mortonfc Posted January 28, 2008 Share Posted January 28, 2008 Hi there i am a newbie to PHP and MYSql, I made some PHP pages and a MYSQL database a couple of years ago to catalogue my mp3 collection on discs i have. It worked fine. While changing computer a while back I didn't save all the data and so when deciding to get back to this project the other day I had to make up a new datsbase structure. Anyway to cut a long story short, the PHP pages now dont seen to add an album -error "Album was not added due to an error", display the search results ( I can add data in PHPMYAdmin or command line ok) Here is the add album verification page.... ----------------------- $dbc=@mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or die ('Could not connect to MySQL: '.mysql_error());; @mysql_select_db(DB_NAME) or die ('Could not select the database: '.mysql_error()); if(isset($_POST['Submit'])){ //If form has been submitted // Album name if(!empty($_POST['title'])){ $alb=$_POST['title']; }else{ $alb=FALSE; echo'<p>Please enter the album name</p>'; } //ok upto here //Artist //Check for new or existing artist if($_POST['artist']=='newartist'){ //If its a new artist check that an artists name has been typed and add that artist to the database if(!empty($_POST['newartist'])){ $newart=$_POST['newartist']; $query="INSERT INTO artist(art_id,artist) VALUES(NULL, '$newart')"; $result=@mysql_query($query); $artists_id=@mysql_insert_id(); }else{ $artists_id=FALSE; echo'<p>Please enter the artists name</p>'; } }elseif($_POST['artist']=='existingartist'){ //If its an existing artist check that an artist has been selected from the pull down menu if($_POST['exsistingartist']>0){ $artists_id=$_POST['exsistingartist']; echo"$artists_id"; }else{ $artists_id=FALSE; echo'Please pick an artist from the pull down menu'; } }else{ //if no artist type is picked $artists_id=FALSE; echo'Please choose an existing or new artist'; } //ok upto here //Disc if($_POST['disc']=='newdisc'){ if(!empty($_POST['newdisc'])){ $newdisc=$_POST['newdisc']; $query="INSERT INTO disc(disc_id,disc) VALUES('$newdisc', '$newdisc')"; $result=@mysql_query($query); $disc_id=@mysql_insert_id(); echo'<p>new disc added</p>'; }else{ $disc_id=FALSE; echo'Please enter the disc number.'; } }elseif($_POST['disc']=='existingdisc'){ if(!empty($_POST['existingdisc'])){ $disc_id=$_POST['existingdisc']; }else{ $disc_id=FALSE; echo'Please enter the disc number'; } } //ok upto here //Genre if($_POST['genre']){ $gen=$_POST['genre']; }else{ $gen=FALSE; echo'Please choose a genre'; } //ok upto here //Year (not necessary) if(!empty($_POST['year'])){ $year=$_POST['year']; echo"<p>$year</p>"; }else{ $year=' '; } //ok upto here //If we have everything we need then add the album to the database if($alb && $artists_id && $disc_id && $gen){ $query="INSERT INTO album(alb_id,art_id,gen_id,disc_id,year,title) VALUES (NULL, '$artists_id', '$gen', '$disc_id',''$year',$alb')"; if($result=@mysql_query($query)){ echo'The album has been added'; }else{ echo'Album was not added due to an error'; } }else{//if we don't have everything we need echo'Please fill in the missing information'; } //if forms not submitted }else{ echo'Please fill in the form'; } ?> <p><a href="add_album.php">Add another album </a></p> <p><a href="../search/search.php">Search </a></p> </BODY> </HTML> ---------- Thank you in advance Quote Link to comment https://forums.phpfreaks.com/topic/88280-php-mysql-help-album-was-not-added-due-to-an-error/ Share on other sites More sharing options...
nethnet Posted January 28, 2008 Share Posted January 28, 2008 Change this: echo'Album was not added due to an error'; To this: echo "Album was not added due to an error. Query: " . $query; That way we can see what the exact query is. Quote Link to comment https://forums.phpfreaks.com/topic/88280-php-mysql-help-album-was-not-added-due-to-an-error/#findComment-451731 Share on other sites More sharing options...
AndyB Posted January 29, 2008 Share Posted January 29, 2008 Quotemarks in the wrong place: $query="INSERT INTO album(alb_id,art_id,gen_id,disc_id,year,title) VALUES (NULL, '$artists_id', '$gen', '$disc_id',''$year',$alb')"; Should be: $query="INSERT INTO album(alb_id,art_id,gen_id,disc_id,year,title) VALUES (NULL, '$artists_id', '$gen', '$disc_id','$year','$alb')"; Quote Link to comment https://forums.phpfreaks.com/topic/88280-php-mysql-help-album-was-not-added-due-to-an-error/#findComment-451767 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.