Convert string with preg_replace in PHP - regex

Question

I have this string
$string = "some words and then #1.7 1.7 1_7 and 1-7";
and I would like that #1.7/1.7/1_7 and 1-7 to be replaced by S1E07.
Of course, instead of "1.7" is just an example, it could be "3.15" for example.
I managed to create the regular expression that would match the above 4 variants
/\#\d{1,2}\.\d{1,2}|\d{1,2}_\d{1,2}|\d{1,2}-\d{1,2}|\d{1,2}\.\d{1,2}/
but I cannot figure out how to use preg_replace (or something similar?) to actually replace the matches so they end up like S1E07

Answer 1

You need to use preg_replace_callback if you need to pad 0 if the number less than 10.
$string = "some words and then #1.7 1.7 1_7 and 1-7";
$string = preg_replace_callback('/#?(\d+)[._-](\d+)/', function($matches) {
return 'S'.$matches[1].'E'.($matches[2] < 10 ? '0'.$matches[2] : $matches[2]);
}, $string);

Answer 2

You could use this simple string replace:
preg_replace('/#?\b(\d{1,2})[-._](\d{1,2})\b/', 'S${1}E${2}', $string);
But it would not yield zero-padded numbers for the episode number:
// some words and then S1E7 S1E7 S1E7 and S1E7
You would have to use the evaluation modifier:
preg_replace('/#?\b(\d{1,2})[-._](\d{1,2})\b/e', '"S".str_pad($1, 2, "0", STR_PAD_LEFT)."E".str_pad($2, 2, "0", STR_PAD_LEFT)', $string);
...and use str_pad to add the zeroes.
// some words and then S01E07 S01E07 S01E07 and S01E07
If you don't want the season number to be padded you can just take out the first str_pad call.

Answer 3

I believe this will do what you want it to...
/\#?([0-9]+)[._-]([0-9]+)/
In other words...
\#? - can start with the #
([0-9]+) - capture at least one digit
[._-] - look for one ., _ or -
([0-9]+) - capture at least one digit
And then you can use this to replace...
S$1E$2
Which will put out S then the first captured group, then E then the second captured group

Answer 4

You need to put brackets around the parts you want to reuse ==> capture them. Then you can access those values in the replacement string with $1 (or ${1} if the groups exceed 9) for the first group, $2 for the second one...
The problem here is that you would end up with $1 - $8, so I would rewrite the expression into something like this:
/#?(\d{1,2})[._-](\d{1,2})/
and replace with
S${1}E${2}
I tested it on writecodeonline.com:
$string = "some words and then #1.7 1.7 1_7 and 1-7";
$result = preg_replace('/#?(\d{1,2})[._-](\d{1,2})/', 'S${1}E${2}', $string);