Question

我有一个显示状态的选择选项我有一个DIV返回所选州的运输成本我的问题是，当我选择一个选项并返回DIV的值时页面会刷新我想显示DIV的值而不刷新整个页面

    <html>
<body>
<div id ="content"> this is the cost of shipping of :
<?php 
    $ship=20;
    if (isset($_POST['city'])){
        $listings = $_POST['city'];
    } else if (isset($_GET['city'])) {
        $listings = $_GET['city'];
    } else {
        $listings = 'city';
    }
    switch ($listings) {
        case 'California':
            $ship=$ship+5; echo $ship;
            break;
        case 'Alabama':
            $ship=$ship+10; echo $ship;
            break;
        case 'Texas':
            $ship=$ship+15; echo $ship;
    }

?>
    </div>
    <form id = "form1" name = "form" >
        <select id ="city" name ="city" method="post"   onchange="this.form.submit()"    >
            <option value="California" <?php echo (isset($_POST['city']) && $_POST['city'] == 'California') ? 'selected="selected"' : ''; ?>>California</option>
            <option value="Alabama" <?php echo (isset($_POST['city']) && $_POST['city'] == 'Alabama') ? 'selected="selected"' : ''; ?>>Alabama</option>             
            <option value="Texas" <?php echo (isset($_POST['city']) && $_POST['city'] == 'Texas') ? 'selected="selected"' : ''; ?>>Texas</option>
    
        </select>
    
    </form>


</body>
</html>

java脚本

   <script type="text/javascript" >
$("#form1").click(function(){
    $("#content").load(" #content");
});
</script>

Answer 1

您需要使用jquery的onchange事件并获取select的值，然后使用ajax将其传递到您的php页面。您需要在代码中进行的更改：

您的 html 代码：

  <!--remove onchange from here-->
  <select id ="city" name ="city">
     <!--same code -->
   </select>

您的 jquery 代码：

//onchange of select
$("#city").change(function() {
  //get select value
  var city = $(this).val();
   console.log(city)
  //ajax call 
  $.ajax({
     url: 'urlofphppage',
    type: 'POST',
    data: {
      'city': city
    },
     success: function(data) {
      alert('Data: ' + data);
      //add data which recieve from server to div
      $("#content").html(data);
    }

  });
});

您的 php 页：

 if (isset($_POST['city'])){
        $listings = $_POST['city'];
   
    switch ($listings) {
        case 'California':
            $ship=$ship+5; 
            echo $ship;//will send back to ajax as response
            break;
        case 'Alabama':
            $ship=$ship+10;
             echo $ship; //will send back to ajax as response
            break;
        case 'Texas':
            $ship=$ship+15;
           echo $ship; //will send back to ajax as response
    }
  }

重新加载div而没有刷新页面

1 个答案: