I need help with write regex pattern for replace my variable.
Input data
strings for parse:
/hello/world/1111/2/3/4
/hello/world/2222/2/3/4
/hello/world/3333/2/3/4
/hello/world/3333
/hello/world/1111
replace value = some
Out
/hello/world/some/2/3/4
/hello/world/some/2/3/4
/hello/world/some/2/3/4
/hello/world/some
/hello/world/some
$out = preg_replace("#^/(\w+)/(\w+)/\d{4}(/.*)?$#", "/$1/$2/some$3", $in);
This will do the correct replacement for you, here a short explanation:
^ start of line
(\w+) >1 literals
\d{4} exactly 4 digits
(/.*)? anything starting with / or nothing at all
$ end of line
If they all have four digits in the part you want to replace, just do this:
$input = preg_replace("/\d{4}/", "some", $input);
If the input number is always four digits, then
$input = preg_replace("~/\d{4}(/.*)?$~", "some$1", $input);
Related
Given an address stored as a single string with newlines delimiting its components like:
1 Street\nCity\nST\n12345
The goal would be to replace all newline characters except the first one with spaces in order to present it like:
1 Street
City ST 12345
I have tried methods like:
[$street, $rest] = explode("\n", $input, 2);
$output = "$street\n" . preg_replace('/\n+/', ' ', $rest);
I have been trying to achieve the same result using a one liner with a regular expression, but could not figure out how.
I would suggest not solving this with complicated regex but keeping it simple like below. You can split the string with a \n, pop out the first split and implode the rest with a space.
<?php
$input = explode("\n","1 Street\nCity\nST\n12345");
$input = array_shift($input) . PHP_EOL . implode(" ", $input);
echo $input;
Online Demo
You could use a regex trick here by reversing the string, and then replacing every occurrence of \n provided that we can lookahead and find at least one other \n:
$input = "1 Street\nCity\nST\n12345";
$output = strrev(preg_replace("/\n(?=.*\n)/", " ", strrev($input)));
echo $output;
This prints:
1 Street
City ST 12345
You can use a lookbehind pattern to ensure that the matching line is preceded with a newline character. Capture the line but not the trailing newline character and replace it with the same line but with a trailing space:
preg_replace('/(?<=\n)(.*)\n/', '$1 ', $input)
Demo: https://onlinephp.io/c/5bd6d
You can use an alternation pattern that matches either the first two lines or a newline character, capture the first two lines without the trailing newline character, and replace the match with what's captured and a space:
preg_replace('/(^.*\n.*)\n|\n/', '$1 ', $input)
Demo: https://onlinephp.io/c/2fb2f
I leave you another method, the regex is correct as long as the conditions are met, in this way it always works
$string=explode("/","1 Street\nCity\nST\n12345");
$string[0]."<br>";
$string[1]." ".$string[2]." ".$string[3]
Maybe it can not be solved this issue as I want, but maybe you can help me guys.
I have a lot of malformed words in the name of my products.
Some of them has leading ( and trailing ) or maybe one of these, it is same for / and " signs.
What I do is that I am explode the name of the product by spaces, and examines these words.
So I want to replace them to nothing. But, a hard drive could be 40GB ATA 3.5" hard drive. I need to process all the word, but I can not use the same method for 3.5" as for () or // because this 3.5" is valid.
So I only need to replace the quotes, when it is at the start of the string AND at end of the string.
$cases = [
'(testone)',
'(testtwo',
'testthree)',
'/otherone/',
'/othertwo',
'otherthree/',
'"anotherone',
'anothertwo"',
'"anotherthree"',
];
$patterns = [
'/^\(/',
'/\)$/',
'~^/~',
'~/$~',
//Here is what I can not imagine, how to add the rule for `"`
];
$result = preg_replace($patterns, '', $cases);
This is works well, but can it be done in one regex_replace()? If yes, somebody can help me out the pattern(s) for the quotes?
Result for quotes should be this:
'"anotherone', //no quote at end leave the leading
'anothertwo"', //no quote at start leave the trailin
'anotherthree', //there are quotes on start and end so remove them.
You may use another approach: rather than define an array of patterns, use one single alternation based regex:
preg_replace('~^[(/]|[/)]$|^"(.*)"$~s', '$1', $s)
See the regex demo
Details:
^[(/] - a literal ( or / at the start of the string
| - or
[/)]$ - a literal ) or / at the end of the string
| - or
^"(.*)"$ - a " at the start of the string, then any 0+ characters (due to /s option, the . matches a linebreak sequence, too) that are captured into Group 1, and " at the end of the string.
The replacement pattern is $1 that is empty when the first 2 alternatives are matched, and contains Group 1 value if the 3rd alternative is matched.
Note: In case you need to replace until no match is found, use a preg_match with preg_replace together (see demo):
$s = '"/some text/"';
$re = '~^[(/]|[/)]$|^"(.*)"$~s';
$tmp = '';
while (preg_match($re, $s) && $tmp != $s) {
$tmp = $s;
$s = preg_replace($re, '$1', $s);
}
echo $s;
This works
preg_replace([[/(]?(.+)[/)]?|/\"(.+)\"/], '$1', $string)
Hi I want to remove a characters using preg_replace in php so i have this code here which i want to remove the whole characters, letters and numbers except the last digit(s) which has dash(-) symbol followed by a digits so here's my code.
echo preg_replace('/(.+)(?=-[0-9])|(.+)/','','asdf1245-10');
I expect the result will be
-10
the problem is above is not working very well. I checked the pattern using http://www.regextester.com/ it seems like it works, but on the other side http://www.phpliveregex.com/ doesn't work at all. I don't know why but anyone who can help to to figure it out?
Thanks a lot
Here is a way to go:
echo preg_replace('/^.+?(-[0-9]+)?$/','$1','asdf1245-10');
Output:
-10
and
echo preg_replace('/^.+?(-[0-9]+)?$/','$1','asdf124510');
Output:
<nothing>
My first thinking is to use explode in this case.. make it simple like the following code.
$string = 'asdf1245-10';
$array = explode('-', $string);
end($array);
$key = key($array);
$result = '-' . $array[$key];
$result => '-10';
An other way:
$result = preg_match('~\A.*\K-\d+\z~', $str, $m) ? $m[0] : '';
pattern details:
\A # start of the string anchor
.* # zero or more characters
\K # discard all on the left from match result
-\d+ # the dash and the digits
\z # end of the string anchor
echo preg_replace('/(\w+)(-\w+)/','$2', 'asdf1245-10');
I Have one string like below.
$string = "2346#$ABSC$%###234567";
Now I want last character from this string that is not numeric or special character, It should be only A-a to Z-z.
Means, I need only "C" from this string.
I have try this formula:
substr($string, -1);
You should look into regular expressions using something like preg_match()
An expression like this would match:
/([a-z])[^a-z]*$/i
It means:
([a-z]) Capture an a-z character (the i at the end makes it case-insensitive)
[^a-z]*$ followed by 0 or more non a-z characters until the end of the string
See an example.
This should work for you:
(Here I just replace everything expect a-zA-Z with an empty string. After this I just access the last character)
<?php
$string = '2346#$ABSC$%###234567';
$string = preg_replace("/[^a-zA-Z]/", "", $string);
echo $string[strlen($string)-1];
?>
output:
C
The proper regex is: ([a-z])[^a-z]*$
I have a string like this:
[numbers]firstword[numbers]mytargetstring
I would like to know how is it possible to extract "targetstring" taking account the following :
a.) Numbers are numerical digits for example, my complete string with numbers:
12firstword21mytargetstring
b.) Numbers can be any digits, for example above are two digits each, but it can be any number of digits like this:
123firstword21567mytargetstring
Regardless of the number of digits, I am only interested in extracting "mytargetstring".
By the way "firstword" is fixed and will not change with any combination.
I am not very good in Regex so I appreciate someone with strong background can suggest how to do this using PHP. Thank you so much.
This will do it (or should do)
$input = '12firstword21mytargetstring';
preg_match('/\d+\w+\d+(\w+)$/', $input, $matches);
echo $matches[1]; // mytargetstring
It breaks down as
\d+\w+\d+(\w+)$
\d+ - One or more numbers
\w+ - followed by 1 or more word characters
\d+ - followed by 1 or more numbers
(\w+)$ - followed by 1 or more word characters that end the string. The brackets mark this as a group you want to extract
preg_match("/[0-9]+[a-z]+[0-9]+([a-z]+)/i", $your_string, $matches);
print_r($matches);
You can do it with preg_match and pattern syntax.
$string ='2firstword21mytargetstring';
if (preg_match ('/\d(\D*)$/', $string, $match)){
// ^ -- end of string
// ^ -- 0 or more
// ^^ -- any non digit character
// ^^ -- any digit character
var_dump($match[1]);}
Try it like,
print_r(preg_split('/\d+/i', "12firstword21mytargetstring"));
echo '<br/>';
echo 'Final string is: '.end(preg_split('/\d+/i', "12firstword21mytargetstring"));
Tested on http://writecodeonline.com/php/
You don't need regex for that:
for ($i=strlen($string)-1; $i; $i--) {
if (is_numeric($string[$i])) break;
}
$extracted_string = substr($string, $i+1);
Above it's probably the faster implementation you can get, certainly faster than using regex, which you don't need for this simple case.
See the working demo
your simple solution is here :-
$keywords = preg_split("/[\d,]+/", "hypertext123language2434programming");
echo($keywords[2]);