I have found out a code below
<?php $string = 'April 15, 2003';
$pattern = '/(\w+) (\d+), (\d+)/i';
$replacement = '${1}1,$3';
I just want to know why this 1 after ${1} is declaired and why the value 3 is assigned before the end of the string symbol ?
分享 If you execute the code as it is, you get:
April1,2003
This is because you are delimiting the variable $1 to separate it from what would be $11. Here's more on that.
If you change it, you get:
,2003
...because backreference $11 doesn't exist.
Why is $3 declared before the ending string symbol? Because it's part of the preg_replace backreference. If you move it outside of the replacement string, you'll see it crash and burn.
Footnote: why you're replacing this with <variable>1,<variable> seems really odd. That 1 is static - in that it never changes. I'd be more inclined to think you'd want a replacement like ${1}${2},$3 which would return April15,2003.
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