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 make a replace in a string with a regex, and I really hope the community can help me.
I have this string :
031,02a,009,a,aaa,AZ,AZE,02B,975,135
And my goal is to remove the opposite of this regex
[09][0-9]{2}|[09][0-9][A-Za-z]
i.e.
a,aaa,AZ,AZE,135
(to see it in action : http://regexr.com?3795f )
My final goal is to preg_replace the first string to only get
031,02a,009,02B,975
(to see it in action : http://regexr.com?3795f )
I'm open to all solution, but I admit that I really like to make this work with a preg_replace if it's possible (It became something like a personnal challenge)
Thanks for all help !
As #Taemyr pointed out in comments, my previous solution (using a lookbehind assertion) was incorrect, as it would consume 3 characters at a time even while substrings weren't always 3 characters.
Let's use a lookahead assertion instead to get around this:
'/(^|,)(?![09][0-9]{2}|[09][0-9][A-Za-z])[^,]*/'
The above matches the beginning of the string or a comma, then checks that what follows does not match one of the two forms you've specified to keep, and given that this condition passes, matches as many non-comma characters as possible.
However, this is identical to #anubhava's solution, meaning it has the same weakness, in that it can leave a leading comma in some cases. See this Ideone demo.
ltriming the comma is the clean way to go there, but then again, if you were looking for the "clean way to go," you wouldn't be trying to use a single preg_replace to begin with, right? Your question is whether it's possible to do this without using any other PHP functions.
The anwer is yes. We can take
'/(^|,)foo/'
and distribute the alternation,
'/^foo|,foo/'
so that we can tack on the extra comma we wish to capture only in the first case, i.e.
'/^foo,|,foo/'
That's going to be one hairy expression when we substitute foo with our actual regex, isn't it. Thankfully, PHP supports recursive patterns, so that we can rewrite the above as
'/^(foo),|,(?1)/'
And there you have it. Substituting foo for what it is, we get
'/^((?![09][0-9]{2}|[09][0-9][A-Za-z])[^,]*),|,(?1)/'
which indeed works, as shown in this second Ideone demo.
Let's take some time here to simplify your expression, though. [0-9] is equivalent to \d, and you can use case-insensitive matching by adding /i, like so:
'/^((?![09]\d{2}|[09]\d[a-z])[^,]*),|,(?1)/i'
You might even compact the inner alternation:
'/^((?![09]\d(\d|[a-z]))[^,]*),|,(?1)/i'
Try it in more steps:
$newList = array();
foreach (explode(',', $list) as $element) {
if (!preg_match('/[09][0-9]{2}|[09][0-9][A-Za-z]/', $element) {
$newList[] = $element;
}
}
$list = implode(',', $newList);
You still have your regex, see! Personnal challenge completed.
Try matching what you want to keep and then joining it with commas:
preg_match_all('/[09][0-9]{2}|[09][0-9][A-Za-z]/', $input, $matches);
$result = implode(',', $matches);
The problem you'll be facing with preg_replace is the extra-commas you'll have to strip, cause you don't just want to remove aaa, you actually want to remove aaa, or ,aaa. Now what when you have things to remove both at the beginning and at the end of the string? You can't just say "I'll just strip the comma before", because that might lead to an extra comma at the beginning of the string, and vice-versa. So basically, unless you want to mess with lookaheads and/or lookbehinds, you'd better do this in two steps.
This should work for you:
$s = '031,02a,009,a,aaa,AZ,AZE,02B,975,135';
echo ltrim(preg_replace('/(^|,)(?![09][0-9]{2}|[09][0-9][A-Za-z])[^,]+/', '', $s), ',');
OUTPUT:
031,02a,009,02B,975
Try this:
preg_replace('/(^|,)[1-8a-z][^,]*/i', '', $string);
this will remove all substrings starting with the start of the string or a comma, followed by a non allowed first character, up to but excluding the following comma.
As per #GeoffreyBachelet suggestion, to remove residual commas, you should do:
trim(preg_replace('/(^|,)[1-8a-z][^,]*/i', '', $string), ',');
I need to parse a string and replace a specific format for tv show names that don't fit my normal format of my media player's queue.
Some examples
Show.Name.2x01.HDTV.x264 should be Show.Name.S02E01.HDTV.x264
Show.Name.10x05.HDTV.XviD should be Show.Name.S10E05.HDTV.XviD
After the show name, there may be 1 or 2 digits before the x, I want the output to always be an S with two digits so add a leading zero if needed. After the x it should always be an E with two digits.
I looked through the manual pages for the preg_replace, split and match functions but couldn't quite figure out what I should do here. I can match the part of the string I want with /\dx\d{2}/ so I was thinking first check if the string has that pattern, then try and figure out how to split the parts out of the match but I didn't get anywhere.
I work best with examples, so if you can point me in the right direction with one that would be great. My only test area right now is a PHP 4 install, so please no PHP 5 specific directions, once I understand whats happening I can probably update it later for PHP 5 if needed :)
A different approach as a solution using #sprintf using PHP4 and below.
$text = preg_replace('/([0-9]{1,2})x([0-9]{2})/ie',
'sprintf("S%02dE%02d", $1, $2)', $text);
Note: The use of the e modifier is depreciated as of PHP5.5, so use preg_replace_callback()
$text = preg_replace_callback('/([0-9]{1,2})x([0-9]{2})/',
function($m) {
return sprintf("S%02dE%02d", $m[1], $m[2]);
}, $text);
Output
Show.Name.S02E01.HDTV.x264
Show.Name.S10E05.HDTV.XviD
See working demo
preg_replace is the function you are looking function.
You have to write a regex pattern that picks correct place.
<?php
$replaced_data = preg_replace("~([0-9]{2})x([0-9]{2})~s", "S$1E$2", $data);
$replaced_data = preg_replace("~S([1-9]{1})E~s", "S0$1E", $replaced_data);
?>
Sorry I could not test it but it should work.
An other way using the preg_replace_callback() function:
$subject = <<<'LOD'
Show.Name.2x01.HDTV.x264 should be Show.Name.S02E01.HDTV.x264
Show.Name.10x05.HDTV.XviD should be Show.Name.S10E05.HDTV.XviD
LOD;
$pattern = '~([0-9]++)x([0-9]++)~i';
$callback = function ($match) {
return sprintf("S%02sE%02s", $match[1], $match[2]);
};
$result = preg_replace_callback($pattern, $callback, $subject);
print_r($result);
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.