PHP error help?

Question

This is my php mysql select query:

$query = "SELECT badge_number, first_name, last_name, email, user_id FROM users WHERE email = '$email' AND activation_code = '$code' ";
$result = @mysql_query ($query);
$row = mysql_fetch_array ($result, MYSQL_NUM);

BUT it comes up with the error

mysql_fetch_array(): supplied argument is not a valid MySQL result resource in...

Whats wrong??

Answer 1

try to replace this line:
$result = @mysql_query ($query);

with this:
$result = mysql_query ($query);

is it helpful?

Answer 2

It probably means your query failed. Remove the @ so you get error logging and check to see if $result is false before using mysql_fetch_array(), if it is print the result of mysql_error() to get a description of the error (and the result of mysql_errno() if you want the error code as well)

So basicly take
$result = @mysql_query ($query);
$row = mysql_fetch_array ($result, MYSQL_NUM);
and replace it with
$result = mysql_query ($query);
if(!$result)
{
echo(mysql_error());
}
else
{
$row = mysql_fetch_array ($result, MYSQL_NUM);
}

Answer 3

as for me that error message comes because of the fault in your query. I think you should check your query again!

mysql_fetch_array ( resource result [, int result_type])

try to change [ int result_type ] to MYSQL_BOTH as the default value of argument mysql_fetch_array.

hope it will help!
good luck!

Answer 4

You are getting worse by the minute!
160 questions up to now, and you have learnt nothing!
Get a life (in another field than programming)

Answer 5

try youngcoders.com they will help