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Call to a member function query() on a non-object

Fatal Error on Line 38 -- at $result

     
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);
?>
10:30 pm on Apr 30, 2012 (gmt 0)



Oh, I think I corrected it now:

$result = $con->query($query);
10:43 pm on Apr 30, 2012 (gmt 0)



No, same problem. Any ideas?
11:29 pm on Apr 30, 2012 (gmt 0)

WebmasterWorld Administrator incredibill is a WebmasterWorld Top Contributor of All Time 10+ Year Member Top Contributors Of The Month



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.
4:13 pm on May 1, 2012 (gmt 0)

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



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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