form values in db

[derksj]

help!!!

i have this code to make a form and the form works

<?

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

echo "<select name=\"test\" MULTIPLE size=\"8\">";

while ($rows=mysql_fetch_array($result))
{
$option = $rows['LandID'];
$Land = $rows['LandNaam'];

echo "<option value\"$option\">$Land</option>";
}

echo "</select>";
?>

This is the code to put the values into the database

<?php

$db = mysql_connect("intranet.hr.nl", "login", "password");
mysql_select_db("vakantiehuizengids", $db);

$sql1 = "insert into tblPlaatsNaamWoning (PlaatsNaamWoning) values ('$PlaatsNaamWoning')";
$result1 = mysql_query($sql1);
$insert_value1 = mysql_insert_id();

$sql2 = "insert into tblAdresWoning

(PlaatsNaamWoningID,LandID,PostcodeWoning,LiggingWoning,HuisNummerWoning,LocatieBeschrijvingWoning,StraatNaamWoning)

values

('$insert_value1','$LandID','$PostcodeWoning','$LiggingWoning','$HuisNummerWoning','$LocatieBeschrijvingWoning','$StraatNaamWoning')";
$result2 = mysql_query($sql2);

print ("de database is gevuld<br>\n");

?>

But the LandID doesnt show into the DB

please help

[The_Chancer]

You have assigned the value of $option to have $LandID, so when the form gets processed $LandID has not been passed, so will have a null value.

If you use the name of the select box ie 'test' then you will get the value of the option selected from the select box.

I.E. code should read...

$sql2 = "insert into tblAdresWoning 

(PlaatsNaamWoningID,LandID,PostcodeWoning,LiggingW
oning,HuisNummerWoning,LocatieBeschrijvingWoning,S
traatNaamWoning) 

values 

('$insert_value1','$test,'$PostcodeWoning','$Li
ggingWoning','$HuisNummerWoning','$LocatieBeschrij
vingWoning','$StraatNaamWoning')"; 
$result2 = mysql_query($sql2);

Should work...

[derksj]

thanx

it works