[SOLVED] Help with MadTechie's Dynamic DropDown PHP/AJAX code

[larsbrimmer]

When i run the code i get:

Database is down

and

Warning: mysql_fetch_assoc(): on line 34

 

My tabel:

`RayMusicCategories` (`ID`, `Parent`, `Title`) VALUES
(1, 0, 'House'),
(2, 1, 'Progressive House'),
(3, 1, 'Acid house'),
(4, 1, 'Deep house'),
(5, 1, 'Electro house'),
(6, 1, 'Tech house'),
(7, 1, 'Tribal house'),
(8, 0, 'Trance'),
(9, 2, 'Acid trance'),
(10, 2, 'Hard trance'),
(11, 2, 'Progressive trance'),
(12, 2, 'Vocal trance');

 

Here is the code:

<?php
   $hostname = "localhost";
   $username = "";
   $password = "";
   $dbh = mysql_connect($hostname, $username, $password)
      or die("Unable to connect to MySQL");
   $selected = mysql_select_db("RayMusicCategories",$dbh);
      if( isset($_POST['Submit']) )
      {
         echo "<pre>";
         print_r($_POST);
      }
      if( isset($_GET['ajax']) )
   {
      //In this if statement
      $query = sprintf("SELECT * FROM RayMusicCategories WHERE PARENT=%d",$_GET['ajax']);
      
      $result = mysql_query($query);
      echo "<option value=''></option>";
      while ($row = mysql_fetch_assoc($result))
      {
         echo "<option value='{$row['ID']}'>{$row['TITLE']}</option>\n";
      }
      mysql_close($dbh);
      exit; //we're finished so exit..
   }
   
   if (!$result = mysql_query("SELECT * FROM RayMusicCategories WHERE PARENT=0"))
   {
      echo "Database is down";
   }
   //for use with my FIRST list box
   $List1 = "";
   while ($row = mysql_fetch_assoc($result))
   {
      $List1 .= "<option value='{$row['ID']}'>{$row['TITLE']}</option>\n";
   }
?>

 

For some reason this looks wrong to me.

<?php
   if (!$result = mysql_query("SELECT * FROM RayMusicCategories WHERE PARENT=0"))
   {
      echo "Database is down";
   }
?>

 

-Lars

 

 

[Philip]

Change:

if (!$result = mysql_query("SELECT * FROM RayMusicCategories WHERE PARENT=0"))
   {
      echo "Database is down";
   }

to:

$result = mysql_query("SELECT * FROM RayMusicCategories WHERE PARENT=0") or die(mysql_error());

 

[trq]

Your field is called Parent, not PARENT, there is a difference.

[larsbrimmer]

I replaced code with both suggestions and get No database selected error

 

<?php
   $hostname = "localhost";
   $username = "";
   $password = "";
   $dbh = mysql_connect($hostname, $username, $password)
      or die("Unable to connect to MySQL");
   $selected = mysql_select_db("RayMusicCategories",$dbh);
      if( isset($_POST['Submit']) )
      {
         echo "<pre>";
         print_r($_POST);
      }
      if( isset($_GET['ajax']) )
   {
      //In this if statement
      $query = sprintf("SELECT * FROM RayMusicCategories WHERE Parent=%d",$_GET['ajax']);
      
      $result = mysql_query($query);
      echo "<option value=''></option>";
      while ($row = mysql_fetch_assoc($result))
      {
         echo "<option value='{$row['ID']}'>{$row['TITLE']}</option>\n";
      }
      mysql_close($dbh);
      exit; //we're finished so exit..
   }
   
$result = mysql_query("SELECT * FROM RayMusicCategories WHERE Parent=0") or die(mysql_error());
   //for use with my FIRST list box
   $List1 = "";
   while ($row = mysql_fetch_assoc($result))
   {
      $List1 .= "<option value='{$row['ID']}'>{$row['TITLE']}</option>\n";
   }
?>

 

My connection properties are correct and server is up.

[larsbrimmer]

Sorry, I got it now.

 

I was using the table name as the db name.

 

-lars