Question

我希望用户在首次登录时在技能页面中设置技能，就像一次性设置一样，以及将来登录时，他们不再需要再次设置它进入主页，因为他们已经设置了它，如果他们想要在他们的个人资料页面中更新它。 我怎样才能实现这一目标？我应该使用哪些查询？

任何建议都会有所帮助。谢谢！

这是我尝试但未能实现的代码。

import threading


def main():
    # Initialise thread object
    flag = threading.Event()
    for i in range(2):
        if i == 0:
            t1 = threading.Thread(target=t1_func, args=(flag,))
            t1.start()
        else:
            t2 = threading.Thread(target=t2_func, args=(flag,))
            t2.start()


def t1_func(flag):
    for i in range(100000000):
        if i == 100000000 - 1:
            flag.set()


def t2_func(flag):
    while not flag.is_set():
        print('not flag yet')
    print('GOT FLAG !')


if __name__ == '__main__':
    main()

Answer 1

试试这个：

我假设user_skill.user_id和 job_seeker.user_id具有主键外键关系。

if (count($errors) == 0) {
    $password = md5($password);
    $query  = "SELECT * FROM job_seeker WHERE username = '$username' AND password = '$password'";
    $results  = mysqli_query($db, $query);

    // Below, I am getting the user_id from the user table and adding it in the WHERE Clause (So, there will rows ONLY when given user has defined any skills of his/her )        
    $user = mysqli_fetch_assoc($results);
    $userId = $user['user_id'];


    $querx = "SELECT user_skill.user_id AS id_one, job_seeker.user_id 
                 AS id_two
                 FROM user_skill 
                 INNER JOIN job_seeker ON (user_skill.user_id = job_seeker.user_id) 
                 WHERE user_skill.user_id = $userId";

    $result   = mysqli_query($db, $querx);

    if (mysqli_num_rows($results) == 1) {
        session_regenerate_id();
        $user = mysqli_fetch_assoc($results);
        $job  = mysqli_fetch_assoc($result);
        $_SESSION['SESS_USER_ID']           = $user['user_id'];
        $_SESSION['SESS_USER_IDs']          = $job ['user_id'];
        $_SESSION['SESS_USER_NAME']         = $user['username'];
        $_SESSION['SESS_FIRST_NAME']        = $user['f_name'];
        $_SESSION['SESS_LAST_NAME']         = $user['l_name'];
        $_SESSION['SESS_USER_EMAIL']        = $user['email'];
        $_SESSION['SESS_BIRTH_DATE']        = $user['birthdate'];
        $_SESSION['SESS_USER_GENDER']       = $user['gender'];
        $_SESSION['SESS_USER_CONTACT']      = $user['contact'];
        $_SESSION['SESS_USER_ADDRESS']      = $user['address'];
        $_SESSION['SESS_USER_NATIONALITY']  = $user['nationality'];

            if(mysqli_num_rows($result) > 0) {
                header('location: userhome.php');
                     // Now, you don't need the below IF code
                    // Since, the rows will be only be returned if that
                    // particular user's skills exists in the table.
               /*  while($fetch = mysqli_fetch_assoc($result)) { 

                   if($fetch['id_one'] === $fetch['id_two']) {

                        header('location: userhome.php');

                    }
                     else {

                        header('location: userskill.php');

                    } 

             } */
            }
            else {

                header('location: userskill.php');

                    } 
            session_write_close();
        }
    else {
        array_push($errors, "Wrong username or password!");
    }
}
}

Answer 2

已经解决了这个问题，未来可能对某人有所帮助。

// LOGIN - NO ERRORS
        if (count($errors) == 0) {
        $password = md5($password);
        $query    = "SELECT * FROM job_seeker WHERE username = '$username' AND password = '$password'";
        $results  = mysqli_query($db, $query);

        if (mysqli_num_rows($results) == 1) {
            session_regenerate_id();
            $user = mysqli_fetch_assoc($results);
            $job  = mysqli_fetch_assoc($result);
            $_SESSION['SESS_USER_ID']           = $user['user_id'];
            $_SESSION['SESS_USER_IDs']          = $job ['user_id']; 
            $_SESSION['SESS_USER_NAME']         = $user['username'];
            $_SESSION['SESS_FIRST_NAME']        = $user['f_name'];
            $_SESSION['SESS_LAST_NAME']         = $user['l_name'];
            $_SESSION['SESS_USER_EMAIL']        = $user['email'];
            $_SESSION['SESS_BIRTH_DATE']        = $user['birthdate'];
            $_SESSION['SESS_USER_GENDER']       = $user['gender'];
            $_SESSION['SESS_USER_CONTACT']      = $user['contact'];
            $_SESSION['SESS_USER_ADDRESS']      = $user['address'];
            $_SESSION['SESS_USER_NATIONALITY']  = $user['nationality'];
            $_SESSION['SESS_DATE_REGISTERED']   = $user['date_registered'];
            $_SESSION['SESS_USER_IMG']          = $user['avatar'];
            $_SESSION['SESS_USER_STAT']         = $user['status'];

            $userId = $user['user_id'];
            $jobId  = $job['user_id'];

            $query    = "SELECT * FROM job_seeker";
            $resultx  = mysqli_query($db, $query);

            $user = mysqli_fetch_assoc($resultx);
            $userId = $user['user_id'];

            $skills  = "SELECT *
                        FROM job_seeker
                        INNER JOIN user_skill ON user_skill.user_id = job_seeker.user_id
                        WHERE user_skill.user_id = '" . $_SESSION['SESS_USER_ID'] . "'";

            $results = mysqli_query($db, $skills);           

            if($results->num_rows) {

                header('location: userhome.php');

            }
            else {

                header('location: userskill.php');

            }
            session_write_close();
        }

PHP MySQLi One Time设置表单

2 个答案: