i can save it to data base using this
<html>
<head>
<title>Upload Party photo</title>
</head>
<body>
<h1>Upload Party photo</h1>
<p>This form allows a user to upload a photo. </p>
<form action=<?php print $PHP_SELF; ?> method="post" enctype="multipart/form-data">
Picture to load: <input type="file" name="imageFile" size="50" />
<br/>Name ID <input type="text" name="nameID" size="4"/>
<br/><input type="submit" value='Upload'>
</form>
<?php
// print_r($_REQUEST);
// print_r($_FILES);
// should check that this is a valid nameID
if (!isset($_REQUEST['nameID'])) die("");
$nameID = $_REQUEST['nameID'];
// details of the file stored on the server are in the $_FILES array
$filename = $_FILES['imageFile']['name'];
$tempfilename = $_FILES['imageFile']['tmp_name'];
// get rid of any newline characters in the file
$tempfile = file($tempfilename);
$imageJPEG=implode($tempfile,'');
//base64 is one possible encoding of a binary file into printable characters - it takes more space
//and takes time to code and decode but it can be manipulated as a string
//an alternative would be to save as binary but I have difficulty making that work reliably :-(
$image64 = base64_encode($imageJPEG);
// connect to the MySQL server
include("lib.php");
$dblink =db_connect();
//replace is used here instead of insert to insert or replace
$query = "replace into photo(photoID,image) values($nameID,'$image64')";
if(!$dbresult = mysql_query($query,$dblink) ) {
print "Query failed :";
print(mysql_error() . "<br/>");
exit;
}
// close the database
print("<p>$filename added, size = ". strlen($image64)."</p>");
print("<img src='getphoto.php?photoID=$photoID' />");
mysql_close($dblink);
?>
</body>
</html>
but it wont now come up with the photo.. arbgh!! please help