通过AJAX选择更改提交表单

Question

假设我有这样的形式:

<form action="Change-status.php" method="post">
        <select class="changeStatus" name="changeStatus">
                <option value="0">Starting</option>
                <option value="1">Ongoing</option>
                <option value="2">Over</option>
        </select>
        <input class="projectId" type="hidden" name="projectId" value="<?php echo $data['id'];?>"/>
</form>

我目前正在使用此脚本提交表单,但它意味着刷新:

$('select').change(function ()
{
    $(this).closest('form').submit();
});

我想要做的是在不刷新页面的情况下在选择更改时发送表单.我知道我必须使用AJAX这样做,但我无法弄清楚如何实现它.

你能指导我怎么做吗？

谢谢你的帮助.

编辑:

在考虑了意见后,我最终得到了以下代码:

Html:

<form action="" method="post">
        <select class="changeStatus" name="changeStatus">
                <option value="0">Starting</option>
                <option value="1">Ongoing</option>
                <option value="2">Over</option>
        </select>
        <input class="projectId" type="hidden" name="projectId" value="<?php echo $data['id'];?>"/>
</form>

JS:

$(document).ready(function() {
    $('select.changeStatus').change(function(){
        $.ajax({
                type: 'POST',
                url: 'Change-status.php',
                data: {selectFieldValue: $('select.changeStatus').val(), projectId: $('input[name$="projectId"]').val()},
                dataType: 'html'
         });
    });
});

PHP:

<?php
    include('../Include/Connect.php');

    $changeStatus=$_POST['selectFieldValue'];
    $id=$_POST['projectId'];

    $sql='UPDATE project SET progress="'.$changeStatus.'" WHERE id="'.$id.'"';

    mysql_query($sql) or die("Erreur: ".mysql_error());
?>

Answer 1

获取跨浏览器onchange事件和AJAX请求的工作并非易事.我建议你使用某种类型的javascript框架,它抽象出所有的跨浏览器问题,所以你不必担心它们.

试试js框架

Jquery只是一个这样的框架,它有一些方法,例如.change()将处理程序附加到类似的元素的更改事件,<select>并.get()执行GET请求.

这里有一些代码可以帮助您入门: -

// The $ is the shorthand for a jquery function, you can then use jquery 
// selectors which are essentially the same as css selectors, so here
// we select your select field and then bind a function to 
// it's change event handler
$('select.changeStatus').change(function(){

    // You can access the value of your select field using the .val() method
    alert('Select field value has changed to' + $('select.changeStatus').val());

   // You can perform an ajax request using the .ajax() method
   $.ajax({
       type: 'GET',
      url: 'changeStatus.php', // This is the url that will be requested

      // This is an object of values that will be passed as GET variables and 
      // available inside changeStatus.php as $_GET['selectFieldValue'] etc...
      data: {selectFieldValue: $('select.changeStatus').val()},

      // This is what to do once a successful request has been completed - if 
      // you want to do nothing then simply don't include it. But I suggest you 
      // add something so that your use knows the db has been updated
      success: function(html){ Do something with the response },
      dataType: 'html'
    });

});

一些参考文献会比我的解释更好

请注意,要使此代码生效,您需要在页面上包含带有<script>标记的jquery库.

有关使用jquery的快速入门指南,请参见此处

这里有一个关于如何使用jquery ajax()方法的初学者教程