Here is the sample PHP code:
<?php
$str = '10,000.1 $100,000.1';
$pattern = '/(?!\$)\d+(,\d{3})*\.?\d*/';
$replacement_str = 'Without$sign';
echo preg_replace($pattern, $replacement_str, $str);?>
Target is to replace numbers only (i.e. "$100,000.1" should not be replaced). But the above code replaces both 10,000.1 and $100,000.1. How to achieve the exclusion?
This assertion is always true (?!\$)\d+ as you match a digit which can not be a $
As the . and the digits at the end of the pattern are optional, it could also match ending on a dot like for example 0,000.
Instead you can assert a whitespace boundary to the left, and optionally match a dot followed by 1 or more digits:
(?<!\S)\d+(?:,\d{3})*(?:\.\d+)?\b
Regex demo
Example:
$str = '10,000.1 $100,000.1';
$pattern = '/(?<!\S)\d+(?:,\d{3})*(?:\.\d+)?\b/';
$replacement_str = 'Without$sign';
echo preg_replace($pattern, $replacement_str, $str);
Output (If you remove the numbers, the text "Without$sign" is not correct)
Without$sign $100,000.1
Related
I have the following:
$pattern = "/^([\w_]{1})(.+)([\w_]{1}#)/u";
$replacement = "$1*$3***$4";
$email = "testa#weste.de";
echo "obfuscated: ".preg_replace($pattern, $replacement, $email).RT;
The result is: t*a#***weste.de
But I would like to have: t*#w***.de
How to grab the letter after the # and not before. And how does it work with the .de part?
For the replacement in the example data, you might use a match with \K to forget what is matched after the first character and keep it.
To keep the first character after the # sign, you can use a capture group and use that in the replacement.
^\w\K[^\s#]+#(\w)[^\s.#]+
^ Start of string
\w Match a single word char (That will also match _)
\K Forget what is matched so far
[^\s#]+ Match 1+ chars other than # or a whitespace char
# Match the # char
(\w) Capture group 1, match a word char (to keep)
[^\s.#]+ Match 1+ chars other than #, a whitespace char or dot
Regex demo | Php demo
In the replacement use a single capture group *#$1***
$email = "testa#weste.de";
$pattern = "/^\w\K[^\s#]+#(\w)[^\s.#]+/";
$replacement = "*#$1***";
echo preg_replace($pattern, $replacement, $email);
Output
t*#w***.de
You can make the pattern as specific as you would like. If there should for example be a dot followed by at least 2 chars a-z at the end of the string, and you don't want to stop matching at the first dot after the #
^\w\K[^\s#]+#(\w)[^\s#]+(?=\.[a-z]{2,}$)
Regex demo
I found this way to do it:
$email = 'someemail#domain.com'
[$firstPart, $lastPart] = explode('#', $email);
$maskedEmail = str_replace(substr($firstPart, 0, 7), str_repeat('*', 7), $email);
Uses PHP native functions and works just fine!
The code below works perfectly:
$string = '(test1)';
$new = preg_replace('/^\(+.+\)+$/','word',$string);
echo $new;
Output:
word
If the code is this:
$string = '(test1) (test2) (test3)';
How to generate output:
word word word?
Why my regex do not work ?
^ and $ are anchors which means match should start from start of string and expand upto end of string
. means match anything except newline, + means one or more, by default regex is greedy in nature so it tries to match as much as possible where as we want to match ( ) so we need to change the pattern a bit
You can use
\([^)]+\)
$string = '(test1) (test2) (test3)';
$new = preg_replace('/\([^)]+\)/','word',$string);
echo $new;
Regex Demo
Hi I'm new to php and I need a little help
I need to change the text that is between ** in php string and put it between html tag
$text = "this is an *example*";
But I really don't know how and i need help
personally I would use explode, you can then piece the sentence back together if the example appears in the middle of a sentence
<?php
$text = "this is an *example*";
$pieces = explode("*", $text);
echo $pieces[0];
?>
Edit:
Since you're looking for what basically amounts to custom BB Code use this
$text = "this is an *example*";
$find = '~[\*](.*?)[\*]~s';
$replace = '<span style="color: green">$1</span>';
echo preg_replace($find,$replace,$text);
You can add this to a function and have it parse any text that gets passed to it, you can also make the find and replace variables into arrays and add more codes to it
You really should use a DOM parser for things like this, but if you can guaratee it will always be the * character you can use some regex:
$text = "this is an *example*";
$regex = '/(?<=\*)(.*?)(?=\*)/';
$replacement = 'ostrich';
$new_text = preg_replace($regex, $replacement, $text);
echo $new_text;
Returns
this is an *ostrich*
Here is how the regex works:
Positive Lookbehind (?<=\*)
\* matches the character * literally (case sensitive)
1st Capturing Group (.*?)
.*? matches any character (except for line terminators)
*? Quantifier — Matches between zero and unlimited times, as few times as possible, expanding as needed (lazy)
Positive Lookahead (?=\*)
\* matches the character * literally (case sensitive)
This regex essentially starts and ends by looking at what is ahead of and behind the search character you specified and leaves those characters intact during the replacement with preg_replace().
This codes convert any url to clickable link:
$str = preg_replace('/(http[s]?:\/\/[^\s]*)/i', '$1', $str);
How to make it not convert when url starts with [ character? Like this:
[http://google.com
Use a negative lookbehind:
$str = preg_replace('/(?<!\[)(http[s]?:\/\/[^\s]*)/i', '$1', $str);
^^^^^^^
Then, the http... substring that is preceded with [ won't be matched.
You may enhance the pattern as
preg_replace('/(?<!\[)https?:\/\/\S*/i', '$0', $str);
that is: remove the ( and ) (the capturing group) and replace the backreferences from $1 with $0 in the replacement pattern, and mind that [^\s] = \S, but shorter. Also, [s]? = s?.
I have a regex which does all matches except one match.The PHP Code for the word match is:
$string = preg_replace("/\b".$wordToMatch."\b/","<span class='sp_err' style='background-color:yellow;'>".$wordToMatch."</span>",$string);
Here in the above regex when the $wordToMatch variable value becomes "-abc" and the $string value is "The word -abc should match and abc-abc should not match".With above regex it fails to catch "-abc".
I want enhancement in the above regex so that it can catch "-abc" in $string,but if it tries to match "-abc" in "abc-abc" of $string it should not.
In case your keywords can have non-word characters on both ends you can rely on lookarounds for a whole word match:
"/(?<!\\w)".$wordToMatch."(?!\\w)/"
Here, (?<!\w) will make sure there is no word character before the word to match, and (?!\w) negative lookahead will make sure there is no word character after the word to match. These are unambiguous subpatterns, while \b meaning depends on the context.
See regex demo showing that -abc is not matched in abc-abc and matches if it is not enclosed with word characters.
PHP demo:
$wordToMatch = "-abc";
$re = "/(?<!\\w)" . $wordToMatch . "(?!\\w)/";
$str = "abc-abc -abc";
$subst = "!$0!";
$result = preg_replace($re, $subst, $str);
echo $result; // => abc-abc !-abc!