So basically I'm trying to select all content that is in /thank-you/hello/, so this can be /thank-you/hello/x/, /thank-you/hello/y/, /thank-you/hello/z/, etc.
This is what I'm using right now:
preg_match ('#^/thank-you/hello/#', $_SERVER['REQUEST_URI']
This block of code only works for stuff that is in /thank-you/hello/.
How should I change this snippet to include all the other folders that are after /hello/?
I suggest you read more about regex
I also recommend regex101 to test and study the site
In the desired pattern you can replace the desired word from .*?
.: Matches any character other than newline (or including line terminators with the /s flag)
a*: Matches zero or more consecutive a characters.
a?: Matches an a character or nothing.
They may seem a little incomplete without their examples
I suggest you see their examples on regex101
example:
preg_match('#^/thank-you/hello/.*?/#', $_SERVER['REQUEST_URI']);
It may not be exactly what you want
Or something may increase or decrease later and you may want to make a change
I think everyone should learn regex so that they can implement what they want according to their own desires.
I do not think it is a good idea to use patterns that you do not know what they mean
Related
Input:
http://foo/bar/baz/../../qux/
Desired Output:
http://foo/qux/
This can be achieved using regular expression (unless someone can suggest a more efficient alternative).
If it was a forward look-up, it would be as simple as:
/\.\.\/[^\/]+/
Though I am not familiar with with how to make a backward look up for the first "/" (ie. not doing /[a-z0-9-_]+\/\.\./).
One of the solutions I thought of is to use strrev then apply forward look up regex (first example) and then do strrev. Though I am sure there is a more efficient way.
Not the clearest question I've ever seen, but if I understand what you're asking, I think you only need to switch around what you have like this:
/[^\/]+/\.\./
...then replace that with a /
Do that until no replacements are made and you should have what you want
EDIT
Your attempt seems to try to match a forward slash / and two dots \.\. followed by a slash / (or \/ - they should both match the same thing), then one or more non-slash characters[^/]+, terminated by a slash /. Flipping it around, you want to find a slash followed by one or more non-slash characters and a terminating slash, then two dots and a final slash.
You may be confused into thinking that the regex engine parses and consumes things as it goes (so you wouldn't want to consume a directory name that is not followed by the correct number of dots), but that's not how it typically works - a regex engine matches the entire expression before it replaces or returns anything. So, you can have two dots followed by a directory name, or a directory name followed by two dots - it doesn't make a difference to the engine.
If your attempt is using the slash-enclosed Perl-style syntax, then you would of course need to use \/ for any slashes you're trying to match such as the middle one, but I would also recommend matching and replacing the enclosing slashes in the url as well: I think the PHP would be something like
preg_replace('/\/[^\/]+\/\.\.\//', '/', $input)
(??)
Technically what do you want is replace segments of '/path1/path2/../../' by '/' what is needed to do that is match 'pathx/'^n'../'^n that is definetly NOT a regular expression (Context Free Lenguaje) ... but most of Regex libraries supports some non regular lenguajes and can (with a lot of effort) manage those kind of lenguajes.
An easy way to solve it is stay in Regular Expressions and cycle several times, replacing '/[^./]+/../' by ''
if you still to do it in a single step, Lookahead and grouping is needed, but it will be hard to write it, (I'm not so used on, but I will try)
EDIT:
I've found the solution in only 1 REGEX... but should use PCRE Regex
([^/.]+/(?1)?\.\./)
I've based my solution on the folowing link:
Match a^n b^n c^n (e.g. "aaabbbccc") using regular expressions (PCRE)
(note that dots are "forbidden" in the first section, you cannot have path.1/path.2/ if you whant to is quite more complex because you should admit them but forbid '../' as valid in the first section
this sub expression is for admiting the path names like 'path1/'
[^/.]+/
this sub expression is for admiting the double dots.
\.\./
you can test the regexp in
https://www.debuggex.com/
(remember to set it in PCRE mode)
Here is a working copy:
https://eval.in/52675
I'm building this regex with a positive look ahead in it. Basically it must select all text in the line up to last period that precedes a ":" and add a "|" to the end to delimit it. Some sample text below. I am testing this in gskinner and editpadpro which has full grep regex support apparently so if I could get the answers in that for I'd appreciate it.
The regex below works to a degree but I am unsure if it is correct. Also it falls down if the text contains brackets.
Finally I would like to add another ignore rule like the one that ignores but includes "Co." in the selection. This second ignore rule would ignore but include periods that have a single Capital letter before them. Sample text below too. Thanks for all the help.
^(?:[^|]+\|){3}(.*?)[^(?:Co)]\.(?=[^:]*?\:)
121| Ryan, T.N. |2001. |I like regex. But does it like me (2) 2: 615-631.
122| O' Toole, H.Y. |2004. |(Note on the regex). Pages 90-91 In: Ryan, A. & Toole, B.L. (Editors) Guide to the regex functionality in php. Timmy, Tommy& Stewie, Quohog. * Produced for Family Guy in Quohog.
I don't think I understand what you want to do. But this part [^(?:Co)] is definitely not correct.
With the square brackets you are creating a character class, because of the ^ it is a negated class. That means at this place you don't want to match one of those characters (?:Co), in other words it will match any other character than "?)(:Co".
Update:
I don't think its possible. How should I distinguish between L. Co. or something similar and the end of the sentence?
But I found another error in your regex. The last part (?=[^:]*?\:) should be (?=[^.]*?\:) if you want to match the last dot before the : with your expression it will match on the first dot.
See it here on Regexr
This seems to do what you want.
(.*\.)(?=[^:]*?:)
It quite simply matches all text up to the last full stop that occurs before the colon.
i have this regex code
/^(https?:\/\/+[\w\-]+\.[\w\-]+)/i
it works but there is a problem
you NEED http:// in the url for it to validate, and what i am making, the user will not want to add http:// to the url they want to just have example.com, if its possible i need it to work weather it has http:// or not
i don't know how to make my own regex, and ive searched but cannot find a one that does what i need, unless im just not looking in the right place. (Google :P)
Don't bother with regex. Use parse_url function.
You can just make it optional
/^((?:https?:\/\/+)?[\w\-]+\.[\w\-]+)/i
The (?:) around the part you don't want to have is a non capturing group, the ? afterwards makes it optional.
I'm not sure what the + after the second slash is good for, it says at least one of the preceding character. That means it allows also stuff like http://////////.
I hope you are aware, that this regex is far from matching valid URLs.
For example it will match stuff like
http://////////------------.-
or at least
http://N.O
^ after this position you can write what you want and it will match valid.
Here on Regexr you can see what your regex is matching.
See Purple Coder's answer for a probably better solution.
/^((https?:\/\/+)?[\w-]+.[\w-]+)/i
I'm using this :
// Validate that the string contains at least a dot .
var filterWebsite = /^([a-zA-Z0-9:_\.\-/])+\.([a-zA-Z0-9_\.\-/])+$/;
I have written the following Regex in PHP for use within preg_replace().
/\b\S*(.com|.net|.us|.biz|.org|.info|.xxx|.mx|.ca|.fr|.in|.cn|.hk|.ng|.pr|.ph|.tv|.ru|.ly|.de|.my|.ir)\S*\b/i
This regex removes all URLs from a string pretty effectively this far (though I am sure I can write a better one). I need to be able to add an exclusion though from a specific domain. So the pseudo code will look like this:
IF string contains: .com or .net or. biz etc... and does not contain: foo.com THEN execute condition.
Any idea on how to do this?
Just add a negative lookahead assertion:
/(?<=\s|^)(?!\S*foo\.com)\S*\.(com|net|us|biz|org|info|xxx|mx|ca|fr|in|cn|hk|ng|pr|ph|tv|ru|ly|de|my|ir)\S*\b/im
Also, remember that you need to escape the dot - and that you can move it outside the alternation since each of the alternatives starts with a dot.
Use preg_replace_callback instead.
Let your callback decide whether to replace.
It can give more flexibility if the requirements become too complicated for a simple regex.
I have a load of user-submitted content. It is HTML, and may contain URLs. Some of them will be <a>'s already (if the user is good) but sometimes users are lazy and just type www.something.com or at best http://www.something.com.
I can't find a decent regex to capture URLs but ignore ones that are immediately to the right of either a double quote or '>'. Anyone got one?
Jan Goyvaerts, creator of RegexBuddy, has written a response to Jeff Atwood's blog that addresses the issues Jeff had and provides a nice solution.
\b(?:(?:https?|ftp|file)://|www\.|ftp\.)[-A-Z0-9+&##/%=~_|$?!:,.]*[A-Z0-9+&##/%=~_|$]
In order to ignore matches that occur right next to a " or >, you could add (?<![">]) to the start of the regex, so you get
(?<![">])\b(?:(?:https?|ftp|file)://|www\.|ftp\.)[-A-Z0-9+&##/%=~_|$?!:,.]*[A-Z0-9+&##/%=~_|$]
This will match full addresses (http://...) and addresses that start with www. or ftp. - you're out of luck with addresses like ars.userfriendly.org...
This thread is old as the hills, but I came across it while working on my own problem: That is, convert any urls into links, but leave alone any that are already within anchor tags. After a while, this is what has popped out:
(?!(?!.*?<a)[^<]*<\/a>)(?:(?:https?|ftp|file)://|www\.|ftp\.)[-A-Z0-9+&#/%=~_|$?!:,.]*[A-Z0-9+&#/%=~_|$]
With the following input:
http://www.google.com
http://google.com
www.google.com
<p>http://www.google.com<p>
this is a normal sentence. let's hope it's ok.
www.google.com
This is the output of a preg_replace:
http://www.google.com
http://google.com
www.google.com
<p>http://www.google.com<p>
this is a normal sentence. let's hope it's ok.
www.google.com
Just wanted to contribute back to save somebody some time.
I made a slight modification to the Regex contained in the original answer:
(?<![.*">])\b(?:(?:https?|ftp|file)://|[a-z]\.)[-A-Z0-9+&#/%=~_|$?!:,.]*[A-Z0-9+&#/%=~_|$]
which allows for more subdomains, and also runs a more full check on tags. To apply this to PHP's preg replace, you can use:
$convertedText = preg_replace( '#(?<![.*">])\b(?:(?:https?|ftp|file)://|[a-z]\.)[-A-Z0-9+&#/%=~_|$?!:,.]*[A-Z0-9+&#/%=~_|$]#i', '\0', $originalText );
Note, I removed # from the regex, in order to use it as a delimiter for preg_replace. It's pretty rare that # would be used in a URL anyway.
Obviously, you can modify the replacement text, and remove target="_blank", or add rel="nofollow" etc.
Hope that helps.
To skip existing ones just use a look-behind - add (?<!href=") to the beginning of your regular expression, so it would look something like this:
/(?<!href=")http://\S*/
Obviously this isn't a complete solution for finding all types of URLs, but this should solve your problem of messing with existing ones.
if (preg_match('/\b(?<!=")(https?|ftp|file):\/\/[-A-Z0-9+&##\/%?=~_|!:,.;]*[A-Z0-9+&##\/%=~_|](?!.*".*>)(?!.*<\/a>)/i', $subject)) {
# Successful match
} else {
# Match attempt failed
}
Shameless plug: You can look here (regular expression replace a word by a link) for inspiration.
The question asked to replace some word with a certain link, unless there already was a link. So the problem you have is more or less the same thing.
All you need is a regex that matches a URL (in place of the word). The simplest assumption would be like this: An URL (optionally) starts with "http://", "ftp://" or "mailto:" and lasts as long as there are no white-space characters, line breaks, tag brackets or quotes).
Beware, long regex ahead. Apply case-insensitively.
(href\s*=\s*['"]?)?((?:http://|ftp://|mailto:)?[^.,<>"'\s\r\n\t]+(?:\.(?![.<>"'\s\r\n])[^.,!<>"'\s\r\n\t]+)+)
Be warned - this will also match URLs that are technically invalid, and it will recognize things.formatted.like.this as an URL. It depends on your data if it is too insensitive. I can fine-tune the regex if you have examples where it returns false positives.
The regex will produce two match groups. Group 2 will contain the matched thing, which is most likely an URL. Group 1 will either contain an empty string or an 'href="'. You can use it as an indicator that this match occurred inside a href parameter of an existing link and you don't have to do touch that one.
Once you confirm that this does the right thing for you most of the time (with user supplied data, you can never be sure), you can do the rest in two steps, as I proposed it in the other question:
Make a link around every URL there is (unless there is something in match group 1!) This will produce double nested <a> tags for things that have a link already.
Scan for incorrectly nested <a> tags, removing the innermost one