Question

我对JSON很新，目前我一直在尝试插入数据库，目前我得到正确的消息，所有输入正确的验证通过（验证仍然需要我知道的一些工作）但是我无法得到要插入数据库，任何建议都会很棒。

HTML 权力的游戏社交

    <link rel="stylesheet" href="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/themes/smoothness/jquery-ui.css">
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
    <script src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/jquery-ui.min.js"></script>

<script>
$(document).ready(function() {
    $("form").on("submit", function(event) {
            event.preventDefault();
            $("span.error").empty()
            $("span.success").empty()
            $.getJSON('registerForm.php', $(this).serialize(), function(data) {
                    if (!data.errors) {
                        $(".success").append(data.message) // deal with a no-error response ( all is good)
                    }else{
                        $.each(data.errors,function(i,datum){
                            $("[name='"+datum.name+"']").next().html(datum.error)
                        })
                    }
            });
    });
});
</script>

</head>
    <body>
    <span class="success"></span>
    <form action="" method="POST">
    <div class="formControl">
        <input type="input" name="username" placeholder="Username" value="">
        <span class="error"> </span>
    </div>
    <div class="formControl">
        <input type="text" name="email" placeholder="E-mail"  value="">
        <span class="error"></span>
    </div>
    <div class="formControl">
        <input type="password" name="password" placeholder="Password">
        <span class="error"> </span>
    </div>
    <div class="formControl">
        <input type="password" name="repeatPassword" placeholder="Confirm Password">
        <span class="error"> </span>
    </div>
    <div class="formControl">
        <input type="hidden" name="code" value="<?php echo substr(str_shuffle("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"), 0, 1).substr(md5(time()),1); ?>">
        <span class="error"> </span>
    </div>  
    <input type="submit" value="Submit">
    </form>

    </body>
</html>

PHP

 <?php
require_once'connection.php';
    header('Content-Type: application/json');
    $errors = [];
    $username = trim($_GET['username']);
    $email = trim($_GET['email']);
    $password = trim($_GET['password']);
    $repeatPassword = trim($_GET['repeatPassword']);
    $errors = [];
    if(filter_var($username,  FILTER_VALIDATE_REGEXP,["options"=> [ "regexp" => "/.{6,25}/"]]) === FALSE){
        $errors[]= ["name"=>"username","error"=>"invalid Id (6 to 25 characters)"];
    }
    if(filter_var($email,FILTER_VALIDATE_EMAIL) === FALSE) {
        $errors[]= ["name"=>"email","error"=>"invalid Email"];
    }
    if(filter_var($password,  FILTER_VALIDATE_REGEXP,["options"=> [ "regexp" => "/.{6,25}/"]]) === FALSE){
        $errors[]= ["name"=>"password","error"=>"invalid password (6 to 25 characters)"];
    }
    if($password !== $repeatPassword){
        $errors[]= ["name"=>"repeatPassword","error"=>"passwords don't match"];
    }
    if (count($errors) === 0) {
        $salt= uniqid(mt_rand(), true);
        $options=['salt'=>$salt, 'cost'=>12];


        // everything is OK, the browser should send us to the next page
        $sql = "INSERT INTO username (username,password, eMail ,joinedDate, active, activecode) VALUES (:username, :password, :email ,NOW(), 0, :code)";
        $query = $db->prepare($sql);
        $query->execute();
        $json[] = array("username" => $username, "password" => $password, "email" => $email, "code" => $code);
    //  $json = json_encode($json);
    //  $json = file_get_contents('php://input');
        $obj = json_decode($json,true);

        echo json_encode(["message"=>"Please view your email account to activate your account"]);

    }else{
        echo json_encode(["errors"=>$errors]);
    }
?>

删除前的第一次尝试。

当我尝试使用echo json_encode(["message"=>"Please view your email account to activate your account"]);时，这不起作用我不知道为什么在if (count($errors) === 0) {}内，它无效。

header('Content-Type: application/json');
    $errors = [];
    $username = trim($_GET['username']);
    $email = trim($_GET['email']);
    $password = trim($_GET['password']);
    $repeatPassword = trim($_GET['repeatPassword']);
    $code = $_GET['code'];
     $query = $db->prepare("SELECT username.username FROM username WHERE username.username = :username LIMIT 1");
     $query->bindValue(':username', $username, PDO::PARAM_STR);
     $query->execute();

    if ( $query->rowCount() > 0 ) {
        $response=1;
         $errors[]= ["name"=>"username","error"=>"Username taken"];
    }

    if(filter_var($username,  FILTER_VALIDATE_REGEXP,["options"=> [ "regexp" => "/.{3,25}/"]]) === FALSE){
        $errors[]= ["name"=>"username","error"=>"invalid Id (3 to 25 characters)"];
    }
    if(preg_match('/[^a-z_\-0-9]/i', $username))
        {
            $errors[]= ["name"=>"username","error"=>"invalid Id (Usernames may not contain symbols)"];
        }

    if(filter_var($email,FILTER_VALIDATE_EMAIL) === FALSE) {
        $errors[]= ["name"=>"email","error"=>"invalid Email"];
    }

    $emailQ = $db->prepare("SELECT username.eMail FROM username WHERE username.eMail = :email LIMIT 1");
     $emailQ->bindValue(':email', $email, PDO::PARAM_STR);
     $emailQ->execute();

    if ( $query->rowCount() > 0 ) {
        $response=1;
         $errors[]= ["name"=>"email","error"=>"Email registered"];
    }


    if(filter_var($password,  FILTER_VALIDATE_REGEXP,["options"=> [ "regexp" => "/.{6,25}/"]]) === FALSE){
        $errors[]= ["name"=>"password","error"=>"invalid password (6 to 25 characters)"];
    }

    if(!preg_match("/(?=[a-z]*[0-9])(?=[0-9]*[a-z])([a-z0-9-]+)/i",$password)) {
         $errors[]= ["name"=>"password","error"=>"Password must contain numbers and letters"];
    }

    if($password !== $repeatPassword){
        $errors[]= ["name"=>"repeatPassword","error"=>"passwords don't match"];
    }


$salt= uniqid(mt_rand(), true);
$options=['salt'=>$salt, 'cost'=>12];

    if (count($errors) === 0) {
        // everything is OK, the browser should send us to the next page

        $sql = "INSERT INTO username (username,password, eMail ,joinedDate, active, activecode) VALUES (:username, :password, :email ,NOW(), 0, :code)";
        $query = $db->prepare($sql);

        $query->execute(array(
            ':username'=> $username,
            ':password'=> $cryptpwd=crypt($password,'$2y$12$'.$salt.'$'),
            ':email'=> $email,
            ':code'=> $code
        ));


    echo $message = '
    http://gotsocial.co.uk/active.php?activecode='.$code.'.
    ';
    $to = $email;
    $subject = 'Game of Thrones Social';
    $from = "register@gotsocial.co.uk";

    $result = mail($to, $subject, $message, "From: $from");

     echo json_encode(["message"=>"Please view your email account to activate your account"]);

    }
    echo json_encode($errors);

这部分是我的问题，没有它，我的成功消息会显示出来，我的成功消息不会显示。

echo $message = '
    http://gotsocial.co.uk/active.php?activecode='.$code.'.
    ';
    $to = $email;
    $subject = 'Game of Thrones Social';
    $from = "register@gotsocial.co.uk";

    $result = mail($to, $subject, $message, "From: $from");

Answer 1

此处不涉及JSON。问题是你正在使用SQL绑定参数（这很好），但在执行它时不会传入任何值。

    $sql = "INSERT INTO username (username,password, eMail ,joinedDate, active, activecode) VALUES (:username, :password, :email ,NOW(), 0, :code)";
    $query = $db->prepare($sql);
    $query->execute();

最后$query->execute();是问题所在。执行需要每个:username，:password等的键/值对...否则它将插入NULL。为此，请使用normal PHP array。

$query->execute(array(
    ':username' => $username,
    ':password' => $password,
    ':email'    => $email,
    ':code'     => $code
));

如果$query->execute()没有错误，则表示您的数据库架构在每个字段中都允许NULL值。您可能不希望这样，因为您发现它隐藏了错误。您应该更改架构并将NOT NULL添加到每个字段并删除所有默认值。

JSON插入数据库

1 个答案: