OK, I don't know any php really so i'm really stuck here, I got this code from about.com and it is supposed to handle a file upload and a mysql insert (making a video blog) but I get \"parse error: unexpected T_VARIABLE in line 8\"
can anyone shed some light on this for me please?
line 8, for you lazy ones is the line that starts with $nameCode:<?php //This is the directory where images will be saved $target = \"video/\"; $target = $target . basename( $_FILES['video']['name'] //This gets all the other information from the form $name = $_POST['title']; $caption = $_POST['caption']; $date = $_POST['date']; $video = ($_FILES['video']['name']); // Connects to your Database mysql_connect(\"localhost\", \"admin\", \"password\") or die(mysql_error()); mysql_select_db(\"vlg_videoblog\") or die(mysql_error()); //Writes the information to the database mysql_query(\"INSERT INTO `vlg_article_art` (title_art, caption_art, date_art, video_art) VALUES ('$name', '$caption', '$date', '$video')\"); //Writes the photo to the server if(move_uploaded_file($_FILES['video']['tmp_name'], $target)){ //Tells you if its all ok echo \"The file \". basename( $_FILES['uploadedfile']['name']). \" has been uploaded, and your information has been added to the directory <a href=\"index.php\">Back To Index</A>\"; } else { //Gives and error if its not echo \"Sorry, there was a problem uploading your file. please <a href=\"post2.php\">Try Again</a>\"; } ?>



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