使用AJAX从PHP文件获取响应

Question

所以这是我的问题，我正在使用AJAX（jQuery）将表单发布到，process.php但是页面实际上需要回显诸如apple或的响应plum。我不确定如何从中获取响应process.php并将其存储为变量...

这是我到目前为止的代码：

<script type="text/javascript">

        function returnwasset(){

            alert('return sent');

            $.ajax({

                type: "POST",

                url: "process.php",

                data: somedata;

                success function(){

                    //echo what the server sent back...

                }

            });

        }

    </script>

我还需要process.php在json中回显响应吗？还是纯文本会好？

抱歉，这听起来像是一个愚蠢的问题，这是我第一次在Ajax中进行此类操作。

PS：如何在上述代码中命名POST请求？

Answer 1

好的做法是这样使用：$.ajax({&nbsp; &nbsp; type: "POST",&nbsp; &nbsp; url: "/ajax/request.html",&nbsp; &nbsp; data: {action: 'test'},&nbsp; &nbsp; dataType:'JSON',&nbsp;&nbsp; &nbsp; success: function(response){&nbsp; &nbsp; &nbsp; &nbsp; console.log(response.blablabla);&nbsp; &nbsp; &nbsp; &nbsp; // put on console what server sent back...&nbsp; &nbsp; }});而PHP部分是：<?php&nbsp; &nbsp; if(isset($_POST['action']) && !empty($_POST['action'])) {&nbsp; &nbsp; &nbsp; &nbsp; echo json_encode(array("blablabla"=>$variable));&nbsp; &nbsp; }?>

Answer 2

<script type="text/javascript">&nbsp; &nbsp; &nbsp; &nbsp; function returnwasset(){&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; alert('return sent');&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $.ajax({&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; type: "POST",&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; url: "process.php",&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; data: somedata;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; dataType:'text'; //or HTML, JSON, etc.&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; success: function(response){&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; alert(response);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; //echo what the server sent back...&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; });&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; </script>

Answer 3

<?php echo 'apple'; ?> 实际上，这是服务器上所需的全部内容。至于JS端，服务器端脚本的输出将作为参数传递给成功处理程序函数，因此您需要success: function(data) {&nbsp; &nbsp;alert(data); // apple}