2012-07-23 36 views

回答

0
$fileName = $_POST['fileName']; //posted value of text box 
$userName =$_POST['userName']; //posted value of text box 
$myFile = $fileName.".txt"; 
$fh = fopen($myFile, 'w') or die("can't open file"); 
$stringData = $userName; 
fwrite($fh, $stringData); 
fclose($fh); 
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