mysql_fetch_array not valid mysql result

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.

You must see the line 33 of your script.

I asume that is

"$sql = mysql_query("SELECHT * FROM jaren
WHERE jaar=$jaar"); "

if you see, you write SELECHT and the correct statement es SELECT

Best Regards.

Martin Mantaras

Oh my god... That's imbarissing :D Thanx a lot!

In the other hand.
I see you need know the num of records (in the next line)
This can be obtaining with the mysql_num_rows function.

http://php.net/manual/en/function.mysql-num-rows.php

Best Regards again.

Applying the last comment.
Your code can't be

<?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");

if($jaar!="")
{
$sql = mysql_query("SELECT * FROM jaren WHERE jaar=".$jaar);
$numofrecords=mysql_num_rows($sql);
if($numofrecords == 0)
{
echo "Over het jaar ".$jaar." is nog geen informatie beschikbaar.";
} else {
while($row = mysql_fetch_array($sql)){
echo "Datum: ".$row['datum'];
}
}
}
echo "</div>";
?>


Best Regards again