help with values drop down menu in database

derksj

What am i doing wrong
the values wont appear in the database

#my first dropdown menu
.....

<?
$db = mysql_connect("intranet.hr.nl", "login", "passw");
mysql_select_db("vakantiehuizengids",$db);
$result = mysql_query("SELECT * from tblLanden");

while ($rubriek = mysql_fetch_array($result))
{

echo "<option value = \"".$rubriek['LandID']."\"".$selected.">".$rubriek['LandNaam']."</option>\n";
}

?>
</select>

...
...

#my second dropdown menu

<?
$db = mysql_connect("intranet.hr.nl", "login", "passw");
mysql_select_db("vakantiehuizengids",$db);
$result = mysql_query("SELECT * from tblLanden");

while ($rubriek = mysql_fetch_array($result))
{
echo "<option value = \"".$rubriek['LandID']."\"".$selected.">".$rubriek['LandNaam']."</option>\n";
}

?>
</select>

My formhandler

$sql3 = "insert into tblAdresWoning
(PlaatsWoningID,LandID2,PostcodeWoning,HuisNummerWoning,LocatieBeschrijvingWoning,StraatWoning)
values
('$insert_value1','$LandID','$PostcodeWoning','$HuisNummerWoning','$LocatieBeschrijvingWoning','$StraatWoning')";
$result3 = mysql_query($sql3);
$insert_value3 = mysql_insert_id();

$sql4 = "insert into tblAdresContactPersoon
(PlaatsContactPersoonID,LandID1,PostcodeContactPersoon,HuisNummerContactPersoon,StraatContactPersoon)
values
('$insert_value2','$LandID','$PostcodeContactPersoon','$HuisNummerContactPersoon','$StraatContactPersoon')";
$result4 = mysql_query($sql4);
$insert_value4 = mysql_insert_id();

michaelk

if you want to try and debug it, i always find it easier to seperate all of my items so they are easy to follow, and you can find the errors faster.

also, when making your query, you need to provide the connection. (see $result below)

i would do your drop down box, more like this:

<select name="LandID1" size="1"> 

<? 
$db = mysql_connect("intranet.hr.nl", "login", "passw"); 
$select_db = mysql_select_db("vakantiehuizengids",$db); 

# if you don't need all the values, why select them all?
$sql = "SELECT LandID, LandNaam FROM tblLanden";

$result = mysql_query($sql,$db); 

while ($rubriek = mysql_fetch_array($result)) { 

  $LandID = $rubriek['LandID']; 
  $LandNaam  = $rubriek['LandNaam'];

  echo "<option value=\"$LandID\">$LandNaam</option>\n"; 

} 

?> 

</select>

the result of this drop down box is that after making a selection and submitting a form you will have a variable called $LandID1 and it will have a value of whatever $LandID is for that selected value. Make sure you are submitting LandID1 into your table in the right spot.

for your queries, make sure all of the arguments are being supplied, somewhere else in the form. but with that drop down box, you should get $LandID put into the database, when submitted.

derksj

thanx

mattotto

The code looks good have you got some extra code that would set the default value of the form if the record already existed in the database.

Cheers