我想做一张这样的桌子:
name | Math | Physics | History | Biology | Gym
-------------------------------------------------------
Johnat| 5 | 3 | | |
Sarah | 1 | | 2 | 3 | 2
我有一个这样的SQL结构:
艾因
id | nimetus |
-----------------
1 | Math |
2 | Physics |
3 | History |
4 | Biology |
5 | Gym |
学生
id | name |
-----------------
1 | Johnat |
2 | Sarah |
艾茵嫒学生
id | aine_id | student_id |
---------------------------------
1 | 1 | 1 |
2 | 2 | 1 |
3 | 1 | 2 |
4 | 3 | 2 |
5 | 4 | 2 |
6 | 5 | 2 |
希恩
id | tulemus | aine_tudeng_id|
---------------------------------
1 | 5 | 1 |
2 | 3 | 2 |
3 | 1 | 3 |
4 | 2 | 4 |
5 | 3 | 5 |
6 | 2 | 6 |
现在我做了一个像这样的查询,它包含了我需要的所有数据
SELECT t.name,a.nimetus,h.tulemus FROM aine_student at
INNER JOIN student t ON t.student_id=at.student_id
INNER JOIN aine a ON a.aine_id=at.aine_id
INNER JOIN hinne h ON h.aine_student_id=at.aine_student_id
查询结果
name | nimetus | tulemus|
---------------------------------
Johnat | Math | 5 |
Sarah | Math | 1 |
Johnat | Physics | 3 |
Sarah | History | 2 |
Sarah | Biology | 3 |
Sarah | Gym | 2 |
但是我想这个查询不适合这个表,所以我为separate
aine and name,tulemus
function getNimetus()
{
$query = $this->db->prepare("SELECT distinct a.nimetus FROM aine_student at
INNER JOIN student t ON t.student_id=at.student_id
INNER JOIN aine a ON a.aine_id=at.aine_id
INNER JOIN hinne h ON h.aine_student_id=at.aine_student_id ");
$query->execute();
$exist = $query->fetchAll();
if(!$exist) {
throw new Exception('DB error'); //you can send your exception
}
$nimetus=[];
for ($i=0; $i < count($exist); $i++) {
$nimetus[]=$exist[$i]["nimetus"];
}
return $nimetus;
}
function getTulemus()
{
$query = $this->db->prepare("SELECT t.name,h.tulemus FROM aine_student at
INNER JOIN student t ON t.student_id=at.student_id
INNER JOIN aine a ON a.aine_id=at.aine_id
INNER JOIN hinne h ON h.aine_student_id=at.aine_student_id ");
$query->execute();
$exist = $query->fetchAll();
if(!$exist) {
throw new Exception('DB error'); //you can send your exception
}
$tulemus=[];
for ($i=0; $i < count($exist); $i++) {
$tulemus[]=["Nimi"=>[$exist[$i]["nimi"]=>$exist[$i]["tulemus"]]];
}
return $tulemus;
}
function createTable() {
$nimi=$this->getNames();
$nimetus=$this->getNimetus();
$tulemus=$this->getTulemus();
echo "<table><thead><th>Nimi</th>";
for ($i=0; $i <count($nimetus) ; $i++) {
echo "<th>".$nimetus[$i]."</th>";
}
echo "</thead><tbody>";
echo "<pre>".print_r($tulemus,true)."</pre>";
for ($i=0; $i <count($tulemus) ; $i++) {
foreach ($tulemus[$i]["Nimi"] as $key => $value) {
if($key==)
}
}
}
但这就是我所走的路。我真的很难做这张桌子,所以任何帮助都是值得感谢的。
最佳答案
这个过程称为pivot。
这是通过组合MAX和CASE来完成的
查询
SELECT
student.name
, MAX(
CASE
WHEN Aine.nimetus = 'Math'
THEN Hinne.tulemus
ELSE ''
END
)
AS Math
, MAX(
CASE
WHEN Aine.nimetus = 'Physics'
THEN Hinne.tulemus
ELSE ''
END
)
AS Physics
, MAX(
CASE
WHEN Aine.nimetus = 'History'
THEN Hinne.tulemus
ELSE ''
END
)
AS History
, MAX(
CASE
WHEN Aine.nimetus = 'Biology'
THEN Hinne.tulemus
ELSE ''
END
)
AS Biology
, MAX(
CASE
WHEN Aine.nimetus = 'Gym'
THEN Hinne.tulemus
ELSE ''
END
)
AS Gym
FROM
student
INNER JOIN
Aine_student
ON
student.id = Aine_student.student_id
INNER JOIN
Aine
ON
Aine.id = Aine_student.aine_id
INNER JOIN
Hinne
ON
Aine.id = Hinne.aine_student_id
GROUP BY
student.name
结果
我的结果和你预期的不同。
我认为你在制作例外结果表时犯了一些错误。
| name | Math | Physics | History | Biology | Gym |
|--------|------|---------|---------|---------|-----|
| Johnat | 5 | 3 | | | |
| Sarah | 5 | | 1 | 2 | 3 |
演示http://www.sqlfiddle.com/#!9/62aa4a/1
关于php - 如何根据名称放置单元格值?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/46608144/