我試圖弄清楚我如何從我的questions表中獲取「問題」，同時獲取相應的answers，或者在我的具體情況下，我稱他們爲選擇。從問題表急切加載，相應的答案

目前I'm獲取這樣的：

public static function getQuestion($id) 
{ 

    $sql = "SELECT * FROM questions WHERE id = :id"; 
    $query = $database->prepare($sql); 
    $query->execute(array(':id' => $id)); 

    if($query->rowCount() == 1){ 
     return $query->fetch(); 
    } 
     return false; 
} 

    public static function getChoices($id) 
{ 
    $sql = "SELECT * FROM choices WHERE question_id = :id"; 
    $query = $database->prepare($sql); 
    $query->execute(array(':id' => $id)); 

     return $query->fetchAll(); 
}

所以I'm做兩（2）查詢，首先我取這個問題，那麼我取這個問題的選擇。 JSON格式的結果是這樣的：

{ 
    "question": { 
    "id": "12", 
    "content": "asdasd", 
    "source": "asd", 
    "image_url": "156ebc3206212c_qijpmnklohgfe.jpeg", 
    "lastmodified": "2016-03-18 09:58:08", 
    "quiz_id": "6" 
    }, 
    "answers": [ 
    { 
     "id": "45", 
     "content": "Ja", 
     "correct": "0", 
     "question_id": "12" 
    }, 
    { 
     "id": "46", 
     "content": "nej", 
     "correct": "0", 
     "question_id": "12" 
    }, 
    { 
     "id": "47", 
     "content": "inte", 
     "correct": "0", 
     "question_id": "12" 
    }, 
    { 
     "id": "48", 
     "content": "kanske ", 
     "correct": "1", 
     "question_id": "12" 
    } 
    ] 
}

但他「正確」的方式或許應該是答案（選擇）嵌套在問題裏面：

{ 
    "question": { 
    "id": "12", 
    "content": "asdasd", 
    "source": "asd", 
    "image_url": "156ebc3206212c_qijpmnklohgfe.jpeg", 
    "lastmodified": "2016-03-18 09:58:08", 
    "quiz_id": "6", 
    "answers": [ 
     { 
     "id": "45", 
     "content": "Ja", 
     "correct": "0", 
     "question_id": "12" 
     }, 
     { 
     "id": "46", 
     "content": "nej", 
     "correct": "0", 
     "question_id": "12" 
     }, 
     { 
     "id": "47", 
     "content": "inte", 
     "correct": "0", 
     "question_id": "12" 
     }, 
     { 
     "id": "48", 
     "content": "kanske ", 
     "correct": "1", 
     "question_id": "12" 
     } 
    ] 
    } 
}

我的問題：

我如何加入（或急切地加載）這個問題的選擇？

來源

2016-04-25 Adam

你可以編寫一個第三個函數來使用連接來提取這兩個函數。

$sql = "SELECT 
q.id as question_id, q.content, q.source, q.image_url, q.lastmodified, q.quiz_id, 
c.id AS answer_id, c.content as an_content, c.correct, c.question_id 
FROM questions as q 
INNER JOIN choices AS c ON q.id = c.question_id 
WHERE q.id = :id";

來源

2016-04-25 20:09:33 VikingBlooded

後非常simpel的解決方案，並沒有以預期的方式加載數據。 ' 「問題」：[{ 「ID」：「45」，「內容」：「JA」，「源」：「ASD」，「IMAGE_URL」：「156ebc3206212c_qijpmnklohgfe.jpeg」，「上次更改時間「：」2016-03-18 09:58:08「，」「question_id」：「12」 }， – Adam

$database = DatabaseFactory::getFactory()->getConnection(); 

$question = Self::getFraga($id); 
$answer = Self::getAnswers($id); 
$question->answer = $answer; 
return $question;

產地：

{ 
    "question": { 
    "id": "12", 
    "content": "asdasd", 
    "source": "asd", 
    "image_url": "156ebc3206212c_qijpmnklohgfe.jpeg", 
    "lastmodified": "2016-03-18 09:58:08", 
    "quiz_id": "6", 
    "answer": [ 
     { 
     "id": "45", 
     "content": "Ja", 
     "correct": "0", 
     "question_id": "12" 
     }, 
     { 
     "id": "46", 
     "content": "nej", 
     "correct": "0", 
     "question_id": "12" 
     }, 
     { 
     "id": "47", 
     "content": "inte", 
     "correct": "0", 
     "question_id": "12" 
     }, 
     { 
     "id": "48", 
     "content": "kanske ", 
     "correct": "1", 
     "question_id": "12" 
     } 
    ] 
    } 
}

所有

來源

2016-04-26 19:38:02 Adam

從問題表急切加載，相應的答案

我的問題：

回答

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