PHP的MySQL的选择查询

Question

i have multiple drop list box that contain data from the database so i want to select the second drop list based on the section of the first one
the first table have these three fields district_id(primary key),district_name, governorate_id(foreign key) second table has these three fields village_id(primary key), village_name, district_id(foreign key)

so i want the user to select the district box first then based on the select district name that have an id i want that the second box display all the village name that have a village.district_id = district.district_id. can anyone help me?? i was selecting each table independent from the other one but i need it to be based on the first one

chunk of the code

function districtQuery(){

$distData = mysql_query("SELECT * FROM districts");

  while($recorddist = mysql_fetch_array($distData)){

     echo'<option value="' . $recorddist['district_name'] .  '">' . $recorddist['district_name'] . '</option>';

  }


}
// function for select by village
function villageQuery(){

//$villageData = mysql_query("SELECT * FROM village");

  $villageData = mysql_query("SELECT village_name FROM village WHERE district_id = ('SELECT district_id FROM districts')") or die (mysql_error());

  while($recordvillage = mysql_fetch_array($villageData)){

     echo'<option value="' . $recordvillage['village_name'] .  '">' . $recordvillage['village_name'] . '</option>';

  }


}

Answer 1

You need to use a server side call and populate the 2nd dropdown based on the selectyion of the first.

php + populate drop down menu on the selection of another

Answer 2

Here i described two types of selected query

$query = "select * from table_name";

above one select the whole table

$query = "select * from table_name where id = 1";

above one select the particular record only