Нахождение файла с собственным именем с любым расширением
Обратите внимание на то, что я хочу, чтобы число отсека изменилось.
<?php
$compartment = "1";
/* HERE I NEED SOME SCRIPT TO FIND THE EXTENSION OF THE FILE NAME $compartment AND TO SAVE THAT AS A VARIABLE NAMED 'EXTENSION'.*/
if (file_exists($compartment.$extension)) {
echo "$compartment.$extension exists!
} else {
echo "No file name exists that is called $compartment. Regardless of extension."
}
?>
<?php
$compartment = "2";
/* HERE I NEED SOME SCRIPT TO FIND THE EXTENSION OF THE FILE NAME $compartment AND TO SAVE THAT AS A VARIABLE NAMED 'EXTENSION'.*/
if (file_exists($$compartment.$extension)) {
echo "$compartment.$extension exists!
} else {
echo "No file name exists that is called $compartment. Regardless of extension."
}
?>
Спасибо!
8
задан Andy Cheeseman 8 July 2010 в 09:26
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2 ответа
Вам понадобится glob () .
$compartment = "2";
$files = glob("/path/to/files/$compartment.*"); // Will find 2.txt, 2.php, 2.gif
// Process through each file in the list
// and output its extension
if (count($files) > 0)
foreach ($files as $file)
{
$info = pathinfo($file);
echo "File found: extension ".$info["extension"]."<br>";
}
else
echo "No file name exists called $compartment. Regardless of extension."
кстати, то, что вы делаете выше, требует петли. Не повторяйте блоки кода, а оберните один из них следующим образом:
$compartments = array(1, 3, 6, 9); // or whichever compartments
// you wish to run through
foreach ($compartments as $compartment)
{
..... insert code here .......
}
21
ответ дан 5 December 2019 в 06:36
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