php - 查找具有特定名称和任何扩展名的文件

Question

请注意，我希望更改车厢编号。

<?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."
    }
?>

谢谢你！

Answer 1

你需要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."

顺便说一句，你在上面做的就是在哭一个循环。不要重复你的代码块，但将其中一个包装成这样：

 $compartments = array(1, 3, 6, 9); // or whichever compartments 
                                    // you wish to run through

 foreach ($compartments as $compartment)
  {
   ..... insert code here .......
  }

Answer 2

抬头：

• glob — 查找匹配模式的路径名
• fnmatch — 将文件名与模式匹配
• pathinfo — 返回有关文件路径的信息