I have many strings that all start with #and a pseudo and I want to change these pseudo via regex to the real name.
I haven't many pseudo (maybe 5 to 10) so I can go with a simple regex like:
$find = array('#alex', '#donald');
$replace = array('Alex A.', 'Donald B.' );
$result= preg_replace($find, $replace, $feed->itemTitle);
My problem is that I already have a pre_replace on these string, that removes the link. So far this is my regex:
<?php echo preg_replace('#(https?://([-\w\.]+[-\w])+(:\d+)?(/([\w/_\.#-]*(\?\S+)?[^\.\s])?)?).....#',' ',$feed->itemTitle); ?>
I can't come up with a solution that will mix the two regex. (regex is something I am not confortable with).
To have already a preg_replace for the links isn't a problem, don't bother about that.
If you want you can build a giant pattern with capture groups to be used with preg_replace_callback that allows the callback function to choose which is the replacement string to return according to the capture group number. However, this isn't the good way.
Since, you want to replace fixed strings (#alex, #donald are fixed strings) the best and fastest way is to use strtr (even if you parse the string a second time):
$trans = array('#alex' => 'Alex A.',
'#donald' => 'Donald B.');
$result = strtr($feed->itemTitle, $trans);
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 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.
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'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.