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Call to a member function query() on a non-object
Fatal Error on Line 38 -- at $result
jdial



 
Msg#: 4447711 posted 10:26 pm on Apr 30, 2012 (gmt 0)

Here's my code:

<?php

// create short variable names
$state=$_POST['state'];



if (!$state)
{
echo 'You have not entered all the required details.<br />'
.'Please go back and try again.';
exit;
}
if (!get_magic_quotes_gpc())
{
$state = addslashes($state);
}



$con = mysql_connect("localhost","qfs","abc123","applicants");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

$query = "insert into applicants values
('".$state."')";

$result = $db->query($query);
if ($result)
echo $db->affected_rows.' application inserted into database.';

mysql_close($con);
?>

 

jdial



 
Msg#: 4447711 posted 10:30 pm on Apr 30, 2012 (gmt 0)

Oh, I think I corrected it now:

$result = $con->query($query);

jdial



 
Msg#: 4447711 posted 10:43 pm on Apr 30, 2012 (gmt 0)

No, same problem. Any ideas?

incrediBILL

WebmasterWorld Administrator incredibill us a WebmasterWorld Top Contributor of All Time 5+ Year Member Top Contributors Of The Month



 
Msg#: 4447711 posted 11:29 pm on Apr 30, 2012 (gmt 0)

Why are you trying to use some query function as if it were from some object?

You want mysql_query() which will accept $con as a variable.
[php.net...]

I'm thinking you used some database class library that contained query() in the past and you forgot to include that class library as part of your current code.

rocknbil

WebmasterWorld Senior Member rocknbil us a WebmasterWorld Top Contributor of All Time 10+ Year Member



 
Msg#: 4447711 posted 4:13 pm on May 1, 2012 (gmt 0)

Wouldn't that give an "undefined function:query" error though?

$result = $db->query($query);

What is $db? It should be your database object. You make the connection, but fail to link to it. How does $db become assigned as the database object?

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