如何在查詢中選擇特定的值加上額外的隨機值？

Question

我對此很新，但我正在嘗試5個選擇題的測驗。每個問題有4個選擇。 我在數據庫中有問題和他們各自的答案。我試圖隨機化問題和答案，因此每次用戶參加測驗時都會顯示不同的問題。

這裏是我的代碼：

$qquery = "SELECT question from questions order by rand() limit 1"; 
$question = getvalue($qquery); 


$aquery = "SELECT (SELECT answer from questions where question = '$question') 
as right_ans, (select answer from questions where question != '$question') 
order by rand() limit 4"; 

的的GetValue（查詢）函數返回的問題串。我正在使用它來爲數據庫找到正確的答案。該函數還運行查詢並顯示結果。我的第二個查詢不起作用。我將不勝感激任何意見。謝謝！

Answer 1

試試這個新的，不知道，與MySQL工作或不。

$aquery = 
"SELECT TMP.ANSWER, TMP.RANDOM FROM ( 
SELECT ANSWER AS 'ANSWER', RAND() AS 'RANDOM' FROM QUESTION WHERE QUESTION = '$question' 
LIMIT 1 
UNION ALL 
SELECT ANSWER AS 'ANSWER', RAND() AS 'RANDOM' FROM QUESTION WHERE QUESTION != 'question' 
LIMIT 4 
) AS TMP 
ORDER BY TMP.RANDOM "; 

Answer 2

根據您目前的模式，你可以嘗試使用一個查詢這樣

SELECT r.question, r.answer, q.answer wrong_answer 
    FROM 
(
    SELECT question, answer 
    FROM questions 
    ORDER BY RAND() 
    LIMIT 1 
) r JOIN questions q 
    ON q.question <> r.question 
ORDER BY RAND() 
LIMIT 4; 

輸出示例：

 
| QUESTION | ANSWER | WRONG_ANSWER | 
|-----------|---------|--------------| 
| question4 | answer4 |  answer9 | 
| question4 | answer4 |  answer11 | 
| question4 | answer4 |  answer7 | 

這裏是SQLFiddle演示

很明顯，你看問題和權利回答呃一次（例如從第一行開始），並從wrong_answer列中獲取其他3個錯誤答案，同時在php中遍歷結果集。

---

，或者你還可以去收拾錯誤的答案在分隔列表

SELECT question, answer, GROUP_CONCAT(wrong_answer) wrong_answer 
    FROM 
(
    SELECT r.question, r.answer, q.answer wrong_answer 
    FROM 
    (
    SELECT question, answer 
    FROM questions 
    ORDER BY RAND() 
    LIMIT 1 
) r JOIN questions q 
     ON q.question <> r.question 
    ORDER BY RAND() 
    LIMIT 3 
) q 

輸出示例：

 
| QUESTION | ANSWER |   WRONG_ANSWER | 
|-----------|---------|-------------------------| 
| question9 | answer9 | answer6,answer1,answer7 | 

這裏是SQLFiddle演示

然後在PHP你總是有單行結果集。你可以很容易地explode()wrong_answer值。