1

我正在尝试填写第二个下拉列表，但它没有填充过滤器数据，但它根据第一个下拉列表中选择的国家/地区获取所有州，而不是仅相关州请告诉我代码中有什么问题。找不到 id 的值，因此所有州名都已填写在下拉列表中。

               <script type="text/javascript" src="jquery-ajax/jquery.js"></script>
<script>
$(document).ready(function(){             

    $("#country").change(function(){            

        //var optionValue = $("select[name='country_select']").val();      
       var id=$("#country").val(); 
        //var dataString = 'id='+ id;          
          //alert("datastring"+dataString); 

        $.ajax({
            type: "POST",
            url: "getstate.php",
            data: {Country_Id:id},
            beforeSend: function(){ $("#ajaxLoader").show(); },
            complete: function(){ $("#ajaxLoader").hide(); },
            success: function(response){

                $("#stateAjax").html(response);
                $("#stateAjax").show();

        }

        });        
    });

});
</script>

</head>
<?php
include("connection.php");
?>
<body>
<form method="post">
Select Country:<select name="country_select" id="country">
    <option value="">Select Country</option>
<?php
$cntry = mysql_query("SELECT Country_Id, name FROM country ORDER BY name ASC");
while($row = mysql_fetch_array($cntry))
{
    $id=$row['Country_Id'];
    $name=$row['name'];
echo '<option value="'.$id.'">'.$name.'</option>';
}
?>
    </select>
 <span id="errmsg" style="display:none">There is no matching option</span>
  <div id="ajaxLoader" style="display:none"><img src="jquery-ajax/ajax-loader.gif" alt="loading..."></div>
  <div id="stateAjax" style="display:none">
  <select name="state_select" id="state" style="display:none">
  <option value="">Please Select</option></select></select>
  </div>
</form>   
</body>

        getstate.php


    <?php

    echo $_POST['id'];

    include("connection.php");

    $cntry = mysql_query("SELECT * FROM state where country_id=".$_POST['id']);
    ?>
     State: 
    <select name="state_select" id="state">
    <option value="">Please Select</option>
    <?php
    while($row = mysql_fetch_array($cntry))
    {
        $id=$row['State_id'];
        $name=$row['name'];
    echo '<option value="'.$id.'">'.$name.'</option>';

    }

    ?>
    </select>
4

1 回答 1

1 
$.ajax({
                type: "POST",
                url: "getstate.php",
                data: {country_id:id},
                ...

编辑1：也

var id=$('#country').val();

编辑 2：getstate.php 

include("connection.php");

$country_Id = intval($_POST['Country_Id']);//if country_Id is integer...if not:$country_Id = $_POST['Country_Id'];
$cntry = mysql_query("SELECT * FROM state where country_id=".$country_Id);
于 2013-04-03T03:20:40.320 回答