Question

Here is my database schema:

user
*user_id
*username
*password
*etc

quiz_response
*response_id
*user_id
*question_id
*response
*is_correct
*answer_time

question_choice
*choice_id
*question_id
*is_correct
*choice (VARCHAR)

question
*question_id
*quiz_id
*question (VARCHAR)

quiz
*quiz_id
*title (VARCHAR)

我正在利用PHP建造一个静态的网络应用程序,我正在遇到麻烦。目前,我尝试——没有幸运——这一问询,我知道问题在哪里——我不知道如何解决。因此,我为什么来到这里谈谈S.O.。

// Grab the response data from the database to generate the form
    $query = "SELECT qr.response_id AS r_id, qr.question_id, qr.response, q.question, quiz.title " . 
        "FROM quiz_response AS qr " . 
        "INNER JOIN question AS q USING (question_id) " . 
        "INNER JOIN quiz USING (quiz_id) " . 
        "WHERE qr.user_id =  " . $_SESSION[ user_id ] . " ";
    $data = mysqli_query($dbc, $query) or die("MySQL error: " . mysqli_error($dbc) . "<hr>
Query: $query");

此时,我感觉到我的第二名Inner Join(INNER JOIN quiz USING (quiz_id)是问题。当我不列入这条线并去除qui。标书。因此,我的问题是:,我如何维持原数据库图象,同时仍然根据从表问题问题中得来的静z标题?。任何帮助都值得赞赏!

Answer 1

我认为,问题是,Quz_id不是在quiz_response。我使用了《国家公报》的关键词。 Try:

// Grab the response data from the database to generate the form
$query = "SELECT qr.response_id AS r_id, qr.question_id, qr.response, q.question, 
          quiz.title " . 
         "FROM quiz_response AS qr " . 
         "INNER JOIN question AS q ON (q.question_id = qr.question_id) " . 
         "INNER JOIN quiz ON (quiz.quiz_id = q.quiz_id) " . 
         "WHERE qr.user_id =  " . $_SESSION[ user_id ] . " ";
$data = mysqli_query($dbc, $query) or 
         die("MySQL error: " . mysqli_error($dbc) . "<hr>
Query: $query");

Answer 2

1. 与下列方面有关的事项:

// Grab the response data from the database to generate the form
    $query = "SELECT qr.response_id AS r_id, qr.question_id, qr.response, q.question, quiz.title " . 
        "FROM quiz_response qr " . 
        "INNER JOIN question q USING (question_id) " . 
        "INNER JOIN quiz ON quiz.quiz_id = q.quiz_id " . 
        "WHERE qr.user_id =  " . $_SESSION[ user_id ] . " ";
    $data = mysqli_query($dbc, $query) or die("MySQL error: " . mysqli_error($dbc) . "<hr>
Query: $query");

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