[SOLVED] Unpacking an array results negative

[GreenUser]

Hi all, currently line 30 is not producing anything.  The results are the same if I use 'echo $avgrow->id;'

 

PHP Notice:  Undefined index:  id in

C:\\Program Files\\Apache\\htdocs\\ratemydate\\percent.php on line 6

 

PHP Notice:  Trying to get property of non-object in C:\\ProgramFiles\\Apache\\htdocs\\ratemydate\\percent.php on line 30

 

 

I played around for a while with the sql and in phpmyadmin I do get results I want, I just need to get them to display, not sure what I am doing wrong.

 

<?php

require("config.php");
require("functions.php");

if(pf_check_number($_GET['id']) == TRUE) {
$validid = $_GET['id'];
}
else {
$validid = 21;
}

require("header.php");

$sql = "SELECT * FROM stories WHERE id = " . $validid . ";";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);

echo "<h1>" . $row['location'] . "</h1>";
echo nl2br($row['body']);

$avgsql = "SELECT SUM(fun + soh + con + sma + sex + hyg + mes) FROM stories WHERE id = " . $validid . ";";
$avgresult = mysql_query($avgsql);
$avgrow = mysql_fetch_array($avgresult);

echo "<p>";
echo "<strong>Rating</strong>: ";


echo $avgrow['id'];

echo "</p>";

require("footer.php");

?>

[DarkWater]

You never actually pull "id" out of the database.  And the second error only occurs when you use $avgrow->id because $avgrow isn't an object.

[GreenUser]

Okay, with what $avgrow is set to it won't work then.  Not sure that the 'id' is so important at this step, just trying to get the info out of the db and onto the page.

[DarkWater]

I'd change the query to:

 

$avgsql = "SELECT SUM(fun + soh + con + sma + sex + hyg + mes) AS total FROM stories WHERE id = " . $validid . ";";

 

Then $avgrow['total'] will work as the sum.

[GreenUser]

Thank you Darkwater.  AS more or less sets 'total' as a variable?

 

It was beautiful to see the number 35 finally display!  Very good, thank you.

[DarkWater]

AS is used to create aliases in MySQL.  You basically need to create an alias for any function returns unless you want to use the numerically indexed array.  Also, aliases are good for something like:

 

SELECT u.userid AS uid FROM users as u, friends as f WHERE f.userid = $id AND f.friendid = u.userid;