dounouxi1020 2014-08-19 12:38
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根据名称获取文件名include_once文件

I have multiple files in a directory for pages.

All the pages are the same except the content I enter based on rental inspections.

bedroom1.php bedroom2.php bedroom3.php

But to get them to use the right header I need them to see the correct header based on their own filename.

bedroom1.php to include header1.php

bedroom2.php to include header2.php

bedroom3.php to include header3.php

.......

bedroom10.php to include header10.php

I can get the filename easy enough.

I'm trying to use preg_match(Maybe should use something else?) but with not getting any errors in the logs so I'm not sure what I'm missing and not knowing enough about file comparing I'm lost.

EDIT: ADDED : Forgot to add, this code is in bedroom1.php etc...

Thanks in advance

<?php

$file = $_SERVER["SCRIPT_NAME"];
$break = Explode('/', $file);
$pfile = $break[count($break) - 1];

if (preg_match('/bedroom . $i .php/', $pfile, $i)) {
    $number = $i[1];
    foreach(array('header') as $base) {
        include_once "$base$number.php";
    }
}

?>
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  • dongzhong5756 2014-08-19 12:44
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    It should be:

    if (preg_match('/bedroom(\d+)\.php/', $pfile, $i)) {
    

    You need to use \d+ to match numeric digits, and put it inside parentheses to make it a capture group, so you can access it with $i[1].

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