查詢此問題的最佳方法是什么？

Question

我有看起來像這樣的表：

questions:
id description
1   Q1
2   Q2

answers:
id question_id x y description
1     1        1 2   A1
2     1        3 4   A2
3     2        5 6   A3
4     2        7 8   A4

我想得到的是一個可以 output 這個查詢：

Q1 A1 1,2 A2 3,4
Q2 A3 5,6 A4 7,8

幾天來我一直在拉頭發，試圖弄清楚這一點。 我在 PHP 和 MySQL 中這樣做，所以如果有人可以在那里闡明一些觀點，那就太好了。

編輯：我忘了提到我也在為此使用 CodeIgniter 。 所以，這可能有助於答案。

Answer 1

考慮到每個問題可能有隨機數量的答案，您無法設計一個返回固定數量列的查詢。 您必須為每個問題返回一個結果，然后在代碼中進行一些解析。

GROUP_CONCAT function 可以幫助解決此類問題：

SELECT q.description, GROUP_CONCAT(
    CONCAT(a.description,' ',a.x,',',a.y) ORDER BY a.id
    SEPARATOR ' '
    ) AS answers
FROM questions q
  JOIN answers a ON a.question_id = q.id
GROUP BY q.description;

將返回

+-------------+---------------+
| description | answers       |
+-------------+---------------+
| Q1          | A1 1,2 A2 3,4 |
| Q2          | A3 5,6 A4 7,8 |
+-------------+---------------+
2 rows in set (0.00 sec)

您可以通過要在代碼中解析結果的任何內容來更改SEPARATOR值。 您可以使用GROUP_CONCAT function 的ORDER BY子句對每個答案的返回結果中的答案進行排序（這里我按答案 id 排序）。

編輯：如果您確定每個問題的答案永遠不會超過 4 個，您可以發出以下查詢以將每個答案放在自己的列中：

SELECT description,
  REPLACE(SUBSTRING(SUBSTRING_INDEX(answers, '$', 1), LENGTH(SUBSTRING_INDEX(answers, '$', 1 - 1)) + 1), '$', '') answer_1,
  REPLACE(SUBSTRING(SUBSTRING_INDEX(answers, '$', 2), LENGTH(SUBSTRING_INDEX(answers, '$', 2 - 1)) + 1), '$', '') answer_2,
  REPLACE(SUBSTRING(SUBSTRING_INDEX(answers, '$', 3), LENGTH(SUBSTRING_INDEX(answers, '$', 3 - 1)) + 1), '$', '') answer_3,
  REPLACE(SUBSTRING(SUBSTRING_INDEX(answers, '$', 4), LENGTH(SUBSTRING_INDEX(answers, '$', 4 - 1)) + 1), '$', '') answer_4
FROM (
    SELECT q.description, GROUP_CONCAT(
        CONCAT(a.description,' ',a.x,',',a.y) ORDER BY a.id
        SEPARATOR '$'
        ) AS answers
    FROM questions q
      JOIN answers a ON a.question_id = q.id
    GROUP BY q.description
  ) t;

將返回

+-------------+----------+----------+----------+----------+
| description | answer_1 | answer_2 | answer_3 | answer_4 |
+-------------+----------+----------+----------+----------+
| Q1          | A1 1,2   | A2 3,4   |          |          |
| Q2          | A3 5,6   | A4 7,8   | A5 9,10  |          |
+-------------+----------+----------+----------+----------+
2 rows in set (0.00 sec)

我為插圖添加了第二個問題的答案。

Answer 2

select * from answers ORDER BY question_id

$question_id = 0;
$print_ln = null;
foreach ($result as $row) {
  if ($question_id != $row['question_id']) { 
     echo "<br>";
     $question_id = $row['question_id'];
     $print_ln = "Q" . $row['question_id'] . " " . $row['description'] . " " . $row['x'] . "," . $row['y'];
  } else { 
     $print_ln = $print_ln . " " . $row['description'] . " " . $row['x'] . "," . $row['y'];
  }
  echo $print_ln;
 }

請注意，此代碼仍需要一些工作......它可以讓您了解如何執行此操作。

Answer 3

這個查詢：

SELECT 
      q.id
    , q.description
    , a.description
    , CONCAT(a.x, ',', a.y)
FROM questions AS q
    JOIN answers AS a
        ON a.question_id = q.id
ORDER BY q.id
       , a.id

將會呈現：

| 1 | Q1 | A1 | 1,2 |
| 1 | Q1 | A2 | 3,4 |
| 2 | Q2 | A3 | 5,6 |
| 2 | Q2 | A4 | 7,8 |

您所描述的最終結果稱為旋轉，在 MySQL 中並不容易，這取決於您擁有的數據。 例如，如果一個問題的答案超過 2 個，應該顯示什么？

為什么結果會是這樣：

| Q1 | A1 | 1,2 | A2 | 3,4 | 
| Q2 | A3 | 5,6 | A4 | 7,8 |

不是那樣的嗎？：

| Q1 | A1 | 1,2 | A2 | 3,4 | 
| Q2 | A4 | 7,8 | A3 | 5,6 | 

---

無論如何，對於每個問題最多 4 個答案和每個answers.id排序，這將起作用。 最好使用前面的查詢並在 PHP 中進行旋轉，在這里您可以毫無問題地處理可變數量的列：

SELECT 
      q.id
    , q.description
    , ( SELECT a.description
        FROM answers AS a
        WHERE a.question_id = q.id
        ORDER BY a.id  LIMIT 0,1
      ) AS answer1
    , ( SELECT CONCAT(a.x, ',', a.y)
        FROM answers AS a
        WHERE a.question_id = q.id
        ORDER BY a.id  LIMIT 0,1
      ) AS xy1
    , ( SELECT a.description
        FROM answers AS a
        WHERE a.question_id = q.id
        ORDER BY a.id  LIMIT 1,1
      ) AS answer2
    , ( SELECT CONCAT(a.x, ',', a.y)
        FROM answers AS a
        WHERE a.question_id = q.id
        ORDER BY a.id  LIMIT 1,1
      ) AS xy2
    , ( SELECT a.description
        FROM answers AS a
        WHERE a.question_id = q.id
        ORDER BY a.id  LIMIT 2,1
      ) AS answer3
    , ( SELECT CONCAT(a.x, ',', a.y)
        FROM answers AS a
        WHERE a.question_id = q.id
        ORDER BY a.id  LIMIT 2,1
      ) AS xy3
    , ( SELECT a.description
        FROM answers AS a
        WHERE a.question_id = q.id
        ORDER BY a.id  LIMIT 3,1
      ) AS answer4
    , ( SELECT CONCAT(a.x, ',', a.y)
        FROM answers AS a
        WHERE a.question_id = q.id
        ORDER BY a.id  LIMIT 3,1
      ) AS xy4
FROM questions AS q
ORDER BY q.id