[SOLVED] supplied argument is not a valid MySQL-Link resource

[RichiT81]

Hi,

 

I am new to this, I am using a Head First PHP & MySQL book and am working on my second task which is inserting data entered via a web form into a MySQL database.

 

I have tested the MySQL statement direct in an SQL query and it works fine, but when ever i submit the form I get the following error:

 

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/research/public_html/phptest/aliens/report.php on line 32

Error querying database.

 

 

Here is my php code:

<?php 
$first_name = $_POST['firstname']; 
$last_name = $_POST['lastname']; 
$when_it_happened = $_POST['whenithappened']; 
$how_long = $_POST['howlong']; 
$how_many = $_POST['howmany']; 
$alien_description = $_POST['aliendescription']; 
$what_they_did = $_POST['whattheydid']; 
$fang_spotted = $_POST['fangspotted']; 
$other = $_POST['other']; 
$email = $_POST['email']; 

$dbc = mysql_connect('localhost', 'research_admin', '*****', 'research_aliendatabase') 
or die('Error connecting to MySQL server.'); 

$query = "INSERT INTO aliens_abduction (first_name, last_name, when_it_happened, how_long, " . 
"how_many, alien_description, what_they_did, fang_spotted, other, email) " . 
"VALUES ('$first_name', '$last_name', '$when_it_happened', '$how_long', '$how_many', " . 
"'$alien_description', '$what_they_did', '$fang_spotted', '$other', '$email')"; 

$result = mysql_query($dbc, $query) 
or die('Error querying database.'); 

mysql_close($dbc); 

echo 'Thanks for submitting the form.<br />'; 
echo 'You were abducted ' . $when_it_happened; 
echo ' and were gone for ' . $how_long . '<br />'; 
echo 'Number of aliens: ' . $how_many . '<br />'; 
echo 'Describe them: ' . $alien_description . '<br />'; 
echo 'The aliens did this: ' . $what_they_did . '<br />'; 
echo 'Was Fang there? ' . $fang_spotted . '<br />'; 
echo 'Other comments: ' . $other . '<br />'; 
echo 'Your email address is ' . $email; 
?>

 

Any help would be greatly appreciated as I am at a loss!

 

Thanks in advance.

 

Richi

 

[kickstart]

Hi

 

Looks like the parameters are the wrong way round on the mysql_query:-

 

Try this:-

 

$result = mysql_query($query,$dbc)

or die('Error querying database.');

 

All the best

 

Keith

[RichiT81]

Thanks Keith,

 

I have tried your suggestion and now get the following message:

 

Error querying database.

 

[RichiT81]

Forget that, got it working!

 

Thanks very much for your help!