水平和动态连接表

Question

我需要水平连接表格的帮助，我的表格是

+---------+---------+
| Candidates Table  |
+---------+---------+
| can_id   | Name   | 
+---------+---------+
| 1       | Liza    |
| 2       | Sarah   |
| 3       | Jane    |
|         |         |
+---------+---------+

+---------+---------+
| Judges Table      |
+---------+---------+
| id      | Name    | 
+---------+---------+
| 1       | judge1  |
| 2       | judge2  |
| 3       | judge3  |
+-------------------+

+---------+---------------+--------+-------+
|                Score Table               |
+---------+-------+------------------------|
| sco_id  |  can_id| jud_id |crit_id |score|
+---------+--------+-----------------------+
| 1       |   1    |   2    |    1   |  87 |
| 2       |   1    |   3    |    1   |  89 |
| 3       |   1    |   1    |    1   |  80 |
+------------------------------------------+

我需要类似这样的输出..

+---------+---------------+-------------+
|                Score board            |
+---------+---------+-------------------|
| Name    |  judge1 | judge2  | judge3  |
+---------+---------+-------------------|
| Liza    |   80    |    87   |    89   |  
|some data|some data|some data|some data|  
|some data|some data|some data|some data|  
+---------------------------------------+

注意：crit_id 是标准表中的标准 id。

通常我会使用一些连接和子查询，但我的问题是我需要动态输出，如果我添加一个新的判断它会自动生成一个新列。我需要至少 1 个包含所有评委分数的候选人数据，然后只需在 php 上使用参数循环它即可获取其他候选人数据，例如

php loop start
   <td>name</td> 
   <td>judge1 score</td>
   <td>judge2 score</td> 
php end loop

或者如果我可以让整个候选人表与评委得分更好，我就不会让每个候选人循环他们

我曾尝试研究类似的问题，例如

Concatenate more than two tables horizontally in SQL Server

我曾尝试自己编写代码，但无法加入评委。

SELECT s.sco_id,c.Name,c.Municipalities 
FROM `tbl_scoring` s 
LEFT JOIN tbl_candidates c ON c.`can_id` = s.`can_id` 
WHERE s.can_id = 11 
  AND crit_id = 1 
ORDER BY s.jud_id asc

我需要一个查询，该查询将根据法官的数量动态生成，或者获取具有法官分数的候选数据，然后在 php 上循环，或者如果我在不循环的情况下获取所有数据，则更好

Answer 1

初始化以下数组：

$judges = [];
$scores = [];
$candidates = [];

然后执行您的查询，并循环结果。为每次迭代设置这些值：

$judges[$row['jud_id']] = 1;
$candidates[$row['can_id']] = $row['Name'];
$scores[$row['can_id']][$row['jud_id']] = $row['score'];

现在你想得到参与者评委的名字，所以让我们运行一个 SQL 查询：

$sql = 'SELECT Name FROM judges WHERE id IN (' . implode(',', array_keys($judges)) . ')';

并且在每次迭代中，在 $judges 数组中设置法官的姓名：

$judges[$row['id']] = $row['Name'];

那么对于输出：

echo '<tr>';
echo '<td>Name</td>';
ksort($judges);
foreach ($judges as $name) {
    echo '<td>Judge: ' . $name . '</td>';
}
echo '</tr>';

foreach ($scores as $candidateId => $data) {
    echo '<tr>';
    echo "<td>$candidates[$candidateId]</td>";
    ksort($data);
    foreach ($data as $score) {
        echo "<td>$score</td>";
    }
    echo '</tr>';
}

我在$judges 和$data 上使用了ksort，所以分数适合每位评委。

Answer 2

首先，我们检索评分表中存在的评委的 ID 和姓名。

$judges = [];
$query = "SELECT id, name FROM Judges WHERE id IN ( SELECT DISTINCT jud_id FROM Score )";
// execute the query and store the results in the $judges array.

我们检索分数表中存在的候选人 ID 和姓名。

$candidates = [];
$query = "SELECT id FROM Candidate WHERE id IN ( SELECT DISTINCT can_id FROM Score )";
// execute the query and store the results in the $candidates array.

然后，我们加入候选和分数表。

$candidate_score = [];
$query = "SELECT Candidate.name, Candidate.id as candidate_id , Score.jud_id, Score.score FROM Candidate JOIN Score ON Score.can_id = Candidate.id";
// execute the query and store it in the $candidate_score array.

现在，对于每个候选人，我们将其分数填充到 $score_board 数组中。

$score_board = [];

foreach ( $candidates as $candidat )
{
   $score_board[$candidat] = [];

   foreach ( $judges as $judge )
   {
      $judge_name = $judge['name'];
      $judge_id = $judge['id'];
      $score_board[$candidat][$judge_name] = get_judge_score($candidate_score,$candidat,$judge_id);
   } 
}

get_judge_score 的工作方式如下：

function get_judge_score ( $scores , $candidate , $judge )
{
   $score_filtred = array_filter($scores, function ($score) use ($candidate,$judge) {
      return $score['jud_id'] == $judge && $score['candidate_id'] = $candidate;
   });

   return count($score_filtred) > 0 ? $score_filtred[0]['score'] : 0;
}