Question

这是我的表position：

+----+---------+------+--------+---------+
| id | teacher | cook | doctor | dentist |
+----+---------+------+--------+---------+
|  1 |       3 |    4 |      2 |       1 |
+----+---------+------+--------+---------+

这是我的表people：

+----+-----------+--------+-----+
| id | firstname |  name  | age |
+----+-----------+--------+-----+
|  1 | Fred      | Miller |  42 |
|  2 | Emily     | Rose   |  32 |
|  3 | Ben       | Harper |  38 |
|  4 | Samanta   | Jones  |  35 |
+----+-----------+--------+-----+

我的mySQL数据库请求

$pdo = $db->query('
SELECT *, position.id AS id, people.id AS people_id 
FROM position 
LEFT JOIN people 
ON position.teacher=people.id;
ON position.cook=people.id;
ON position.doctor=people.id;
ON position.dentist=people.id;
');

while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
echo "The teacher is "$row['firstname']." ".$row['name'];
echo "The cook is "$row['firstname']." ".$row['name'];
echo "The doctor is "$row['firstname']." ".$row['name'];
echo "The dentist is "$row['firstname']." ".$row['name'];
}

我的结果是：

The teacher is Ben Harper
The cook is Ben Harper
The doctor is Ben Harper
The dentist is Ben Harper

我需要的结果：

The teacher is Ben Harper
The cook is Samanta Jones
The doctor is Emiliy Rose
The dentist is Fred Miller

Answer 1

如果你重组你的桌子会更好。

+----+------------+
| id | position   |       
+----+------------|
|  1 |  teacher   |
|  2 |  cook      |
|  3 |  doctor    |
|  4 |  dentist   |
+----+------------+


+----+-----------+--------+-----+-------------+
| id | firstname |  name  | age | position_id |
+----+-----------+--------+-----+-------------+
|  1 | Fred      | Miller |  42 |  4          |
|  2 | Emily     | Rose   |  32 |  3          |
|  3 | Ben       | Harper |  38 |  1          |
|  4 | Samanta   | Jones  |  35 |  2          |
+----+-----------+--------+-----+-------------+

这里你可以在people.position_id和position.id上有一个外键引用。这样你就可以让很多人拥有相同的位置。

SELECT position.id id
     , people.id people_id
     , people.firstname
     , people.name
  FROM position 
  LEFT 
  JOIN people 
    ON people.position_id = position.id;

Answer 2

如上面的评论和答案中所述，如果您可以重组您的表格，您绝对应该 - 正如其他人所提到的那样，最好重新构建它以获得额外的列，然后您的查询将更简单，更有效。但是如果你被困在一个你无法做到的位置（如果你只提供了大量的数据，那么效率低下并不是太大的问题）那么你可以使用一个UNION将几个简单的查询粘合在一起：

var newData = data.map(function(item) { return item/scalar }

同样，这不是一种有效的方法 - 如果可以，您应该重新构建数据。但是如果你出于某种原因无法做到这一点，那么这将使你在一组行中找回所需的数据，从而拉回所需的数据。

Answer 3

试试这个，

$pdo = $db->query('
SELECT *, position.id AS id, people.id AS people_id,
(case when position.teacher <> '' then 'teacher'
when position.cook <> '' then 'cook'
when position.doctor <> '' then 'doctor' 
when position.dentist <> '' then 'dentist') position_name
FROM position 
LEFT JOIN people 
ON position.teacher=people.id;
ON position.cook=people.id;
ON position.doctor=people.id;
ON position.dentist=people.id;
');

while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
echo "The ".$row['position_name']." is "$row['firstname']." ".$row['name'];
}

如何从同一个mySQL表中连接多个数据？

3 个答案: