问如何使用AJAX从数据库检索数据并将结果保存在变量中？
EN

Stack Overflow用户

提问于 2020-01-15 22:30:27

回答 2查看 320关注 0票数 0

我是jQuery和AJAX的新手，我正在处理一个登录页面作为一个项目，我需要使用AJAX从数据库中检索数据。我的英语不是100%流利，所以我会尽我最大的努力来解释这个问题(借助谷歌翻译)。下面是我使用的代码：

index.html

<!DOCTYPE html>
<html>
  <head>
  <title>Login</title>
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
  </head>
  <body>
    <form validate="">
      <input type="text" placeholder="Username" id="username" required/><br />
      <input type="password" placeholder="Password" id="password" required/><br />
      <input type="submit" id="submit" value="Login" />
    </form>
    <script type="text/javascript">
    // when document is loaded
    $(document).ready (
      // when submit is clicked
      $("#submit").click (
        // sets test to null
        var test = null;
        // sets username to value of username input
        var username = document.getElementById("username").value;
        // AJAX request 
        $.ajax({
          type: "POST",
          async: true,
          url: test.php,
          data: {username: username},
          success: function (data) {
            test = data;
            console.log(test);
            return test;
          }
        });
      );
    );
    </script>
  </body>
</html>

test.php

<?php
// connects to database
$conn = mysqli_connect('server', 'username', 'password', 'database');

// sets var username to POST username value
$username = $_POST['username'];

// SQL Query
$sql = "SELECT * FROM users WHERE username='" . $username . "'";
$result = mysqli_query($conn, $sql);

// sets result to mysqli_fetch_assoc()
$result = mysqli_fetch_assoc( $result );

// echos $result
echo $result['password'];

// closes database connection
mysqli_close( $conn );
?>

控制台日志

控制台输出：`当前DOM输入元素应具有自动完成属性(建议：“-password”)：(更多信息：https://www.googlesite.com)

未捕获的SyntaxError:意外的令牌变量ajax.html:19

I've looked at the code and I can't seem to find an error.
Thanks in advance! ;)

>P.S.
>It's probably going to end up being some stupid typo.
>Other than that, have a great day!

javascript

php

jquery

ajax

回答 2

Stack Overflow用户

回答已采纳

发布于 2020-01-15 22:54:46

您可以使用submit而不是click事件。

在您的情况下，只需将id给您的form，如下所示-

<form validate="" id="submit">

现在，在您的js脚本中-

$(function() { //shorthand document.ready function
    $('#submit').on('submit', function(e) { 
        e.preventDefault();  //prevent form from submitting
        console.log(data);
        $.ajax({
          type: "POST",
          async: true,
          url: test.php,
          data: $(this).serializeArray(),
          success: function (data) {
            console.log(data);
          }
        });
    });
});

所以检查你的整个代码-

<!DOCTYPE html>
<html>
  <head>
  <title>Login</title>
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
  </head>
  <body>
    <form validate="" id="submit">
      <input type="text" placeholder="Username" id="username" required/><br />
      <input type="password" placeholder="Password" id="password" required/><br />
      <input type="submit" value="Login" />
    </form>
    <script type="text/javascript">
    // when document is loaded
    $(function() { //shorthand document.ready function
    $('#submit').on('submit', function(e) { 
        e.preventDefault();  //prevent form from submitting
        console.log(data);
        $.ajax({
          type: "POST",
          async: true,
          url: test.php,
          data: $(this).serializeArray(),
          success: function (data) {
            console.log(data);
          }
         });
        });
     });
    </script>
  </body>
</html>

希望这能对你有所帮助。

票数 0

Stack Overflow用户

发布于 2020-01-15 22:45:48

您需要将一个函数传递给document.ready()调用和click()调用。

     <script type="text/javascript">

        $(document).ready(function() {
            Your variables here...

            $('#submit').click(function() {
                ... Ajax call here.
            });
        });
    </script>