Question

您好，我在收到ajax请求的回复时遇到了问题，我已多次这样做但我不知道现在的问题是什么，我一直在拉我的头发现在在这一天，我曾尝试使用完整，错误或.done，没有任何作用。我正在做的是通过ajax插入数据，然后在插入数据时得到响应，现在数据插入正常，但我没有得到响应，请你帮我，没有控制台错误。

insert_ajax.php

<?php
?>
<html>
<head>
    <title>Basic Ajax Insert</title>
    <link rel="stylesheet" href="style/style.css" type="text/css"/>
    <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" type="text/css"/>
    <script src="jquery-1.3.1.min.js"></script>
    <script src="custom.js"></script>
</head>
<body>
    <div class="container">
    <div id="succ"></div>
    <h2 class="text-warning">Basic Information</h2>
    <form id="form" action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="post" autocomplete="off">

        <div class="form-group col-md-6">
            <label class="text-warning">First Name: </label>
            <input class="form-control first_name" type="text" name="fname"     />
        </div>

        <div class="form-group col-md-6">
            <label class="text-warning">Last Name: </label>
            <input class="form-control last_name" type="text" name="lname" />
        </div>

        <div class="form-group col-md-6">
            <label class="text-warning">Email: </label>
            <input class="form-control email" type="text" name="email" />
        </div>
        <div class="form-group col-md-6">
            <label class="text-warning">Password: </label>
            <input class="form-control pwd" type="password" name="pass" />
        </div>

        <button class="btn btn-warning" name="btn_sub">Send</button>

    </form>
    </div>
</body>
</html>

ajax.php

<?php

$con=mysqli_connect("localhost","root","","practices");

if(isset($_REQUEST['em'])){
echo'email received..: '.$_REQUEST['em'];
}

$fname=$_REQUEST["fn"];
$lname=$_REQUEST["ln"];
$email=$_REQUEST["em"];
$pass=$_REQUEST["pwd"];
//echo "returned";
$ins="insert into `user` (`first_name`,`last_name`,`email`,`password`) 
values('$fname','$lname','$email','$pass')";
$query=mysqli_query($con,$ins);

if($query){
    echo "Data is entered Love";
}
?>

custom.js

$(document).ready(function(){

$(".btn").click(function(){

    var fname=$(".first_name").val();
    var lname=$(".last_name").val();
    var email=$(".email").val();
    var pass=$(".pwd").val();

    $.ajax({
      method: "POST",
      url : "ajax.php",
      data: {fn:fname,ln:lname,em:email,pwd:pass},
      success: function(r){
        alert(r);
      }

    });



});

});

Answer 1

更改为这些行后尝试

$(".btn").on("click",function(e){
   e.preventDefault();
var fname=$(".first_name").val();
var lname=$(".last_name").val();
var email=$(".email").val();
var pass=$(".pwd").val();

$.ajax({
  method: "POST",
  url : "ajax.php",
  data: {fn:fname,ln:lname,em:email,pwd:pass},
  success: function(ind,val){
    console.log(ind);
    console.log(val);
  }

});

Answer 2

相反如何：

$.post("aqjax.php",{fn:fname,ln:lname,em:email,pwd:pass})
.done(function(data)
{
//Do stuff here
})

$ .ajax未返回回复

2 个答案: