Need to get more fields from database

[jfaust97]

I have downloaded a prepackaged job listing script.
I need some help making it display more of the fields in the database than is in the current code. Since I am new to PHP and coding in general.... I am fuzzy on where to start...

The code in question is.....

#####################

Listing open jobs

#####################

if ($cmd =="List"){

### Get List of Open Jobs ###
    $query = "SELECT DISTINCT id, location 
                from location, listing 
                WHERE location.id = listing.fk_location";
    $result = mysql_db_query($database, $query, $connection) or die ("Error in query: $query. " . mysql_error());

### Check each department ###
    while(list($id, $location) = mysql_fetch_row($result))
    {
    ### Printing department ###
        echo "<b>$location</b>";

    ### Look for jobs in each department and print as list ###
        $query2 = "SELECT jcode, designation
                    from listing 
                    WHERE listing.fk_location = '$id'";
        $result2 = mysql_db_query($database, $query2, $connection) or die ("Error in query: $query2. " . mysql_error());

        echo "";
        while(list($jcode, $dsg) = mysql_fetch_row($result2))
        {
        echo "<li><a href=index_location.php?cmd=Details&jcode=$jcode>$dsg ($jcode)</a>";
        }
        echo "";
        echo "<br><br><br>";
        }

}

The variable names are ....

jcode,
department,
location,
salary,
responsibilities,
qualifications,
cname,
cmail,
posted

The Database is called Jobs and the table is called listing

I am confident I know how to do the echo part of it....
The problem is in my query...
I don't know how to structure the SELECT statement to get the variables correctly....

Thanks for any help....

[voidtrance]

Originally posted by jfaust97
### Look for jobs in each department and print as list ###
$query2 = "SELECT jcode, designation
from listing
WHERE listing.fk_location = '$id'";

If all the fiels are int he same table, just substitue the above with:

$query2 = "SELECT jcode,department, location,salary,responsibilities,qualification,cname,cmail,posted FROM listing WHERE listing.fk_location = '$id';