我有一个文件上传脚本,最终工作,但希望有用户提交的图片到一个文件夹,我想被创建为用户名。我在想,我可以只提交文件,然后它就会创建它,但它没有,那么我该怎么做呢?这是我的脚本: PHP代码:
<?php
// Configuration part
$dbhost = "localhost"; // Database host
$dbname = "users"; // Database name
$dbuser = "dbuser"; // Database username
$dbpass = ""; // Database password
// Connect to database
$db = mysql_connect($dbhost, $dbuser, $dbpass) or die("Error: Couldn't connect to database");
mysql_select_db($dbname, $db) or die("Error: Couldn't select database.");
if($IsLoggedIn)
{ $userid = $_GET["userid"];
if($_POST['Submit'])
{
if($userid)
{ $uploadName = mysql_query("SELECT * FROM users WHERE userId ='"
. $userid . "'");
while ($upName = mysql_fetch_row($uploadName))
{ $userName2 = $upName[1];
$userPic1 = $upName[11];
echo "$userName2"; } }
//If the Submitbutton was pressed do:
if ($_FILES['imagefile']['type'] == 'image/pjpeg')
{
//this is where the pic is put into a folder by the username
copy ($_FILES['imagefile']['tmp_name'],
"pics/profiles/" . $userName2 . "/" . $_FILES['imagefile']['name'])
or die ("Could not copy");
echo "";
echo "Name: ".$_FILES['imagefile']['name']."";
echo "Size: ".$_FILES['imagefile']['size']."";
echo "Type: ".$_FILES['imagefile']['type']."";
echo "Copy Done....";
}
else {
echo "";
echo "Could Not Copy, Wrong Filetype (".$_FILES['imagefile']['name'].")";
}
}
?>
<form name="form1" method="post" action="index.php?page=classupload&userid=<?php echo "$userid"; ?>" enctype="multipart/form-data">
<input type="file" name="imagefile">
<input type="submit" name="Submit" value="Submit">
</form>
<?php } else { ?>How did you get here? You aren't <a href="index.php?page=login">Logged In</a>.
<?php } ?>有没有办法做到这一点?甚至当用户被创建时也是如此?谢谢!
发布于 2010-07-28 17:57:05
这应该是if(!file_exists($targetPath)) mkdir($targetPath);,但您需要具有正确的权限
https://stackoverflow.com/questions/3351670
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