I need help with PHP & MySQL please.

Discussion in 'PHP' started by uploadman1, May 16, 2009.

0
  1. #1
    Hi, Guys i'm still new with PHP & MySQL and still learning, and now i need help to fix some errors on my website,


    Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/d/a/v/davidau/html/CatalogList.php on line 6

    Line 6 :

    while($row=mysql_fetch_array($result_catalog))

    Please tell me what is wrong with it and help me to fix it.

    I hope you guys can help me to figure this out.
    thanks in advance.

    Best regards'
    -Me :eek:

    <?php
	$sq_catalog="Select * from sy_Catalog";
	$result_catalog = mysql_query($sq_catalog);
	$count=0;
	$perRow=12;
	while($row=mysql_fetch_array($result_catalog))
	{
		if(mysql_num_rows($result_catalog)>0)
		{
			if ($count<=$perRow)
			{
				$catalog = $row["Catalog"];
				if ($catalog =='Home Decor')
				{
				echo "<a class='AMenu' href=index.html>" . $catalog. "</a>&nbsp;&nbsp;<font class='hLine'>|</font>&nbsp;&nbsp; ";					
				}
				else if ($catalog =='Zippers')
				{
				echo "<a class='AMenu' href=pacific_zipper.php>" . $catalog. "</a>&nbsp;&nbsp;<font class='hLine'>|</font>&nbsp;&nbsp; ";					
				}
				else
				echo "<a class='AMenu' href=ProductList.php?value=" . $catalog . "&cvalue=All>" . $catalog . "</a>&nbsp;&nbsp;<font class='hLine'>|</font>&nbsp;&nbsp; ";
				$count++;
			}
			if ($count==$perRow)
			{
				echo "<br>";
				$count=0;
			}
		}
		else echo "Record not Found!!!";
		}
	
?>
    PHP:
     
    uploadman1, May 16, 2009 IP
  2. Sky AK47

    Sky AK47 Member

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    #2
    Try;
    $result_catalog = mysql_query($sq_catalog) or die(mysql_error());
    PHP:
     
    Sky AK47, May 16, 2009 IP
  3. uploadman1

    uploadman1 Peon

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    #3
    it says "No Database Selected"
     
    uploadman1, May 16, 2009 IP
  4. Sky AK47

    Sky AK47 Member

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    #4
    Uh...how can you have no problems connecting in your previous posted code while there's no connection setup.
    And also if you didn't understood, I meant you to replace
    $result_catalog = mysql_query($sq_catalog);
    with
    $result_catalog = mysql_query($sq_catalog) or die(mysql_error());
     
    Sky AK47, May 16, 2009 IP
  5. uploadman1

    uploadman1 Peon

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    #5
    i did what you've told me sir and that's what i got.
     
    uploadman1, May 16, 2009 IP
  6. uploadman1

    uploadman1 Peon

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    #6
    maybe the server is offline?
     
    uploadman1, May 16, 2009 IP
  7. Sky AK47

    Sky AK47 Member

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    #7
    That would give a different error.
    Is this page included somewhere?
    Connecting and selecting a MySQL Database goes as;
    //Connecting to serve: server location, username, password
mysql_connect("localhost","username","password");
//Selecting database: database name
mysql_select_db("my_database");
    PHP:
     
    Sky AK47, May 16, 2009 IP
  8. uploadman1

    uploadman1 Peon

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    #8 
       <?php
  $passwd = 'xxxxx';
  $host = 'xxxx.xxxxx.net'; 
  $user = 'xxxxxxx';
  $link = mysql_connect("$host", "$user", "$passwd") or die("Could not connect");
  mysql_select_db($link );
?>
    PHP:
     
    uploadman1, May 16, 2009 IP
  9. Sky AK47

    Sky AK47 Member

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    #9
    Ah, I see, you missed the database name!
    mysql_select_db("DATABASE NAME", $link);
    PHP:
     
    Sky AK47, May 16, 2009 IP
  10. uploadman1

    uploadman1 Peon

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    #10
    AWESOME!!! Thank you so much, mate!!!!

    I hope i can learn more from you, next time.

    Thanks thanks thanks!
     
    uploadman1, May 16, 2009 IP
  11. Sky AK47

    Sky AK47 Member

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    #11
    No problem, I like to help others out :)
     
    Sky AK47, May 17, 2009 IP