I have a feeling that I might be missing something very basic. Anyways heres the scenario:
I'm using preg_replace to convert ===inputA===inputB=== to inputA
This is what I'm using
$new = preg_replace('/===(.*?)===(.*?)===/', '$1', $old);
Its working fine alright, but I also need to further restrict inputB so its like this
preg_replace('/[^\w]/', '', every Link or inputB);
So basically, in the first code, where you see $2 over there I need to perform operations on that $2 so that it only contains \w as you can see in the second code. So the final result should be like this:
Convert ===The link===link's page=== to The link
I have no idea how to do this, what should I do?
Although there already is an accepted answer: this is what the /e modifier or preg_replace_callback() are for:
echo preg_replace(
'/===(.*?)===(.*?)===/e',
'"$1"',
'===inputA===in^^putB===');
//Output: inputA
Or:
function _my_url_func($vars){
return ''.$vars[2].'';
}
echo preg_replace_callback(
'/===(.*?)===(.*?)===/',
'_my_url_func',
'===inputA===inputB===');
//Output: inputB
Try preg_match on the first one to get the 2 matches into variables, and then use preg_replace() on the one you want further checks on?
Why don't you do extract the matches from the first regex (preg_match) and treat thoses results and then put them back in a HTML form ?
Related
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 a regex-noobie, so sorry for this "simple" question:
I've got an URL like following:
http://stellenanzeige.monster.de/COST-ENGINEER-AUTOMOTIVE-m-w-Job-Mainz-Rheinland-Pfalz-Deutschland-146370543.aspx
what I'm going to archieve is getting the number-sequence (aka Job-ID) right before the ".aspx" with preg_replace.
I've already figured out that the regex for finding it could be
(?!.*-).*(?=\.)
Now preg_replace needs the opposite of that regular expression. How can I archieve that? Also worth mentioning:
The URL can have multiple numbers in it. I only need the sequence right before ".aspx". Also, there could be some php attributes behind the ".aspx" like "&mobile=true"
Thank you for your answers!
You can use:
$re = '/[^-.]+(?=\.aspx)/i';
preg_match($re, $input, $matches);
//=> 146370543
This will match text not a hyphen and not a dot and that is followed by .aspx using a lookahead (?=\.aspx).
RegEx Demo
You can just use preg_match (you don't need preg_replace, as you don't want to change the original string) and capture the number before the .aspx, which is always at the end, so the simplest way, I could think of is:
<?php
$string = "http://stellenanzeige.monster.de/COST-ENGINEER-AUTOMOTIVE-m-w-Job-Mainz-Rheinland-Pfalz-Deutschland-146370543.aspx";
$regex = '/([0-9]+)\.aspx$/';
preg_match($regex, $string, $results);
print $results[1];
?>
A short explanation:
$result contains an array of results; as the whole string, that is searched for is the complete regex, the first element contains this match, so it would be 146370543.aspx in this example. The second element contains the group captured by using the parentheeses around [0-9]+.
You can get the opposite by using this regex:
(\D*)\d+(.*)
Working demo
MATCH 1
1. [0-100] `http://stellenanzeige.monster.de/COST-ENGINEER-AUTOMOTIVE-m-w-Job-Mainz-Rheinland-Pfalz-Deutschland-`
2. [109-114] `.aspx`
Even if you just want the number for that url you can use this regex:
(\d+)
so there's a string,
<?php
$string = <<<STR
/\!##$%^&*()?.,djasijdiwqpk,=-c./zcxzo123154897kp02ldz.,world90iops02&&&8ks
STR;
I want to replace everything to NULL, except word "world" and number 1 and 3,
I just want to get "world13" or "world31" from that string USING regular expressions
I have already implemented basic solution,
via strpos() and substr() and this is works as excepted. But I need to do this via RegExp
The question is:
Is it possible to extract that word using RegEx?
~(world(?:(31|13))~i. The 'i' makes the regex case insensitive. The ?: is there so it doesn't put it in the matches array in a separate result. Wouldn't say it's very complex, by the way :) If you want every 1 and 3 in there, you can use ~(world|1|3)~i.
Is it possible to extract that word using RegEx?
Yes. You can use this regular expression:
(world)
I know, that, But I can't extract world13 or world31
Ah, I understand! You can use:
$string = preg_replace('/.*/s', 'world13', $string);
A simple solution is to find things you need and then join them to a string.
preg_match_all('/world|[13]/', $string, $matches);
$ret = join($matches[0]);
I have a string $test='23487°';
How can I remove all the instances of the little circle that appears in the string using preg replace?
What to I enter for the regex to remove it?
EDIT - as Pekka says, str_replace is better I am now using that. But the little circle is still not recognized by PHP...
You don't need regex, just str_replace:
$test = str_replace('°', '', $test);
The first parameter is the search term – the bit that will be found. The second parameter is the replacement string – the text that will be inserted instead. A blank string means "replace it with nothing", i.e. "remove it". The third parameter is the string on which to operate.
try with:
$test = preg_replace('/[^(\x20-\x7F)]*/','', $test);
this will replace all your non ascii characters from your string.
If you want to use preg_replace, you can do it like this:
$test = preg_replace('[°]', '', $test);
Also, for reference, here is a great site to test your regex:
http://www.solmetra.com/scripts/regex/index.php
i'v got such string <>1 <>2 <>3
i want remove all '<>' and symbols after '<>' i want replace with such expression like www.test.com/1.jpg, www.test.com/2.jpg, www.test.com/3.jpg
is it possible to do with regex? i only know to find '/<>.?/'
preg_replace('/<>(\d+)/g', 'www.test.com/bla/$1.jpg', $input);
(assuming your replaced elements are just numbers. If they are more general, you'll need to replace '\d+' by something else).
str_replace('<>', 'www.test.com/', $input);
// pseudo code
pre_replace_all('~<>([0-9]+)~', 'www.test.com/$1.jpg', $input);
$string = '<>1 <>2 <>3';
$temp = explode(' ',preg_replace('/<>(\d)/','www.test.com/\1.jpg',$string));
$newString = implode(', ',$temp);
echo $newString;
Based on your example, I don’t think you need regex at all.
$str = '<>1 <>2 <>3';
print_r(str_replace('<>', 'www.test.com/', $str));
Regex's allow you to manipulate a string in any fashion you desire, to modify the string in the fashion you desire you would use the following regex:
<>(\d)
and you would use regex back referencing to keep the values you have captured in your grouping brackets, in this case a single digit. The back reference is typically signified by the $ symbol and then the number of the group you are referencing. As follows:
www.test.com/$1
this would be used in a regex replace scenario which would be implemented in different ways depending on the language you are implementing your regex replace method in.