Question

以下代码返回对ajax的错误部分的响应。我做错了什么？

JS代码：

    var firstName = $('input[name=first]').val(),
        lastName = $('input[name=last]').val();

    $.ajax({
        url: "add-user.php",
        type: "POST",
        data: {
            first: firstName,
            last: lastName

        },
        success: function (response) {
            console.log(response);
        },
        error: function (err) {
            console.log(err)
        }
    });

PHP代码：

<?php
    $dbConnect = mysqli_connect("localhost", "root", "password", "db");
    $stmt = $dbConnect->prepare("INSERT INTO pokemon (firstName, lastName) VALUES (?, ?)");
    $stmt->bind_param('ss', $firstName, $lastName);

    $firstName = $_POST['first'];
    $lastName = $_POST['last'];

    $stmt->execute();

?>

HTML：

<form action="" method="POST" name="addUser" class="user-input">
    <label>First Name: <input type="text" name="first"></label> <br>
    <label>Last Name: <input type="text" name="last"></label> <br>

    <input type="submit" value="Submit" class="user-submit">
</form>

提交数据库正常，但响应返回错误：

Object {readyState: 0, status: 0, statusText: "error"}

Answer 1

你必须首先初始化变量。

 <?php
  $firstName = $_POST['first'];
  $lastName = $_POST['last'];
  $dbConnect = mysqli_connect("localhost", "root", "password", "db");
  $stmt = $dbConnect->prepare("INSERT INTO pokemon (firstName, lastName) VALUES (?, ?)");
  $stmt->bind_param('ss', $firstName, $lastName);



  $stmt->execute();

?>

像这样制作ajax

$("#userForm").submit(function(e){
    url: "add-user.php",
    type: "POST",
    data: $(this).serialize(),
    success:function(data){console.log(data);}
});

HTML

<form action="" method="POST" name="addUser" id="userForm" class="user-input">
<label>First Name: <input type="text" name="first"></label> <br>
<label>Last Name: <input type="text" name="last"></label> <br>

<input type="submit" value="Submit" class="user-submit">
</form>

Answer 2

你假设php明智地理解JSON - 它没有。

您需要转换ajax以发送“url-encoded”数据，或者让php解析正文（包含JSON字符串）。

成功时Ajax返回错误

2 个答案: