Question

PHP - MySql problem connecting

Asked by: melvynmyers

Hi,

I am getting an error and I can't understand why?  Any help is much appreciated.  The PHP code connects to a MySQL database.

Thank you in advance.

Mel



The error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in xyz.php on line 12


The code:
<?php

$db = mysql_connect("localhost", "user", "password");

@mysql_select_db("database",$db);

$query = "SELECT * FROM products WHERE category = 'accounting_finance' ORDER BY field2;";

$result = mysql_query($query);


if ($row = mysql_fetch_array($result)) {

      do {
            echo "
            <tr><td>$row[field1]</td>
            <td>$row[field2]</td>
            <td>$row[field3]</td>
            <td>$row[field4]</td></tr>
            ";

      } while ($row = mysql_fetch_array($result));

}

?>

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Asked On
2003-09-16 at 02:24:38ID20739939
Topic

PHP Scripting Language

Participating Experts
4
Points
50
Comments
5

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Answers

 

by: prady_21Posted on 2003-09-16 at 02:37:48ID: 9369130

after ur first line add this
<?php
error_reporting(E_ALL);

and then for the mysql connections, try the following

$connection = mysql_connect("server","user","password") or die ("Cannot connect to Mysql\n");
$selectdb = mysql_select_db("$Database", $connection) or die ("could not select the database");
$result = mysql_query($sql_statement, $connection);

post  the errors if any now

also try this in mysql command line and see if the result is not zero rows
SELECT * FROM products WHERE category = 'accounting_finance' ORDER BY field2;

 

by: KaritzPosted on 2003-09-16 at 02:50:04ID: 9369185



$db = mysql_connect("localhost", "user", "password");

@mysql_select_db("database",$db);

$query = "SELECT * FROM products WHERE category = 'accounting_finance' ORDER BY field2";

$result = mysql_query($query) or die(mysql_error().$query);


while ($row = mysql_fetch_array($result))
{

           echo "
         <tr><td>$row[field1]</td>
         <td>$row[field2]</td>
         <td>$row[field3]</td>
         <td>$row[field4]</td></tr>
         ";

}


this will give you a good error message that you will be able to debug

 

by: ThGPosted on 2003-09-16 at 02:51:10ID: 9369190


you are getting an error because your SQL query fails:

$query = "SELECT * FROM products WHERE category = 'accounting_finance' ORDER BY field2;";
$result = mysql_query($query);

try changing the second line with:
$result = mysql_query($query) or die("MySQL query failed: " . mysql_error());

 

by: ICM3Posted on 2003-09-16 at 09:05:41ID: 9371447

remove the ; after field2

field2;";


so that it is a valid query



 

by: melvynmyersPosted on 2003-09-16 at 16:36:06ID: 9374388

Thank you all for your comments.  It was an error in my part - the database name was spelt incorrectly.  The or die ("could not select the database"); statement picked it up.  Karitz's post was also very relevant, but prady_21 has to get the points as the first to post the solution.

Warm Regards.

20120131-EE-VQP-002

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