preg_match 15 or 15.4

skip

Hi,

lets say I am looking for a number to the left of inch or inches

so if we have
1). 15inch
2). 15.4inch
3). 15 inch
4). 15.4 inch

this:

preg_match('/((\d{1,9})(?:inch|in|inches|Inch|"))/', strtolower($name), $matches_text); 
$screen = $matches_text[2];

would return:
1). 15
2). 4 (instead of 15.4)
3). NULL (becuse of the space between 15 and inch)
4). NULL (as above)

the addition of +s reverses the space problem it then does not pic up the ones without a space.

is there a pattern that will cover the decimal,no decimal, space and no space strings?

Thanks all in advance

ClarkF1

I would simply use:

$number = str_replace("inch","",$string);

where $string is the original and $number is just the number returned

If you don't know whether it's actually "inch" that's after the figure use:

$replace = array("inch","inches","in","Inch");
$number = str_replace($replace,"",$string);

skip

Hi thanks for posting

I should have mentioned that the strings are not just 15 inch they could be anything but some will include what I am looking for

e.g
Acer laptop 1024mb ram pentium dual core 1.6 15.4 inch tft screen

Installer

If you don't mind some kludge-y code:

$str = 'Acer laptop 1024mb ram pentium dual core 1.6 15.4 inch tft screen';
$str = str_replace(' inch', 'inch', $str);
$pos = strpos($str, 'inch');
$extract = ' ';
while (($char = substr($str, ($pos -= 1), 1)) != ' ') {
    $extract = $char . $extract;
}
echo $extract;  // 15.4

NogDog

Another possibility:

<?php
$testData = <<<EOD
1). 15inch
2). 15.4inch
3). 15 inch
4). 15.4 inch 
EOD;
// Grab the values:
preg_match_all('/((\d*\.)?\d+)\s*inch/i', $testData, $matches);
// Show the results:
echo "<pre>".print_r($matches[1], TRUE)."</pre>\n";
?>