coder. The negative lookahead prevents a litteral www.youtube match (the lookahead content can be adapted if you need a more complex pattern).
There's nothing js-specific here to my knowledge, but I don't know the ereg regex syntax. with preg functions, you would just need not to escape the slashes, the word boundaries \b and negative lookahead (?!*pattern*) are the same. The /g flag is for a global replacement, that is, not stopping on the first match, I suppose you have a kind of replaceAll function in your toolbox.
Also, I'm not sure about the global flag in php, I guess you can just call a kind of replaceAll function.
You've made several mistakes about valid URI components. The scheme is defined as ALPHA *( ALPHA / DIGIT / "+" / "-" / "." ), not [[:alpha:]]+.
The part after the : of the scheme need not start with //, that's particular to http: and a few other file-oriented schemes. But the [[:alpha:]]+: start of your regex shows you weren't aiming to restrict yourself to http:. In that case, all printable ASCII characters are valid. I.e. everything from ! to ~, or [\x21-x7E]* as a regex.
To summarize: [[:alpha:]][A-Za-z0-9+-.]*:[\x21-x7E]*.
10 websites need to be cached. When caching: photos, css, js, etc are not displayed properly because the base domain isn't attached to the directory. I need a regex to add the base domain to the directory. examples below
base domain: http://www.example.com
the problem occurs when reading cached pages with img src="thumb/123.jpg" or src="/inc/123.js".
they would display correctly if it was img src="http://www.example.com/thumb/123.jpg" or src="http://www.example.com/inc/123.js".
regex something like: if (src=") isn't followed by the base domain then add the base domain
without knowing the language, you can use the (maybe most portable) substitute modifier:
s/^(src=")([^"]+")$/$1www\.example\.com\/$2/
This should do the following:
1. the string 'src="' (and capture it in variable $1)
2. one or more non-double-quote (") character followed by " (and capture it in variable $2)
3. Substitutes 'www.example.com/' in between the two capture groups.
Depending on the language, you can wrap this in a conditional that checks for the existence of the domain and substitutes if it isn't found.
to check for domain: /www\.example\.com/i should do.
EDIT: See comments:
For PHP, I would do this a bit differently. I would probably use simplexml. I don't think that will translate well, though, so here's a regex one...
$html = file_get_contents('/path/to/file.html');
$regex_match = '/(src="|href=")[^(?:www.example.com\/)]([^"]+")/gi';
$regex_substitute = '$1www.example.com/$2';
preg_replace($regex_match, $regex_substitute, $html);
Note: I haven't actually run this to debug it, it's just off the cuff. I would be concerned about 3 things. first, I am unsure how preg_replace will handle the / character. I don't think you're concerned with this, though, unless VB has a similar problem. Second, If there's a chance that line breaks would get in the way, I might change the regex. Third, I added the [^(?:www\.example\.com)] bit. This should change the match to any src or href that doesn't have www.example.com/ there, but this depends on the type of regex being used (POSIX/PCRE).
The rest of the changes should be fine (I added href=" and also made it case-insensitive (\i) and there's a requirement to make it global (\g) otherwise, it will just match once).
I hope that helps.
Matching regular expression:
(?:src|href)="(http://www\.example\.com/)?.+