Question

我尝试用ajax提交我的表单而没有重新加载整个页面，但是ajax不能正常工作，这里是我使用的代码

$("#sendingform").submit(function(){

        var mess  = $("#urmess").val();
        var mid = "1";
        alert (mid);
        $.ajax({
            url: 'ajax/send.php',
            data: { mid: mid, mess: mess},
            success: function (data){
                alert(data);
            }
        });

        return false;
    });

html是

<hr/>
<form action="#" id="sendingform" method="post">
   <textarea id="urmess" class="conposer" name="messtxt"></textarea>
   <input type="submit" class="sendbtn" name="go" class="send" value="Send"/>

 </form>

Answer 1

试试这个

$(function(){ /* Execute when the DOM is ready */
     $(document).on("submit","#sendingform",function(e){
        e.preventDefault(); /* Prevent the default action of the form */
        var mess  = $("#urmess").val(); /* Get the textarea value */
        var mid = "1";
        alert (mid); /* Alert 1 */
        $.ajax({
            url: 'ajax/send.php',
            data: { 
                 mid: mid, 
                 mess: mess
            }, /* Define AJAX data */
            success: function (data){
                alert(data); /* Alert return */
            }
        });
    }); 
});

Answer 2

Here's The Code Hows Your PHP Code Must Look Like : 
<?php
$a=json_decode($_POST["Jdata"]);
$con=mysql_connect("localhost","root","");
mysql_select_db("json",$con);
$t=$a->mess;
$u=$a->mid;
$x=mysql_query("insert into test(data1,data2) values('$t','$u')");
?>

And Here's HTML+JQuery Code 
<html>
<head>
<script src="http://code.jquery.com/jquery-1.10.2.min.js"></script>
</head>
<body>
   <form action="#" id="sendingform" method="post">
   <textarea id="urmess" class="conposer" name="messtxt"></textarea>
   <input type="submit" class="sendbtn" name="go" class="send" value="Send"/>
   </form>
   <script>
        $("#sendingform").submit(function(){
    var mess  = $("#urmess").val();
        var mid = "1";
        alert (mid);
        $.ajax({
             type: "POST",
             url: 'ajax/send.php',
             dataType: "json",
             data:{Jdata:JSON.stringify({'mid': mid,'mess': mess})},
             success: function (data){
             console.log(data);
            }
        });

        return false;
    }); 

   </script>
</body>
</html>

自动表单提交不工作Ajax

2 个答案: