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?
>Solution :
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
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