我如何从php下拉列表中选择并将其转换为变量 - how do I take selection from my php drop down list and turn it into a variable?

how do I take selection from my php drop down list and turn it into a variable?

我试图采取下拉框在PHP和选择一个项目从第一个框。然后，我希望从第一个下拉框中选择项目，并将其放入$lineIDSelection变量中，以便将其插入到下一个下拉框的查询中。我试图过去的主要事情是，当我点击我的下拉框中的一个选项，我需要能够提交它，或者更好的是，有它动态更新到一个变量。这是我正在使用的一段代码。

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die("could not connect to the databse!");
$select_db = mysql_select_db('camdb') or die ('could not select camdb database!!');
    $lineID = "SELECT * FROM camTable;";
    $IDresult = mysql_query($lineID);
            echo"line" . "<br/>";
            echo "<select name='"line'">";
            while ($row = mysql_fetch_array($IDresult)) {
                echo "<option value='" . $row['line'] . "'>" . $row['line'] . "</option>";
            }
            echo "</select>" . "<br/>" . "<br/>";

    $query = "SELECT DISTINCT Length FROM camTable;";
    $result = mysql_query($query);      
            echo"Cam Length" . "<br/>";
            echo "<select name='"Length'">";
            while ($row = mysql_fetch_array($result)) {
                echo "<option value='" . $row['Length'] . "'>" . $row['Length'] . "</option>";
            }
            echo "</select>" . "<br/>" ."<br/>";

Q.I then want the selected item from my first drop down box, and placed it into 
my $lineIDSelection variable so that I can insert it into a query for the next 
drop down box

可以通过Ajax实现:-

on -change在第一个下拉菜单上您将获得值，并基于第一个下拉菜单值，可以生成第二个下拉列表。

PHP:- assign id给下拉列表

 echo "<select name='line' id='dropDown1'>";

Jquery:-

$('#dropDown1').change(function() {    
    var sel_item=$(this).val();
     // sel_item is first dropdown selected value.
     // pass it to ajax request 
         // in you ajax request get sel_item and assign it to $lineIDSelection

});