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Urgent error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL re

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Join Date: Oct 2008

Posts: 3

Reputation: gmaster1440 is an unknown quantity at this point

Solved Threads: 0

gmaster1440 gmaster1440 is offline

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Urgent error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL re

Oct 27th, 2008

Code:

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<?php
 
define( "DATABASE_SERVER", "blah" );
 
define( "DATABASE_USERNAME", "blah" );
 
define( "DATABASE_PASSWORD", "blah" );
 
define( "DATABASE_NAME", "blah" );
 
//connect to the database
 
$mysql = mysql_connect(DATABASE_SERVER, DATABASE_USERNAME, DATABASE_PASSWORD) or die(mysql_error());
 
mysql_select_db( DATABASE_NAME );
 
$sql = 'SELECT * FROM `users`';
$result = mysql_query($sql) or die ('Error: '.mysql_error ());
 
while($row = mysql_fetch_array($result))
{
 
$hPassword = hash ( sha256, $row['password']);
$query = "INSERT INTO new_users (username, password) 
VALUES('".$row['username']."','$hPassword')";
 
$result = mysql_query($query);
}
 
?>
<?php

define( "DATABASE_SERVER", "blah" );

define( "DATABASE_USERNAME", "blah" );

define( "DATABASE_PASSWORD", "blah" );

define( "DATABASE_NAME", "blah" );

//connect to the database

$mysql = mysql_connect(DATABASE_SERVER, DATABASE_USERNAME, DATABASE_PASSWORD) or die(mysql_error());

mysql_select_db( DATABASE_NAME );

$sql = 'SELECT * FROM `users`';
$result = mysql_query($sql) or die ('Error: '.mysql_error ());

while($row = mysql_fetch_array($result))
{
	
$hPassword = hash ( sha256, $row['password']);
$query = "INSERT INTO new_users (username, password) 
VALUES('".$row['username']."','$hPassword')";

$result = mysql_query($query);
}

?>

I keep getting this error:

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 PHP Syntax (Toggle Plain Text) 
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/g/m/a/gmaster1440/html/vote/convertdb.php on line 20
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/g/m/a/gmaster1440/html/vote/convertdb.php on line 20

Yes, credentials are correct, yes 'users' table exists.

Please guys, this is due for a big project tomorrow. I'll appreciate any help I can get.

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Join Date: Jun 2007

Posts: 1,227

Reputation: kkeith29 has a spectacular aura about

Solved Threads: 167

kkeith29

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Nearly a Posting Virtuoso

Re: Urgent error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL re

Oct 27th, 2008

try:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<?php
 
define( "DATABASE_SERVER", "blah" );
 
define( "DATABASE_USERNAME", "blah" );
 
define( "DATABASE_PASSWORD", "blah" );
 
define( "DATABASE_NAME", "blah" );
 
//connect to the database
 
$con = mysql_connect(DATABASE_SERVER, DATABASE_USERNAME, DATABASE_PASSWORD) or die(mysql_error());
 
mysql_select_db( DATABASE_NAME );
 
$sql = 'SELECT * FROM `users`';
$result = mysql_query($sql,$con) or die ('Error: '.mysql_error());
 
while($row = mysql_fetch_array($result,MYSQL_ASSOC))
{
 
$hPassword = hash ( sha256, $row['password']);
$query = "INSERT INTO new_users (username, password) 
VALUES('".$row['username']."','$hPassword')";
 
$result = mysql_query($query);
}
 
?>
<?php

define( "DATABASE_SERVER", "blah" );

define( "DATABASE_USERNAME", "blah" );

define( "DATABASE_PASSWORD", "blah" );

define( "DATABASE_NAME", "blah" );

//connect to the database

$con = mysql_connect(DATABASE_SERVER, DATABASE_USERNAME, DATABASE_PASSWORD) or die(mysql_error());

mysql_select_db( DATABASE_NAME );

$sql = 'SELECT * FROM `users`';
$result = mysql_query($sql,$con) or die ('Error: '.mysql_error());

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

$hPassword = hash ( sha256, $row['password']);
$query = "INSERT INTO new_users (username, password) 
VALUES('".$row['username']."','$hPassword')";

$result = mysql_query($query);
}

?>

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Join Date: Oct 2005

Posts: 93

Reputation: humbug is an unknown quantity at this point

Solved Threads: 13

humbug

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Junior Poster in Training

Re: Urgent error: Warning: mysql_fetch_array()

Oct 28th, 2008

Why the $result = mysql_query($query); on the fourth last line? You don't need the $result = on that line as the mysql_query will return TRUE on success. You are then treating this as a MySQL result resource in mysql_fetch_array($result) (because the expression in the why loop is executed every time the while loop tries to run).
See http://au.php.net/manual/en/function.mysql-query.php

You should get rid of the $result = assignment (or use a different variable name if it is needed) and it should work.