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]);
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+)
I'm trying to write a very simple markup language in PHP that contains tags like [x=123], and I need to be able to match that tag and extract only the value of x.
I'm assuming the answer involves regex but maybe I'm wrong.
So if we had a string:
$str = "F9F[x=]]^$^$[x=123]#3j3E]]#J";
And a regular expression to match:
/^\[x=.+\]$/
How would we get only the ".+" portion of the matching string into a variable?
You can use preg_match to search a string for a regular expression.
Check out the documentation here: http://www.php.net/manual/en/function.preg-match.php for more information on how to use it (as well as some examples). You might also want to take a look at preg_grep.
Following code should work for you:
$str = "F9F[x=]]^$^$[x=123]#3j3E]]#J";
if (preg_match('~\[x=(?<valX>\d+)\]~', $str, $match))
echo $match['valX'] . "\n";
OUTPUT:
123
How can I replace a string starting with 'a' and ending with 'z'?
basically I want to be able to do the same thing as str_replace but be indifferent to the values in between two strings in a 'haystack'.
Is there a built in function for this? If not, how would i go about efficiently making a function that accomplishes it?
That can be done with Regular Expression (RegEx for short).
Here is a simple example:
$string = 'coolAfrackZInLife';
$replacement = 'Stuff';
$result = preg_replace('/A.*Z/', $replacement, $string);
echo $result;
The above example will return coolStuffInLife
A little explanation on the givven RegEx /A.*Z/:
- The slashes indicate the beginning and end of the Regex;
- A and Z are the start and end characters between which you need to replace;
- . matches any single charecter
- * Zero or more of the given character (in our case - all of them)
- You can optionally want to use + instead of * which will match only if there is something in between
Take a look at Rubular.com for a simple way to test your RegExs. It also provides short RegEx reference
$string = "I really want to replace aFGHJKz with booo";
$new_string = preg_replace('/a[a-zA-z]+z/', 'boo', $string);
echo $new_string;
Be wary of the regex, are you wanting to find the first z or last z? Is it only letters that can be between? Alphanumeric? There are various scenarios you'd need to explain before I could expand on the regex.
use preg_replace so you can use regex patterns.
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.