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Oct 27th, 2008

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

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Code:

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
 
$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:

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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mysql_fetch_array warning urgent plz in PHP

gmaster1440

Reputation Points: 10

Solved Threads: 0

Newbie Poster

Offline

3 posts
since Oct 2008

Permalink

Oct 27th, 2008

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

try:

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);
}

?>

kkeith29

Reputation Points: 235

Solved Threads: 193

Nearly a Posting Virtuoso

Offline

1,315 posts
since Jun 2007

Permalink

Oct 28th, 2008

Re: Urgent error: Warning: mysql_fetch_array()

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.

humbug

Reputation Points: 20

Solved Threads: 13

Junior Poster in Training

Offline

93 posts
since Oct 2005

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Oct 29th, 2008

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

wow, i can't believe i missed that. yes humbug is right. when you rename that variable it will work fine.

kkeith29

Reputation Points: 235

Solved Threads: 193

Nearly a Posting Virtuoso

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1,315 posts
since Jun 2007

This thread is more than three months old

No one has posted to this discussion for at least three months. Please let old threads die and do not reply to them unless you feel you have something new and valuable to contribute that absolutely must be added to make the discussion complete. Otherwise, please start a new thread in this forum instead.

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