Regular Expressions

[Roger Price]

Well it didn't take long to come up with another query!

The varable $HTTP_ENV_VARS['HTTP_USER_AGEN T']
returns on my browser:

'Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; Q312461)'

I have to write a PHP script that check to see if it contains MISE and
confirm whether the browser being used is Internet Explorer and what
version.

The first part is easy.

I now need to isolate "6.0" from the string. I have tried many combinations
of "regexp" but none have come even close to finding the string of
characters that start at the beginning and end with MSIE, followed by a
space, thus:

"'Mozilla/4.0 (compatible; MSIE "

The code I am trying to use is:

$search = $HTTP_ENV_VARS['HTTP_USER_AGEN T'];

if ( ereg( "^([a-zA-Z0-9\(\)\.\/\;\'\s])*4.0[[:>:]]", $search,$match ) ) {
print( "String found ending in 'IE': " .$match[ 1 ] . "<br />" );
}


I would then need to replace this with "" and select all characters up to
but not including ";"

I would welcome any advice you can offer.

TIA

Roger

[Pedro Graca]

Re: Regular Expressions

Roger Price wrote:[color=blue]
> I now need to isolate "6.0" from the string.[/color]
(snip)[color=blue]
> The code I am trying to use is:
>
> $search = $HTTP_ENV_VARS['HTTP_USER_AGEN T'];
>
> if ( ereg( "^([a-zA-Z0-9\(\)\.\/\;\'\s])*4.0[[:>:]]", $search,$match ) ) {
> print( "String found ending in 'IE': " .$match[ 1 ] . "<br />" );
> }
>
>
> I would then need to replace this with "" and select all characters up to
> but not including ";"[/color]

Switch to preg_* functions
They're faster amd more powerful.

Try this:

<?php
$search = 'Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; Q312461)';
$version = preg_replace('/^.+MSIE ([^;]+);.*$/', '$1', $search);
?>
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