I am having a hard time to understand the preg_replace().
This is what I have got so far.
preg_replace($patter, $replacement, $string)
But, It confuses me when there is capturing groups in replacement.
But still I get some of it. Like below
preg_replace('/(\w+)/', 'hello', 'say hello!')
I know it will result in 'hello hello!'.
I can do something more with capturing group.
preg_replace('/(\w+)/','\1 Hello', 'Say World!',1)
It will result in 'Say Hello World'.
This is what I don't get.
<?php
$string = 'April 15, 2003';
$pattern = '/(\w+) (\d+), (\d+)/i';
$replacement = '${1}1,$3';
echo preg_replace($pattern, $replacement, $string);
?>
Output:
April1,2003
What ${1}1,$3 means?
What's the difference between using $1 \1 and <\1>
What does this ${1}1 mean?
Great question!
My guess is that <\1> and $1 ${1} are pretty much the same, addressing capturing groups, except one of which has an extra <>.
I think we are using ${1}1 since if we don't, we would have had $11, which does not simply exist, since we don't have that 11 capturing groups, we only have 3 of those, similar story for \11.
${1}1,$3 means the first capturing group, ${1}, and the number 1, and third capturing group, $3.
When we use
\1or\2inside a regular expression, not as a substitute, it would reference back the prior capturing groups$1and$2, which is the same as\1and2in the substitute section, which I also think that might create reasonable and valid confusions.