So everything up to the insert statement works perfectly. I know the database is connecting because I can select information from the database with the first two statements. I also know that the execute_statment3 works because no errors are being printed off and when it is put into the sql the statement is inserted the way it should be. Therefore the problem lies somewhere with the communication between the script and phpmyadmin. Please help I have been staring at this problem for two days and am going rather crazy.
<?php
session_start();
$hostname = 'localhost';
$username = '####';
$password = '####';
$connection = mysql_connect($hostname, $username, $password)
or die ('Connection error!!!');
$database = '####';
mysql_select_db($database);
$uid = $_SESSION['ID'];
$album = $_POST['albumname'];
$description = $_POST['description'];
$filename = $_FILES["upload_file"]["name"];
$filetype = $_FILES["upload_file"]["type"];
$filesize = $_FILES["upload_file"]["size"];
$file_on_server = $_FILES["upload_file"]["tmp_name"];
if ($filetype == "image/jpeg") {
$file_copy_name = date(m.d.y_H.i.s) . ".jpg";
copy($file_on_server, "uploads/" . $file_copy_name);
print "<br>";
print "<img src = \"uploads/$file_copy_name\">";
print "<br>";
$ret = system("pwd");
$picture = "uploads/$file_copy_name";
}
$execute_statement = "SELECT * FROM ImageAlbums WHERE Album = '$album'";
$results = mysql_query($execute_statement) or die ('Error executing SQL statement!!!');
while($item = mysql_fetch_array($results))
{
$album2 = $item['Album'];
}
if ($album2 == $album)
{
$execute_statement2 = "SELECT * FROM ImageAlbums WHERE Album = '$album'";
$results2 = mysql_query($execute_statement2) or die ('Error executing SQL statement2!!!');
while ($row2 = mysql_fetch_array($results2)) {
$AID = $row2["AlbumID"];
}
$execute_statement3 = "INSERT INTO Images (`ImageID`, `AlbumID`, `Description`, `Extensions`) VALUES ('NULL', '$AID', '$description', '$file_copy_name')";
($execute_statement3) or die ('Error executing SQL statement3!!!');
}
print "<br>";
print "<br>";
print $execute_statement3;
print "<br>";
print "<br>";
print $AID;
print "<br>";
print "<br>";
print $picture;
?>
I am using two databases for this script one of the databases is called ImageAlbums and has two columns called AlbumID and Album (AlbumID being a primary key). The second table is called Images and has four columns ImageID (primary key), AlbumID (foreign key), Description and Extensions.