say I have this
searchpage-20/11111111111?node=15
how would I setup a regex to replace the entire string without worrying about the
11111111111
in the middle as long as the rest matches.
I tried
searchallmp3-20/(.+?)\?node=
You should post the code, to see how you are calling the regex and you need also to describe, what is not working.
I assume you are getting some error message because you are using / without escaping it in your regex and your delimiter is also the /
Two possibilities:
Escape the /
/searchallmp3-20\/(.+?)\?node=/
Use another delimiter
~searchallmp3-20/(.+?)\?node=~
See Delimiters on php.net
I still don't get the replace part of your question. If you want to remove the digits before the ?, you should capture the other parts of the string
~(searchallmp3-20/).+?(\?node=~)
and replace with
`$1$2`
this will result in
searchpage-20/?node=15
Is it this what you want?
How about
<?php
$search = 'searchpage-20/11111111111?node=15';
$reg = '#(searchpage-\d+/)\d+(\?node=\d+)#';
echo preg_replace($reg, '${1}blah${2}', $search);
OUTPUT
searchpage-20/blah?node=15
You should be able to use this:
/searchpage-(\d+)\/(\d+)\?node=(\d+)/
Example usage:
preg_replace('/searchpage-(\d+)\/(\d+)\?node=(\d+)/', '', 'searchpage-20/11111111111?node=15');
Related
I have the following content in a string (query from the DB), example:
$fulltext = "Thank you so much, {gallery}art-by-stephen{/gallery}. As you know I fell in love with it from the moment I saw it and I couldn’t wait to have it in my home!"
So I only want to extract what it is between the {gallery} tags, I'm doing the following but it does not work:
$regexPatternGallery= '{gallery}([^"]*){/gallery}';
preg_match($regexPatternGallery, $fulltext, $matchesGallery);
if (!empty($matchesGallery[1])) {
echo ('<p>matchesGallery: '.$matchesGallery[1].'</p>');
}
Any suggestions?
Try this:
$regexPatternGallery= '/\{gallery\}(.*)\{\/gallery\}/';
You need to escape / and { with a \ before it. And you where missing start and end / of the pattern.
http://www.phpliveregex.com/p/fn1
Similar to Andreas answer but differ in ([^"]*?)
$regexPatternGallery= '/\{gallery\}([^"]*?)\{\/gallery\}/';
Don't forget to put / at the beginning and the end of the Regex string. That's a must in PHP, different from other programming languages.
{,},/ are characters that can be confused as a Regex logic, so you have to escape it using \ like \{.
Use ? to make the string to non-greedy, thus saves memory. It avoids error when facing this kind of string "blabla {galery}you should only get this{/gallery} but you also got this instead.{/gallery} Rarely happens but be careful anyway".
Try this RegEx:
\{gallery\}(.*?)\{\/gallery\}
The problem with your RegEx was that you did not escape the / in the closing {gallery}. You also need to escape { and }.
You should use .*? for a lazy match, otherwise if there are 2 tags in one string, it will combine them. I.e. {gallery}by-joe{/gallery} and {gallery}by-tim{/gallery} would end up as:
by-joe{/gallery} and {gallery}by-tim
However, using a lazy match, you would get 2 results:
by-joe
by-tim
Live Demo on Regex101
I'm trying to retrieve the followed by count on my instagram page. I can't seem to get the Regex right and would very much appreciate some help.
Here's what I'm looking for:
y":{"count":
That's the beginning of the string, and I want the 4 numbers after that.
$string = preg_replace("{y"\"count":([0-9]+)\}","",$code);
Someone suggested this ^ but I can't get the formatting right...
You haven't posted your strings so it is a guess to what the regex should be... so I'll answer on why your codes fail.
preg_replace('"followed_by":{"count":\d')
This is very far from the correct preg_replace usage. You need to give it the replacement string and the string to search on. See http://php.net/manual/en/function.preg-replace.php
Your second usage:
$string = preg_replace(/^y":{"count[0-9]/","",$code);
Is closer but preg_replace is global so this is searching your whole file (or it would if not for the anchor) and will replace the found value with nothing. What your really want (I think) is to use preg_match.
$string = preg_match('/y":\{"count(\d{4})/"', $code, $match);
$counted = $match[1];
This presumes your regex was kind of correct already.
Per your update:
Demo: https://regex101.com/r/aR2iU2/1
$code = 'y":{"count:1234';
$string = preg_match('/y":\{"count:(\d{4})/', $code, $match);
$counted = $match[1];
echo $counted;
PHP Demo: https://eval.in/489436
I removed the ^ which requires the regex starts at the start of your string, escaped the { and made the\d be 4 characters long. The () is a capture group and stores whatever is found inside of it, in this case the 4 numbers.
Also if this isn't just for learning you should be prepared for this to stop working at some point as the service provider may change the format. The API is a safer route to go.
This regexp should capture value you're looking for in the first group:
\{"count":([0-9]+)\}
Use it with preg_match_all function to easily capture what you want into array (you're using preg_replace which isn't for retrieving data but for... well replacing it).
Your regexp isn't working because you didn't escaped curly brackets. And also you didn't put count quantifier (plus sign in my example) so it would only capture first digit anyway.
I'm having this issue with a regular expression in PHP that I can't seem to crack. I've spent hours searching to find out how to get it to work, but nothing seems to have the desired effect.
I have a file that contains lines similar to the one below:
Total','"127','004"','"118','116"','"129','754"','"126','184"','"129','778"','"128','341"','"127','477"','0','0','0','0','0','0
These lines are inserted into INSERT queries. The problem is that values like "127','004" are actually supposed to be 127,004, or without any formatting: 127004. The latter is the actual value I need to insert into the database table, so I figured I'd use preg_replace() to detect values like "127','004" and replace them with 127004.
I played around with a Regular Expression designer and found that I could use the following to get my desired results:
Regular Expression
"(\d+)','(\d{3})"
Replace Expression
$1$2
The line on the top of this post would end up like this: (which is what I am after)
Total','127004','118116','129754','126184','129778','128341','127477','0','0','0','0','0','0
This, however, does not work in PHP. Nothing is being replaced at all.
The code I am using is:
$line = preg_replace("\"(\d+)','(\d{3})\"", '$1$2', $line);
Any help would be greatly appreciated!
There are no delimiters in your regex. Delimiters are required in order for PHP to know what is the pattern to match and what is a pattern modifier (e.g. i - case-insensitive, U - ungreedy, ...). Use a character that doesn't occur in your pattern, typically you'll see a slash '/' used.
Try this:
$line = preg_replace("/\"(\d+)','(\d{3})\"/", '$1$2', $line);
You forgot to wrap your regular expression in front-slashes. Try this instead:
"/\"(\d+)','(\d{3})\"/"
use preg_replace("#\"(\d+)','(\d+)\"#", '$1$2', $s); instead of yours
In PHP, I need to be able to figure out if a string contains a URL. If there is a URL, I need to isolate it as another separate string.
For example: "SESAC showin the Love! http://twitpic.com/1uk7fi"
I need to be able to isolate the URL in that string into a new string. At the same time the URL needs to be kept intact in the original string. Follow?
I know this is probably really simple but it's killing me.
Something like
preg_match('/[a-zA-Z]+:\/\/[0-9a-zA-Z;.\/?:#=_#&%~,+$]+/', $string, $matches);
$matches[0] will hold the result.
(Note: this regex is certainly not RFC compliant; it may fetch malformed (per the spec) URLs. See http://www.faqs.org/rfcs/rfc1738.html).
this doesn't account for dashes -. needed to add -
preg_match('/[a-zA-Z]+:\/\/[0-9a-zA-Z;.\/\-?:#=_#&%~,+$]+/', $_POST['string'], $matches);
URLs can't contain spaces, so...
\b(?:https?|ftp)://\S+
Should match any URL-like thing in a string.
The above is the pure regex. PHP preg_* and string escaping rules apply before you can use it.
$test = "SESAC showin the Love! http://twitpic.com/1uk7fi";
$myURL= strstr ($test, "http");
echo $myURL; // prints http://twitpic.com/1uk7fi
I'm trying to extract only certain elements of a string using regular expressions and I want to end up with only the captured groups.
For example, I'd like to run something like (is|a) on a string like "This is a test" and be able to return only "is is a". The only way I can partially do it now is if I find the entire beginning and end of the string but don't capture it:
.*?(is|a).*? replaced with $1
However, when I do this, only the characters preceding the final found/captured group are eliminated--everything after the last found group remains.
is is a test.
How can I isolate and replace only the captured strings (so that I end up with "is is a"), in both PHP and Perl?
Thanks!
Edit:
I see now that it's better to use m// rather than s///, but how can I apply that to PHP's preg_match? In my real regex I have several captured group, resulting in $1, $2, $3 etc -- preg_match only deals with one captured group, right?
If all you want are the matches, the there is no need for the s/// operator. You should use m//. You might want to expand on your explanation a little if the example below does not meet your needs:
#!/usr/bin/perl
use strict;
use warnings;
my $text = 'This is a test';
my #matches = ( $text =~ /(is|a)/g );
print "#matches\n";
__END__
C:\Temp> t.pl
is is a
EDIT: For PHP, you should use preg_match_all and specify an array to hold the match results as shown in the documentation.
You can't replace only captures. s/// always replaces everything included in the match. You need to either capture the additional items and include them in the replacement or use assertions to require things that aren't included in the match.
That said, I don't think that's what you're really asking. Is Sinan's answer what you're after?
You put everything into captures and then replaces only the ones you want.
(.*?)(is|a)(.*?)