PHP在mysql表中创建用户

Question

if ($stmt = $connection->prepare('INSERT INTO users (name, id, password, email, city, avatar, about, activation_code) VALUES (?, ?, ?, ?, ?, ?, ?, ?)')) {
        $password = password_hash($_POST['password'], PASSWORD_DEFAULT);
        $uniqid = uniqid();
        $stmt->bind_param('ssssssss', $_POST['name'], $_POST['id'], $password, $email, $_POST['city'], $_POST['avatar'], $_POST['about'], $uniqid);
        $stmt->execute();
        $stmt->store_result();
        echo 'Account's created';
    } else {
        echo 'Error';
    }

这部分代码不会在myqsl db中创建用户。但是如果我此代码：

$stmt->bind_param('ssssssss', $_POST['name'], $_POST['id'], $password, $email, $_POST['city'], $_POST['avatar'], $_POST['about'], $uniqid);

替换为：

$stmt->bind_param('ssssssss', $_POST['name'], $_POST['id'], $email, $password, $_POST['city'], $_POST['avatar'], $_POST['about'], $uniqid);

它创建了用户，但电子邮件表中有密码，而密码-电子邮件中。我该如何解决？？？

if（$ stmt = $ connection-> prepare（'INSERT INTO users（name，id，password，email，city，avatar，about，activation_code）VALUES（？，？，？，？，？，？，？，？， ？）'））{$ password = password_hash（$ _ POST ['...

Answer 1

问题在于您的prepare插入语句和绑定语句之间存在冲突陈述一