this is what the script looks like
<div style="background-color:#efefef;width:306px;border:5px solid #00c000;padding:1px;text-align:left;">
<?php
echo "control:<p>";
echo "original-name: $upfile_name<br>";
echo "size: $upfile_size Byte<br>";
echo "type: $upfile_type<br>";
$dname = explode(".",$upfile_name);
$dnamesize = count($dname);
$upfile_ext = $dname[$dnamesize-1];
echo "data-extension: " . $upfile_ext . "<p>";
if($upfile_size>0)
{
copy($upfile,$upfile_name);
echo "data has been copied in $upfile_name<p>";
echo "<img src= $upfile_name, width= "301", alt= "$titel">";
}
else
{
echo "copy-error: data does not exist";
echo " or false data-type";
}
?>
</div>
when this site loads, this error appears
Parse error: syntax error, unexpected T_LNUMBER, expecting ‘,’ or ‘;’ in C:\apachefriends\xampp\htdocs\upload\UC19.PHP on line 18
when I only write the source and not width, and alt, it works.
i d also like to know, how i can keep the properties of a pic, because i need a certain width of the pic, but dont know how big its gonna be
thx, in case u help me ^^
