水平的數據表中的分配

Question

所以我有兩個表：

project.sql 
projectNo| title | 
---------+-------+ 
17020002 | test | 
17020003 | test2 | 

process.sql 
projectNo| process | studio | 
---------+---------+---------+ 
17020002 | ANM  | STELLAR | 
17020002 | BLD  |   | 
17020003 | ANM KEY | APEX | 
17020003 | BLD  | PALETTE | 

我想要做的就是創建MySQL查詢，這將水平顯示工作室的名稱作爲表的標題，並把「X」，其中它匹配：

所以我想表如下所示：

| title   | APEX | BASECAMP | CANVAS | HORIZON | LAUNCHPAD | NEBULA | ORBIT | PALETTE | SANDBOX | STELLAR | THE CLIMB | TOONIGAMI | TREEHOUSE | 
+---------------+------+----------+--------+---------+-----------+--------+-------+---------+---------+---------+-----------+-----------+-----------+ 
| Project test |  |   |  |   |   |  |  |   |   |   |   |   |   | 
| ANM   |  |   |  |   |   |  |  |   |   | x  |   |   |   | 
| BLD   |  |   |  |   |   |  |  |   |   |   |   |   |   | 
| Project test2 |  |   |  |   |   |  |  |   |   |   |   |   |   | 
| ANM KEY  | x |   |  |   |   |  |  |   |   |   |   |   |   | 
| BLD   |  |   |  |   |   |  |  | x  |   |   |   |   |   | 

我試圖使用的查詢：

SELECT proc.projectNo, proc.title, 
    max(case when studio = "APEX" then "x" else "" end) as APEX, 
    max(case when studio = "BASECAMP" then "x" else "" end) as BASECAMP, 
    max(case when studio = "CANVAS" then "x" else "" end) as CANVAS, 
    max(case when studio = "HORIZON" then "x" else "" end) as HORIZON, 
    max(case when studio = "LAUNCHPAD" then "x" else "" end) as LAUNCHPAD, 
    max(case when studio = "NEBULA" then "x" else "" end) as NEBULA, 
    max(case when studio = "ORBIT" then "x" else "" end) as ORBIT, 
    max(case when studio = "PALETTE" then "x" else "" end) as PALETTE, 
    max(case when studio = "SANDBOX" then "x" else "" end) as SANDBOX, 
    max(case when studio = "STELLAR" then "x" else "" end) as STELLAR, 
    max(case when studio = "THE CLIMB" then "x" else "" end) as THECLIMB, 
    max(case when studio = "TOONIGAMI" then "x" else "" end) as TOONIGAMI, 
    max(case when studio = "TREEHOUSE" then "x" else "" end) as TREEHOUSE 
    FROM process p1 
    LEFT JOIN 
    (
     SELECT projectNo, title 
     FROM 
     (
      SELECT projectNo, CONCAT('Project ', title) as title, 0 AS a FROM project 
      UNION ALL 
      SELECT projectNo, process, 1 AS a FROM process 
     ) t 
     ORDER BY projectNo, a, title 
    )proc 
    ON p1.projectNo = proc.projectNo 
    GROUP BY proc.projectNO, proc.title 

，並通過使用此查詢，我得到：

| title   | APEX | BASECAMP | CANVAS | HORIZON | LAUNCHPAD | NEBULA | ORBIT | PALETTE | SANDBOX | STELLAR | THE CLIMB | TOONIGAMI | TREEHOUSE | 
+---------------+------+----------+--------+---------+-----------+--------+-------+---------+---------+---------+-----------+-----------+-----------+ 
| ANM   |  |   |  |   |   |  |  |   |   | x  |   |   |   | 
| BLD   |  |   |  |   |   |  |  |   |   | x  |   |   |   | 
| Project test |  |   |  |   |   |  |  |   |   | x  |   |   |   | 
| ANM KEY  | x |   |  |   |   |  |  | x  |   |   |   |   |   | 
| BLD   | x |   |  |   |   |  |  | x  |   |   |   |   |   | 
| Project test2 | x |   |  |   |   |  |  | x  |   |   |   |   |   | 

我如何能實現我想要的結果呢？感謝您的幫助

Answer 1

你應該使用左連接
如果您需要兩個LEVE層次，那麼你可以使用

select * from from (
    SELECT 
      0 as level, 
      pj.projectNo as projectNo, 
      CONCAT('Project ', pj.title) as title, 
      null as APEX, 
      null as BASECAMP, 
      null as CANVAS, 
      null as HORIZON, 
      null as LAUNCHPAD, 
      null as NEBULA, 
      null as ORBIT, 
      null as PALETTE, 
      null as SANDBOX, 
      null as STELLAR, 
      null as THECLIMB, 
      null as TOONIGAMI, 
      null as TREEHOUSE 
    FROM project as pj 

    UNION ALL 
    SELECT 
      max(1), 
      pj.projectNo, 
      pc.process, 
      max(case when pc.studio = "APEX" then "x" else "" end) , 
      max(case when pc.studio = "BASECAMP" then "x" else "" end), 
      max(case when pc.studio = "CANVAS" then "x" else "" end) , 
      max(case when pc.studio = "HORIZON" then "x" else "" end) , 
      max(case when pc.studio = "LAUNCHPAD" then "x" else "" end) , 
      max(case when pc.studio = "NEBULA" then "x" else "" end) , 
      max(case when pc.studio = "ORBIT" then "x" else "" end) , 
      max(case when pc.studio = "PALETTE" then "x" else "" end), 
      max(case when pc.studio = "SANDBOX" then "x" else "" end) , 
      max(case when pc.studio = "STELLAR" then "x" else "" end) , 
      max(case when pc.studio = "THE CLIMB" then "x" else "" end), 
      max(case when pc.studio = "TOONIGAMI" then "x" else "" end) , 
      max(case when pc.studio = "TREEHOUSE" then "x" else "" end) 
    FROM project as pj 
    left join process as pc on pj.projectNo = pc.projectNo 
    GROUP BY pj.projectNO, pj.title) t 
    order by t.level, t.projectNo, t.title 

Answer 2

我找到了另一種方式來做到這一點：

/** start for getting titles**/ 
$sql = ' 
     SELECT proj_title, CONCAT(projectNo, " ", proj_title) AS title 
     FROM (
      SELECT projectNo, CONCAT("Project ", title) AS proj_title, 0 AS a FROM project p1 
      UNION ALL 
      SELECT DISTINCT projectNo, process, 1 AS a FROM process p2) t 
     ORDER BY projectNo, a, title 
     '; 
$projects = []; 
$query = mysqli_query($conn, $sql); 
// for each title 
while ($data = mysqli_fetch_assoc($query)) 
{ 
    $projects[] = $data; 
    $titles = array(); 
    $sql2 = '   
      SELECT a.* FROM 
      (
       (
        SELECT upr.projectNo, upr.process, p.stdid, p.studio_name, COALESCE(upr.num_appearences, 0) AS count 
        FROM studios p 
        LEFT JOIN 
        (
         SELECT CONCAT(upr.projectNo, " ", upr.process) AS process, projectNo, studio, COUNT(DISTINCT studio) AS num_appearences 
         FROM process upr 
         GROUP BY projectNo, process, studio 
        )upr ON p.studio_name = upr.studio 
        AND upr.process = "' . mysqli_real_escape_string($conn, $data['title']) . '" 
       ) 
      ) AS a 
      ORDER BY a.projectNo, a.studio_name, a.process'; 
    $rows = []; 
    $query2 = mysqli_query($conn, $sql2); 
    while ($data2 = mysqli_fetch_assoc($query2)) { 
     $rows[] = $data2; 
     $projectNo = $data['title']; 
     $studioName = $data2['studio_name']; 
     $titles[] = $studioName; 
     $count = $data2['count']; 

     if (!isset($allProcess[$projectNo])) 
      $allProcess[$projectNo] = []; 

     if ($count) 
      $allProcess[$projectNo][$studioName] = $count;  
    } 
    $titles = array_unique($titles); 
} 

而表：

<table class="paginated" style=" margin-right:10%;"> 
<?php 
/** now do the printing of data **/ 
if ($allProcess) 
{ 
    $header = 
     '<th>Title</th>' . 
     array_reduce(array_values($titles), function ($p, $n) { 
      return $p . '<th>Studio ' . htmlspecialchars($n) . '</th>'; 
     }); 

    $body = ''; 
    /** loop through projects first instead **/ 
    foreach ($projects as $p) { 
     $body .= '<tr>'; 
     $body .= '<td>' . htmlspecialchars($p['proj_title']) . '</td>'; 

     /** loop through titles or all process **/ 
     foreach ($titles as $t) { 
      $row = $allProcess[$p['title']]; // e.g. Array ([process] => Array ([170001]... 
      $r = isset($row) && isset($row[$t]) ? $row[$t] : ' '; 
      $body .= '<td>' . htmlspecialchars($r) . '</td>'; 
     } 
     $body .= '</tr>'; 
    } 
    echo "<thead>$header</thead><tbody>$body</tbody>"; 
} 
?>  
</table>