Question

<div>
<script type="text/javascript">
$(document).ready(function(){
$('#selectstate').change(function() {
window.location = "dropexmpl.php?stateval=" + $(this).val();
});
$('#selectcity').change(function() {
        window.location = "dropexmpl.php?cityval=" + $(this).val();
    });
});
</div>
</script>
</head>
<body >
<div>
<?php
//state drop down
    echo "<td"." id="."id_sel_state".">";
    $con=mysql_connect("localhost","root","");
        mysql_select_db("data_filter",$con);
    $query=("SELECT id,state_name FROM state_details ");
    $result = mysql_query ($query); 
    echo "<label id='statelab'>Select state : </label>"."<select name='filterstate'id='selectstate' onchange='getstateval()'>";

    while ($nt = mysql_fetch_array($result)){
    echo "<option value='".$nt['id']."'>".$nt['state_name']."</option>";
    }
    echo "</select>";
//city drop down
    $statestrval= $_GET['stateval'];
    echo "<td id='id_sel_city'>";
    $query1="SELECT id,state_id,city_name FROM city_details WHERE state_id=".$statestrval."";                                                                               $result1 = mysql_query ($query1);
    echo "<label>Select city : </label>"."<select name='filtercity' id='selectcity' >";
    while ($nt = mysql_fetch_array($result1)){
    echo "<option value='".$nt['id']."'>".$nt['city_name']."</option>";
    }
    echo "</select>";

//Zone drop down

    $citystrval= $_GET['cityval'];
    echo "<td id="."id_sel_zone".">";
    $query2="SELECT id,city_id,zone_area FROM zone_details WHERE city_id=".$citystrval."";                          

    $result2 = mysql_query ($query2);
    echo "<label>Select industry zone : </label>"."<select name='filterzone' id='selectzone'>";
    while ($nt = mysql_fetch_array($result2)){
    echo "<option value='".$nt['id']."'>".$nt['zone_area']."</option>";
    }
    echo "</select>";
    mysql_close($con);
?>
</div>
</form>
</body>
</html>

这里我使用3下拉状态city和zone.it工作，但是当页面重新加载window.location代码时，它向我显示具有刷新效果的页面如何使用ajax解决它。它应该像我选择状态时应该显示与该状态相关的城市列表。当城市被选中时，它应该显示与该城市相关的区域。但问题是当我去选择城市它改变状态列表当我选择区域时，它会更改城市列表，因为重新加载页面我可以解决这个问题。

Answer 1

我建议你使用AJAX，jQuery并避免页面刷新查看jQuery文档：http://api.jquery.com/jQuery.getJSON/

如何使用mysql数据库在同一个php页面中获取下拉值

1 个答案: