Error message

[Tezread]

I have an error stating:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #3' at line 1

Im not sure how to deal with this- here is my first set of code:

  <?php 

$user = "root";
$host = "localhost";
$password = "";
$connection = mysql_connect($host, $root, $password) or die ("Couldn't connect to server.");
$database = "courses";
$db = mysql_select_db($database) or die ("Couldn't select database.");

$sql = "SELECT coursecode FROM trainingtopics";
//echo $sql."<br>";

$result = mysql_query($sql) or die(mysql_error());
echo $result."=result<br>";

while($row=mysql_fetch_array($result))
{ // NOTE this one ABOVE the echo
//echo "result found!";
//echo $row[0];
echo "<a href=\"details.php?course=".$row[0]."\">".$row[0]."</a><br>";

}
?>

and here is my processing page called details.php

<?php
$user = "root";
$host = "localhost";
$password = "";
$connection = mysql_connect($host, $root, $password) or die ("Couldn't connect to server.");
$database = "courses";
$db = mysql_select_db($database) or die ("Couldn't select database.");

if (isset($_GET['course'])) {

$course = mysql_real_escape_string($_GET['course']);

$res = mysql_query("SELECT * FROM trainingtopics WHERE coursename = '$course'");

$result = mysql_query($res) or die(mysql_error());
echo $result."=result<br>";

if ($res && mysql_num_rows($res) != 0) {
// get results from query
// output to browser with echo
} else {
//course name not found
//note: the query may have failed to so maybe check for that.
}
} else {
//no course specified
}

?>

any experienced assistance would be greatly appreciated- I have come on leaps and bounds thx to you guys and im nearly there!!

[mrhappiness]

Tezread wrote:

$res = mysql_query("SELECT * FROM trainingtopics WHERE coursename = '$course'");

$result = mysql_query($res) or die(mysql_error());

you pass the result of the 1st mysql-query as new query to the 2nd call of mysql_query?

why?

[Tezread]

not sure why :-)

[Tezread]

here is my table design (just for the record)

coursecode (int 11) primary key
coursename varchar (50)
accreditingbody varchar (50)
aimedat blob
level1 char (50)
level2 char (50)
level3 char (50)

I removed the

$result = mysql_query($res) or die(mysql_error());

and I get no result at all now just a blank page

[kburger]

mrhappiness wrote:

you pass the result of the 1st mysql-query as new query to the 2nd call of mysql_query?

why?

What mrhappiness means is

"Don't do that."

It should be something more along the lines of

$res = "SELECT * FROM trainingtopics WHERE coursename = '$course'";
$result = mysql_query($res) or die(mysql_error());