Drop Down Box Error

[artisticre]

I am having a problem with drop down boxes in mysql

here is my code for the select box it works.....

<?
// script to display all the Employees in the Employees table

// connection information
$hostName = "xxxxxxx";
$userName = "xxxxxxx";
$password = "xxxxxxx";
$dbName = "xxxxxxxxxxxxxxxxx";

// make connection to database
mysql_connect($hostName, $userName, $password) or die("Unable to connect to host $hostName");

mysql_select_db($dbName) or die("Unable to select database $dbName");

// Select all the fields in all the records of the Employees table
$query = "SELECT distinct type, typeid
FROM Type
ORDER BY type";
$result = mysql_query($query);

// Determine the number of employees
$number = mysql_num_rows($result);

// Create drop-down menu of employee names

print "View inventory records for an employee:<p>
<form action=\"winerydisplay.php\" method=\"post\">
<select name=\"typeID\">
<option value=\"\">Select a wine</option>";

for ($i=0; $i<$number; $i++) {
$typeID = mysql_result($result,$i,"typeID");
$type = mysql_result($result,$i,"type");
print "<option value=\"$typeID\">$type</option>";
}

print "</select><input type=\"submit\" value=\"submit\"
name=\"submit\"></form>";

// Close the database connection
mysql_close();

?>

here is the code for my result page that doesn't work
<?
// script to display who has which computers

// connection information
$hostName = "localhost";
$userName = "xxxxxx";
$password = "xxxxxx";
$dbName = "xxxxxxxxxxxxxxxx";

// make connection to database
mysql_connect($hostName, $userName, $password) or die("Unable to connect to host $hostName");

mysql_select_db($dbName) or die( "Unable to select database $dbName");

// Select the fields from the appropriate tables
$query =
"SELECT * FROM Specialty, WineName, Label, Varietal, Type, Producer, Region, Country, Rating, Available, CasePrice, BottlePrice
WHERE
(Specialty.typeid = $typeID) and
(Specialty.winenameid = WineName.winenameid) and
(Specialty.labelid = Label.labelid) and
(Specialty.varietalid = Varietal.varietalid) and
(Specialty.typeid = Type.typeid) and
(Specialty.producerid = Producer.producerid) and
(Specialty.regionid = Region.regionid) and
(Specialty.countryid = Country.countryid) and
(Specialty.ratingid = Rating.ratingid) and
(Specialty.availableid = Available.availableid) and
(Specialty.casepriceid = CasePrice.casepriceid) and
(Specialty.bottlepriceid = BottlePrice.bottlepriceid";

$result = mysql_query($query);

// Determine the number of records returned
$number = mysql_numrows($result);

// Print the relevant information

print "There are $number records in the inventory:<p>";

for($i=0; $i<$number; $i++) {
$producer = mysql_result($result, $i, "producer");
$varietal = mysql_result($result, $i, "varietal");
print "$producer, $varietal<br>";
}

// Close the database connection
mysql_close();
?>

[sarahk]

You don't make it clear where you have a problem

looking at your query you are missing a ) on the last table join.

If you echo out the query and paste into phpMyAdmin (or similar) does the query run? If not it should give you an idea of how to fix it.

Please repost with an idea of where the problem lies. You give alot of code but no idea where it falls over.

[jpmoriarty]

and also please use the php tags [php ] xxx [/php ] to surround your code so that it's easy to read

(just dont put in the spaces after the final p - i'm still trying to figure out how to put them in without having it turn my text into php text!)