Problem on generating mysql tree

[php_student2]

When I tried to generate a tree using javascritps code, every things was good but some times I got a strange message which is:

Warning: mysql_query(): 32 is not a valid MySQL-Link resource in /home/virtual/site1/fst/var/www/html/audio/tap/view2/sql.php on line 22

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/virtual/site1/fst/var/www/html/audio/tap/view2/sql.php on line 23

What do you thing the problem is ??

My code is included.

<?
function call($part_id,$i=0,$user_id)
{

include "../variable.php";
$db = mysql_connect("$server", "$db_user", "$db_pass");
mysql_select_db("$database", $db);

   $sql = "select * from favorite where parentid=$part_id and user_id=$user_id";
   $result=mysql_query($sql);
     while($row=mysql_fetch_array($result))
      {
         echo "var Tree".$row[id]." = new WebFXTreeItem('  ".$row[name]."','javascript:DisplayNode(\'Tree".$row[id]."\');return false;');
               Tree".$part_id.".add(Tree".$row[id].");";

            $sql = "select anasheed.id as nasheed_id,anasheed.nasheed_name,fav_anasheed.fav_id as id1
                          from anasheed,fav_anasheed  where anasheed.id = fav_anasheed.nasheed_id and
                          fav_anasheed.fav_id = $row[id] and fav_anasheed.user_id=$user_id";

		echo $sql."<br>";

            $get_n = mysql_query($sql);
            while($row2=mysql_fetch_array($get_n)) {
			?>
		    var Tree<? echo $row[id].$row2[id1] ?> = new WebFXTreeItem('  <? echo $row2[nasheed_name] ?>','javascript:gohref(\'hear.php?id=<? echo $row2[nasheed_id] ?>\');return false;');
		    Tree<? echo $row[id] ?>.add(Tree<? echo $row[id].$row2[id1] ?>);
			<?
		  }

		 call($row[id],$i,$user_id);
     }

}

call(0,0,1);
?>

[designguy79]

Add some error handling after the MySQL query statement. Never just assume that everything went without a hitch.

So, after this line:

$result=mysql_query($sql);

Add:

if(!$result) {
echo "Database error in query $sql<br><br>";
echo "MySQL said: ".mysql_error()."<br>";
}

Then, you can see what is really causing MySQL to puke.

Have fun!

[dbchip2000]

Try replacing your SQL with these:

$sql = "select * from favorite where parentid='".$part_id."' and user_id='".$user_id."'";

$sql = "select anasheed.id as nasheed_id,anasheed.nasheed_name,fav_anasheed.fav_id as id1 
from anasheed,fav_anasheed where anasheed.id = fav_anasheed.nasheed_id and 
fav_anasheed.fav_id = '".$row[id]."' and fav_anasheed.user_id='".$user_id."'";

I just added single quotes before values, then if there is a null value you will not get an error. When you receive a "not valid resource" MySQL means that you tried to do a semi-invalid query. What you can do is add

echo(mysql_error());

after your queries and it will give you more details

[php_student2]

Thanks a lot guys, i'll tried now 🙂.