Wat gaat er niet goed?

Door Rene Zwolsman op 21-04-2007 18:10

277 views

Ik heb de volgende code:

<?php
include("db_connect.php");

function display_MeanTemp($date)
{
$MeanTemp = mysql_query("SELECT AVG(Tempout) AS GemTemp FROM weerdata WHERE Date='$date'", $db);
if ($MeanTemp) {
while ($MeanTemp_result = mysql_fetch_array($MeanTemp)){
$MeanTemp_result = round($MeanTemp_result['GemTemp'],1);
echo "<td>". $MeanTemp_result . "</td>";
}
}
else {
echo "$error";
}
}

$date = "2007-04-17";
display_MeanTemp($date);
?>

Bij het uitvoeren ervan krijg ik de foutmelding:
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\wamp\www\pages\upload\testtabel.php on line 6

Waar zit de fout?

Jan Koehoorn

21-04-2007 18:16

Je variabele $db is geen geldige.
[edit]Zo zal hij het wel doen:[/edit]
<?php
require 'db_connect.php';

function display_MeanTemp ($date) {
$MeanTemp = mysql_query("SELECT AVG(Tempout) AS GemTemp FROM weerdata WHERE Date='" . $date . "'");
if ($MeanTemp) {
while ($MeanTemp_result = mysql_fetch_array($MeanTemp)) {
$MeanTemp_result = round($MeanTemp_result['GemTemp'],1);
echo "<td>". $MeanTemp_result . "</td>";
}
}
else {
echo "$error";
}
}

$date = "2007-04-17";
display_MeanTemp($date);
?>

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