PHP/MYQSL brings back wrong value (resource ID #3)

**izzysanime** · 09-26-2008, 04:10 PM

HI,

I'm trying to get the highest number from a table, but it brings back "resource ID #3". I have no idea where that comes from, because there is no resource id #3 or even 3 colums. Here is my code

PHP Code:


mysql_select_db("mvy_fomcreds") or die(mysql_error());

$data = mysql_query("SELECT annon FROM `annon` ORDER BY `annon` DESC LIMIT 1")

or die(mysql_error());

$new = $annon + $data;

mysql_select_db("mvy_fomcreds") or die(mysql_error());

mysql_query("INSERT INTO `annon` (`annon`) VALUES ('$new');

");

Print "Thank you for your submission.";

echo "$data";

?>

Thanks,
Josh

**techietim** · 09-26-2008, 04:17 PM

You cannot echo your raw query. You must first grab the information it is returning with a function such as mysql_fetch_array()

**izzysanime** · 09-26-2008, 05:19 PM

OK, i still get the same response. Is this what you meant?

PHP Code:


mysql_select_db("mvy_fomcreds") or die(mysql_error());

$data = mysql_query("SELECT annon FROM `annon` ORDER BY `annon` DESC LIMIT 1")

or die(mysql_error());

$new = $annon + $info['annon'];

mysql_select_db("mvy_fomcreds") or die(mysql_error());

mysql_query("INSERT INTO `annon` (`annon`) VALUES ('$new');

");

Print "Thank you for your submission.";



while($info = mysql_fetch_array( $data ))

{

echo "</head>

<body>";



Print $info['annon'] ;

} 



?>

**izzysanime** · 09-26-2008, 06:03 PM

Hmm, got it working. not really sure where it got #3, I just renamed a few things and it works.

PHP Code:


mysql_select_db("mvy_fomcreds") or die(mysql_error());

$data = mysql_query("SELECT * FROM `annon` ORDER BY `annonCount` DESC LIMIT 1")

or die(mysql_error());





Print "Thank you for your submission.";



while($info = mysql_fetch_array( $data ))

{



echo "</head>

<body> <br />";

$new = 1 + $info['annonCount'];



Print $new ;

} 

mysql_select_db("mvy_fomcreds") or die(mysql_error());

mysql_query("INSERT INTO `annon` (`annonCount`) VALUES ('$new')")

or die(mysql_error());

?>

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