if (isset($image))
{
$filedir = "/home/username/public_html/userphotos";
$maxfilesize = "50000";
$userfile_name = $_FILES['fileupload']['name'];
$userfile_tmp = $_FILES['fileupload']['tmp_name'];
$userfile_size = $_FILES['fileupload']['size'];
$userfile_type = $_FILES['fileupload']['type'];
if ($userfile_name != "") {
if ($userfile_size > $maxfile)
{
echo "The Image you uploaded is to big please make<br>sure the size of the file is less then 50k.";
exit;
}
$newfile = rename($userfile_tmp, $filedir.'/'.$username.'_'.$userfile_name);
$imagesize = getimagesize($newfile);
if ($userfile_type == "image/gif" || $userfile_type == "image/jpeg" || $userfile_type == "image/png")
{
$sqlpath = 'http://www.StudentsView.com/userphotos';
$sqlimage = "<img src=\"$newfile\" border=\"0\" height=\"$imagesize['1']\" width=\"$imagesize['0']\" ><br>";
$sql = mysql_query("UPDATE sv_users SET user_image = '$sqlimage' WHERE user_id = '$id'");
}
else{
echo 'The image you chose to upload was not a valid type.<br>Only JPEG, GIF, and PNG are allowed.';
exit;
}
}
else{
echo 'There was a problem while trying to upload your image. <br> Please try again later date';
}
}
that is my file upload section of a script. I am getting this error:
Parse error: parse error, expecting T_STRING' orT_VARIABLE' or `T_NUM_STRING' in /home/students/public_html/profile.php on line 262
that line is:
$sqlimage = "<img src=\"$newfile\" border=\"0\" height=\"$imagesize['1']\" width=\"$imagesize['0']\" ><br>";
any idea why it is giving me that error?