Problem inserting data using checkboxes and input type=text

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  • AlexanderDeLarge
    New Member
    • Aug 2008
    • 3

    Problem inserting data using checkboxes and input type=text

    Hi!

    I got a problem that's driving me crazy and I'm desperately in need of help.

    I'll explain my scenario:
    I'm doing a database driven site for a band, I got these tables for their discography section:

    Discography
    ---------------------
    DiscID
    DiscName

    Song
    ---------------
    SongID
    SongName

    As this is a many to many relation I got the following table

    DiscoSong
    ------------------
    DiscID
    SongID
    Orden

    I already filled the Discography table with all their records to date, so no problem there. Now what wanna do is that every time I insert a song I want to associate it with the records it appears on as well as its order of appearance on each record, that's why I got the attribute Orden on my DiscoSong table.

    I'm selecting the records with checkboxes in a form and inserting the song number on input type=text fields.

    This is my php code for it:

    [PHP]
    $SongID = $_POST['SongID'];

    if (isset($_POST['DiscID'])) {
    $discs = $_POST['DiscID'];
    } else {
    $discs = array();
    }


    if(isset($_POST['Orden'])){
    $Orden = $_POST['Orden'];
    }else{
    $Orden = array();
    }


    foreach ($discs as $DiscID) {

    foreach($Orden as $ordena){

    $sql = "INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('$SongID', '$DiscID', '$ordena')";
    $ok = @mysql_query($s ql);
    if (!$ok) {
    echo $sql;
    echo "<p>Error inserting Song into Disc $DiscID: " .
    mysql_error() . '</p>';
    } else{
    echo $sql;
    echo "<p> Song $SongID inserted into Disc $DiscID in order $ordena</p>";
    }

    }

    }

    [/PHP]

    My code for the checkboxes and input fields:

    Code:
    $records = mysql_query("SELECT DiscID, DiscName FROM Discography");
    
    while($options = mysql_fetch_array($records)){
    		$DiscID = $options['DiscID'];
    		$DiscName = htmlspecialchars($options['DiscName']);
    		echo "
    		<table align='center'>
    			<tr>
    				<td width='250'>
    				$DiscName
    				</td>
    				<td width='250'>
     				<input name='DiscID[]' type='checkbox'/>
     				</td>
     			</tr>
     				<td width='250'> Song Number: 
     				</td>
     				<td width='250'>
     				<input name='Orden[]' type='text' size='3' />
     				</td>
     			</tr>
     		</table>
     	<br />";
    }
    ?>
    I got 6 records, so when I fill the form selecting 3 records and writing their song number on their respective input fields I get this:

    Code:
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '2', '')
    
    Song 17 inserted into Disc 2 in order
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '2', '2')
    
    Error inserting Song into Disc 2: Duplicate entry '2-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '2', '10')
    
    Error inserting Song into Disc 2: Duplicate entry '2-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '2', '')
    
    Error inserting Song into Disc 2: Duplicate entry '2-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '2', '10')
    
    Error inserting Song into Disc 2: Duplicate entry '2-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '2', '')
    
    Error inserting Song into Disc 2: Duplicate entry '2-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '3', '')
    
    Song 17 inserted into Disc 3 in order
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '3', '2')
    
    Error inserting Song into Disc 3: Duplicate entry '3-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '3', '10')
    
    Error inserting Song into Disc 3: Duplicate entry '3-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '3', '')
    
    Error inserting Song into Disc 3: Duplicate entry '3-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '3', '10')
    
    Error inserting Song into Disc 3: Duplicate entry '3-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '3', '')
    
    Error inserting Song into Disc 3: Duplicate entry '3-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '5', '')
    
    Song 17 inserted into Disc 5 in order
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '5', '2')
    
    Error inserting Song into Disc 5: Duplicate entry '5-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '5', '10')
    
    Error inserting Song into Disc 5: Duplicate entry '5-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '5', '')
    
    Error inserting Song into Disc 5: Duplicate entry '5-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '5', '10')
    
    Error inserting Song into Disc 5: Duplicate entry '5-17' for key 1
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '5', '')
    
    Error inserting Song into Disc 5: Duplicate entry '5-17' for key 1
    And I want to get this

    Code:
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '2', '2')
    
    Song 17 inserted into Disc 2 in order 2
    
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '3', '10')
    
    Song 17 inserted into Disc 3 in order 10
    
    INSERT INTO DiscoSong (SongID, DiscID, Orden) VALUES ('17', '5', '10')
    
    Song 17 inserted into Disc 5 in order 10
    I know my problem is on the second foreach, but I don't know what to do, I've been thinking for almost week and all my attempts have failed.

    Hope someone can help me.

    Thanks in advance.

    PS. Sorry for making this thread so extensive, I be as much clear as possible with my problem
  • Atli
    Recognized Expert Expert
    • Nov 2006
    • 5062

    #2
    Hi.

    Your second loop, the one looping through the Order, shouldn't be a loop.

    You should be adding each Disk and the Orden that is meant for that Disk. That is; you should be looping through the Disks, and using the Orden at the same index as the Disk.

    Consider this:
    [code=php]
    $songID = $_POST['SongID']
    $diskID = $_POST['DiskID']; // This would be an array from a DiskID[] input
    $orden = $_POST['Orden'] // And this from a Orden[] input. (one for each DIskID)

    foreach($DiskID as $_index => $_diskID) {
    $sql = "
    INSERT INTO tbl(DiskID, SongID, Orden)
    VALUES($_diskID , $songID, {$order[$_index]}";
    }
    [/code]

    Comment

    • AlexanderDeLarge
      New Member
      • Aug 2008
      • 3

      #3
      Thank you very much man, my problem is solved :D

      Comment

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