从表单获取用户名，密码和电子邮件，并使用它创建数据库行

Grymfogel1 发表于 Dev

Grymfogel1

我一直在尝试解决有关jQuery，php和mySQL之间通信的问题。

长话短说，我想通过PHP将“注册表格”中的信息插入到我的数据库中。

这是我的表单，位于login.php中：

<form action="login.php" method="post">
    <input type="text" name="username" id="username">
<br><br>
    <input type="password" name="password">
<br><br>
    <input type="text" name="email">
<br><br>
    <input type="submit" value="Register" class="submitRegistration"/>
</form>

我希望在按下“注册”按钮时触发它，它位于head.php中。

$(document).ready(function(){
    $(".submitRegistration").click(function(){
    var clickBtnValue = $(this).val();
    var username = $("#username").val();
    var password = $("#password").val();
    var email = $("#email").val();
    var ajaxurl = 'ajaxdisk.php';
    data = {'action': clickBtnValue};
    $.post(ajaxurl, data, function (response) {
        alert(response);
        });
    });

});

这是ajax.php中的数据库连接以及我要运行的功能

$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "swag";


// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
//I have suspicions that this is where things go wrong, because it checks if 
//$_POST['action'] != NULL before it is defined, I want it to 
//do this WHEN the button is hit, not when the page loads.
if (isset($_POST['action'])) {
    switch ($_POST['action']) {
        case 'Submit':
            submit();
            break;
        case 'Register':
            register($conn);
            break;
    }
}
else {
    print("Error");
}

function submit() {
    echo "The submit function is called.(unfinished)";
    exit;
}

function register($cpnn) {
    echo "The register function is called.";
    $sql = "INSERT INTO MyGuests (username, password, email)
    -- Here i want the values from the textboxes to be, so it gets put     into the database
    VALUES (username, password, email)";
    if ($cpnn->query($sql) === TRUE) {
        echo "New record created successfully";
    }   
    else {
        echo "Error: " . $sql . "<br>" . $conn->error;
    }
}


   $conn->close();

为了解决这个问题，我可能花了无数的时间弄乱了它，但我真的希望你们中的任何一个都能从中弄清一切，并认识到错误。

谢谢。

生死

更改通过注释说明：-

您的html文件代码将如下所示：-

<form method="post">
    <input type="text" name="username" id="username">
<br><br>
    <input type="password" name="password" id="password"> <!-- id required -->
<br><br>
    <input type="text" name="email" id="email"> <!-- id required -->
<br><br>
    <input type="submit" value="Register" class="submitRegistration"/>
</form>
<script src="https://code.jquery.com/jquery-1.12.0.min.js"></script>
<script type = "text/javascript">
$(document).ready(function(){
    $(".submitRegistration").click(function(e){// e for event
    e.preventDefault(); // to stop form submission
    var clickBtnValue = $(this).val();
    var username = $("#username").val();
    var password = $("#password").val();
    var email = $("#email").val();
    $.post('ajaxdisk.php', {'action': clickBtnValue,"username":username,"password":password,"email":email}, function (response) {
        alert(response);
        });
    });
});
</script>

您的php文件（ajaxdisk.php）代码如下所示：-

<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "swag";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
if (isset($_POST['action'])) {
    $username = $_POST['username'];
    $password = $_POST['password'];
    $email = $_POST['email'];
    echo "The register function is called.";
    $sql = 'INSERT INTO `MyGuests` (username, password, email) VALUES ("'.(string)$username.'","'.(string)$password.'","'.(string)$email.'")';
    if ($conn->query($sql)) {
        echo "New record created successfully";
    }   
    else {
        echo "Error: ". $conn->error;
    }
}
?>