Hey,

I have a problem:

I have this code in my website:
PHP Code:
<?php

$jaar 
$_GET["jaar"];

echo 
"<div id='header'>Vriendengroep Oerdegelijk | ".$jaar."</div>";
echo 
"<div id='main'>";

mysql_connect("fdb3.agilityhoster.com","784975_years","oerdegelijk");
mysql_select_db("784975_years");

$sql mysql_query("SELECHT * FROM jaren
WHERE jaar=
$jaar");
$result mysql_query("SELECT count(*) as bestaat FROM jaren WHERE jaar=$jaar");
$rij mysql_fetch_assoc($result);

if(
$rij['bestaat'] != 1){
echo 
"Over het jaar ".$jaar." is nog geen informatie beschikbaar.";
} else {
while(
$row mysql_fetch_array($sql)){
echo 
"Datum: ".$row['datum'];
}
}
echo 
"</div>";
?>
What I want the code to do is to check in the database if there's a row with the year(jaar) as the variable $jaar, when it does it has to

PHP Code:
echo "Over het jaar ".$jaar." is nog geen informatie beschikbaar."
and else it has to give the date that's in the same row as the year.

But I continue getting this error:

Code:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /srv/disk7/784975/www/oerdegelijk.agilityhoster.com/jaren.html on line 33
Can anyone explain me what I'm doing wrong, and give the solution?

p.s. I'm dutch, so when you need any translations, just say.