Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource

Question

i have this script :

<?php
        // remplacer par vos valeurs de connexion
        $host="*****";
        $user="*****";
        $pwd="******";
        $base="*****";
        $mysql_link = mysql_connect($host,$user,$pwd);
        mysql_select_db($base);

       $dbsite="arfooo"; // prefixe de votre table sans "_" par exemple : "arfooo"
        echo "TRUNCATE TABLE `".$dbsite."_categoryparents`;<br>\n";
        $sql = "SELECT * FROM ".$dbsite."_categories";
        $result=mysql_query($sql);
        while ($row=mysql_fetch_object($result)) {
           $depth=0;
           echo "INSERT INTO `".$dbsite."_categoryparents` (`parentId`, `childId`, `depth`) VALUES ($row->categoryId,$row->categoryId,$depth);<br>\n";
           $parentId = $row->parentCategoryId;
           while ($parentId<>0){
              $sql2 = "SELECT * FROM ".$dbsite."_categories WHERE categoryId = $parentId";
              $result2=mysql_query($sql2);
              while ($row2=mysql_fetch_object($result2)) {
                 $depth=$depth+1;
                 echo "INSERT INTO `".$dbsite."_categoryparents` (`parentId`, `childId`, `depth`) VALUES ($row2->categoryId,$row->categoryId,$depth);<br>\n";
                 $parentId = $row2->parentCategoryId;
              }
           }
           $depth=$depth+1;
           echo "INSERT INTO `".$dbsite."_categoryparents` (`parentId`, `childId`, `depth`) VALUES (0,$row->categoryId,$depth);<br>\n";
        }
        ?>

I have the following error:

TRUNCATE TABLE `arfooo_categoryparents`;

PHP Error Message

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/******/public_html/genereParentChildDepth.php on line 14

While you echo you're truncating categoryparents, actually you tried to select categories. What does the mysql_error return? is the categories table valid? — BugFinder, May 24 '11 at 11:44
There's probably an error in the SQL statement, that would be the reason why mysql_query doesn't return a resource identifier. Are you sure your table names are correct? Since it's a very simple select statement you execute — klennepette, May 24 '11 at 11:46

Stephen · Accepted Answer · 2011-05-24T11:51:57.563

1

You're getting a mysql error on the first select statement,

"SELECT * FROM ".$dbsite."_categories"

The table name must be malformed. Verify that arfooo_categories exists.

If it does, try this:

$l = mysql_connect($host,$user,$pwd);
mysql_select_db($base);
$sql = "SELECT * FROM arfooo_categories";
$result = mysql_query($sql);
echo $result ? 'all good' : mysql_errno($l) . ': ' . mysql_error($l) . "\n";

edited May 24 '11 at 11:51

answered May 24 '11 at 11:46

Stephen

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thank you have right prefix is Arfoo instead of arfoo – grigione May 24 '11 at 11:58

score 0 · Answer 2 · answered May 24 '11 at 11:46

You need to check whether you connection has been made

$mysql_link = mysql_connect($host,$user,$pwd);
if (!$mysql_link) {
   echo "Couldn't connect to the database. Check your credentials";
   die();
}

//code

$result = mysql_query($sql);
if (!$result) {
   echo "Oops, something seems to be wrong.";
   die();
}

score 0 · Answer 3 · answered May 24 '11 at 11:57

I would recommend trying this. You need to nail down exactly where the error is coming from.

<?php
        // remplacer par vos valeurs de connexion
        $host="*****";
        $user="*****";
        $pwd="******";
        $base="*****";
        $mysql_link = mysql_connect($host,$user,$pwd);

        if (!$mysql_link) {
            die(mysql_error());
        }

        $db_name = mysql_select_db($base);
        if (!$db_name) {
            die(mysql_error());
        }
       $dbsite="arfooo"; // prefixe de votre table sans "_" par exemple : "arfooo"
       $sql = "SELECT * FROM ".$dbsite."_categories";
       $result=mysql_query($sql) or die(mysql_error());
       while ($row=mysql_fetch_object($result)) {
          $depth=0;
          echo "INSERT INTO `".$dbsite."_categoryparents` (`parentId`, `childId`, `depth`) VALUES ($row->categoryId,$row->categoryId,$depth);<br>\n";
           $parentId = $row->parentCategoryId;
           while ($parentId<>0){
              $sql2 = "SELECT * FROM ".$dbsite."_categories WHERE categoryId = $parentId";
              $result2=mysql_query($sql2);
              while ($row2=mysql_fetch_object($result2)) {
                 $depth=$depth+1;
                 echo "INSERT INTO `".$dbsite."_categoryparents` (`parentId`, `childId`, `depth`) VALUES ($row2->categoryId,$row->categoryId,$depth);<br>\n";
                 $parentId = $row2->parentCategoryId;
              }
           }
           $depth=$depth+1;
           echo "INSERT INTO `".$dbsite."_categoryparents` (`parentId`, `childId`, `depth`) VALUES (0,$row->categoryId,$depth);<br>\n";
        }
        ?>

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource

3 Answers3

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