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View Full Version : a selection in a list to populate another list on the same form



husnamina
02-25-2007, 08:21 PM
Hi there every one,
Please i have a form with 3 dropdown list.
the first is for a canditate, 2nd is for party, 3rd is for state and 4th is for the local areas in the state.
I want if the user selects a state, the dropdown list for the local area to be populated automatically.
But since they are on the same form i am not sure how.here is my code

<td><select name="state" tabindex="3">
<?php #this is to get listing for roles in a drop dowm menu
#the sql query
$sql1 = "SELECT distinct(state_name) FROM state order by state_name ";
#execute query
$rs1 = mysql_query($sql1, $myConn) or die ("could not execute query $sql");

if ($rs1){
while ($row1 = mysql_fetch_array ($rs1)){
echo("<option>" . $row1['state_name'] . "</option>");
}
}
else{
echo "<option> No states have been registered - Add Please </option>";
}

mysql_free_result($rs);


?>
</select></td>
</tr>
<?php $stid= $row1['state_name']; ?>* here is what i thot of
<tr>
<td> <label> Select Local Goverment </label></td>
<td><select name="localgov" tabindex="4">
<?php
$sql2 = "SELECT distinct(name) FROM lg order by name where state_id=(select state_id from state where state_name=\"$stid\")";
#execute query
$rs2 = mysql_query($sql2, $myConn) or die ("could not execute query $sql");

if ($rs2){
while ($row2 = mysql_fetch_array ($rs2)){
echo("<option>" . $row2['name'] . "</option>");
}
}
else{
echo "<option> No local governemnts have been registered - Add Please </option>";
}

mysql_free_result($rs);


?>
</select></td>
</tr>

Thank you.
:confused:



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