我有一个由3个下拉列表组成的页面：代理商网络，代理商名称和客户名称。当用户选择网络时，它将在查询中用于创建代理商下拉列表。当用户选择代理时，它将用于查询以创建客户下拉列表。由于我前几天收到的一些帮助，创建下拉列表似乎工作正常。我现在的问题是持续选择，以便我可以根据网络，代理商和客户创建查询。
这是我目前的代码。
if (!isset($_POST['network'])){
echo "Form error";
} else
{
$message = "form was successfully submitted";
//echo "{$message} <br />";
$network = $_POST['network'];
// echo "<br /> the network is :{$network}";
// echo "<br />";
}
?>
<?php
//**********************************************
// query DB for NETWORK list
//**********************************************
$net = "";
$agency = "";
$query = "SELECT network_id, network_code FROM pps_agency_network";
$result = mysqli_query($connection, $query);
// create the html for the drop down menu for NETWORK
echo "<body>";
echo "<div id='network_name' class='col-md-3'>";
echo "<h2> Agency Network </h2>";
echo "<form action='droplistpop6b.php' method='post'>";
echo "<select name='network'>";
echo '<option value="*">ALL</option>';
while($result1 = mysqli_fetch_assoc($result)) {
// Populate the network dropdown list
unset($network_id, $network_name);
$network_id = $result1['network_id'];
$network_name = $result1['network_code'];
echo $network_name;
echo '<option
value="'.$network_id.'">'.$network_name.'</option>';
}
echo "</select>";
// send the value that was selected from the dropdown list
echo "<input name='submit' type='submit' value='Send' />";
//echo "</form>";
echo "</div>";
if(!isset($_POST['network'])){
//echo "network key was not set <br />";
} else {
$net = $_POST['network'];
?>
$ net变量包含为网络选择的值。在下面的代码片段中，我尝试将此值保存到Javascript中的本地存储。但是，它似乎没有工作
<script>
// ADDED JAVASCRIPT TO SAVE TO LOCAL STORAGE
window.onload = function() {
var net = "<?php echo $net ?>"; // "A string here"
if (localStorage) {
// Save the contents of net in locaStorage
localStorage.setItem('net', net);
alert("net is" + net);
});
else {
// local storage not supported
alert "local storage not supported";
}
}
</script>