Question

我只是想知道如何使用变量名来在PHP中设置文件名？当我运行以下代码时：

<?php
if ($_POST) {
    $filename = $_POST['firstName'];

    header("Content-Type: application/txt");        
    header('Content-Disposition: attachment; filename="$filename.txt"');

    echo "Welcome, ";
    echo $_POST['firstName']. " " . $_POST['lastName'];

    exit;
} else {

?>

<form action="" method="post">
    First Name: <input type="text" name="firstName" /><br />
    Last Name: <input type="text" name="lastName" /><br />
    <input type="submit" name="submit" value="Submit me!" />
</form>

<?php } ?>

文件名始终设置为“$filename.txt”，但我希望它为Adam.txt或Brian.txt等，具体取决于用户输入。

Answer 1

将''替换为'“，以便变量替换起作用

header("Content-Disposition: attachment; filename=\"$filename.txt\"");

或者如果你想使用''

header('Content-Disposition: attachment; filename="'.$filename.'.txt"');

Answer 2

只有双引号允许您插入变量：

$a = "some text"
$b = "another part of $a" //works, results in *another part of some text*
$b = 'another part of $a' //will not work, result *in another part of $a*

有关详细信息，请参阅http://php.net/manual/en/language.types.string.php#language.types.string.parsing

Answer 3

这是因为您使用单引号作为字符串，单引号中的字符串无法解析 - 请参阅the documentation。

要解决此问题，您可以执行此操作：

header('Content-Disposition: attachment; filename="'.$filename.'.txt"');

Answer 4

  <?php  
   if ($_POST) { 
   $filename = isset($_POST['firstName'])? $_POST['firstName'] :'general';
   header("Content-Type: application/txt"); 
   header('Content-Disposition: attachment;  filename='.$filename.'.txt'); 
   echo "Welcome, "; 
   echo
   $_POST['firstName']. " " . $_POST['lastName'];   exit;
  } else
   {

   ?>

   <form action="" method="post"> 
 First Name: <input type="text"
   name="firstName" /><br /> 
 Last Name: <input type="text"
   name="lastName" /><br /> <input type="submit" name="submit"
   value="Submit me!" /> </form>

   <?php } ?>

Answer 5

使用此：

header('Content-Disposition: attachment; filename="'.$filename.'.txt"');

使用PHP中的用户输入变量设置文件名

5 个答案: