所以我有两张桌子：
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 |
我想要做的是创建mysql查询，它将工作室名称水平显示为表格的标题，并将“x”放在匹配的位置：
所以我希望表格看起来像这样：
| 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 | | | | | |
我试图使用的查询：
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
通过使用此查询，我得到了：
| 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 | | | | | |
如何达到我想要的效果？谢谢你的帮助
答案 0 :(得分：0)
你应该使用左连接
如果您需要两个级别，那么您可以使用
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
答案 1 :(得分：0)
我找到了另一种方法：
/** 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);
}
表格：
<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>