How to cut off a fixed "word" in a string

[CrazyMod]

Ok FOr example, so I have my domain, something like mytestyou.com. I have a few domain that also end with "you" such as Itestyou.com and wetestyou.com

I want to show the domain names, but without the extentions (such as.com .net whatever and also cut off the "you" part) How would I do that?

Please help, I am a student, not much experiences.

[longneck]

$s = 'itestyou.org';
echo preg_replace('/you[.][a-z]{2,4}/i','',$s); 

this regular expression looks for you, followed by a period (which must be in brackets because period is a special character in a regular expression), followed by 2 to 4 letters in the range of a-z. the replacement is '', which means that any matches will be replaced with nothing. the i at the end makes it case insensitive.

[earl-grey]

$word = 'you';
// Remove the part after dot
$pos = strrpos($domain, '.');
$domain = substr($domain, $pos);
// Remove the $word ending, if it's present
if (substr($domain, -strlen($word)) == $word) {
  $domain = substr($domain, 0, -strlen($word));
} 

[CrazyMod]

$word = 'you';
// Remove the part after dot
$pos = strrpos($domain, '.');
$domain = substr($domain, $pos);
// Remove the $word ending, if it's present
if (substr($domain, -strlen($word)) == $word) {
  $domain = substr($domain, 0, -strlen($word));
} 

I am little confused, but I already have a var that removed the extension after the dot, called "domain_remove" , so do I use that or i use var that actually list full domain name?

[Mittineague]

@longneck I thought an unescaped "." meant "any character" (including a ".")
AFAIK it should be

echo preg_replace('/you\.[a-z]{2,4}/i','',$s); 

to be "." specific.

[bokehman]

I thought an unescaped "." meant "any character"

A dot within a character class is literal, not wild, hence it matches a dot and nothing else.

[clamcrusher]

. is litteral when inside of a character class. it wouldnt really make sense to match any character in a character class

although yes, i think its more common to forgo the character class and just escape the .

[Mittineague]

Yes, right you are. I didn't "see" it all by it's lonesome like that.

[bokehman]

i think its more common to forgo the character class and just escape the .

It makes no sense whatsoever to have a positive character class with only a single character in it. My guess would be the overhead is enormous when compared to just an escaped dot.

[earl-grey]

I am little confused, but I already have a var that removed the extension after the dot, called "domain_remove" , so do I use that or i use var that actually list full domain name?

In that case using this should be enough:

$word = 'you'; 
// Remove the $word ending, if it's present 
if (substr($domain, -strlen($word)) == $word) { 
  $domain = substr($domain, 0, -strlen($word)); 
} 

[longneck]

It makes no sense whatsoever to have a positive character class with only a single character in it. My guess would be the overhead is enormous when compared to just an escaped dot.

i've never really thought about the performance implications of that construct. i find it more difficult to read backslashes than the single character class.

[CrazyMod]

In that case using this should be enough:

$word = 'you'; 
// Remove the $word ending, if it's present 
if (substr($domain, -strlen($word)) == $word) { 
  $domain = substr($domain, 0, -strlen($word)); 
} 

isnt it suppose to be ?

$word = 'you'; 
// Remove the $word ending, if it's present 
if (substr($domain_remove, -strlen($word)) == $word) { 
  $domain_remove = substr($domain_remove, 0, -strlen($word)); 
} 

[earl-grey]

isnt it suppose to be ?

$word = 'you'; 
// Remove the $word ending, if it's present 
if (substr($domain_remove, -strlen($word)) == $word) { 
  $domain_remove = substr($domain_remove, 0, -strlen($word)); 
} 

Yes, I assumed that $domain contains domain name without the part after dot and dot itself.

[Salkcin]

Or you could just split the string on "you" using explode(), ex:

<?php

$domain = "http://www.wetestyou.com";
list($var) = explode('you', $domain);

echo $var; // Output:  http://www.wetest

?>