Question

我对MySQL很新，我在解决这个问题时遇到了一些麻烦。

我有两个名为doctors和doctor_tags

的表格

doctors的字段：doctor_id，doctor_name

doctor_tags的字段：id，doctor_id，tag_name

现在针对特定的doctor_id，我需要找到相关的医生。如果医生具有相同的tag_name，则相关。一旦我找到这种关系，我需要让所有这些医生doctor_name回来。

我几乎完全失去了我已经走了这么远（我确定这是错的）

SELECT doctors.doctor_name, doctors.doctor_id FROM doctors INNER JOIN doctors_tags ON doctors.doctor_id = doctors_tags.doctor_id ...

我认为这个查询很没用，但我知道我们需要某种类型的连接。

对于那些善于提出某种查询的人，如果你能在这个过程中解释它的每个部分，我将非常感激。谢谢:)

编辑：第2部分

如果我们说我们引入另一个在doctor_id和tag_id之间具有N：N关系的表

表字段 doctor_tags_joins：doctor_id，tag_id

因此标记表的字段变为

doctors_tags：tag_id，tag_name等等......

我们怎么能用混合中包含的附加表做同样的事情。

Answer 1

您需要两次加入表格：

SELECT DISTINCT d1.doctor_id, d2.doctor_name
FROM doctors AS d1
-- Get first doctor's tags
JOIN doctor_tags AS dt1 ON d1.doctor_id = dt1.doctor_id
-- Get all the other doctor_tags rows with the same tags
JOIN doctor_tags AS dt2 ON dt1.tag_name = dt2.tag_name 
    -- Don't list doctors as related to themselves
    AND dt1.doctor_id != dt2.doctor_id
-- Get those doctors' names
JOIN doctors AS d2 ON dt2.doctor_id = d2.doctor_id
ORDER BY d1.doctor_id, d2.doctor_name

我使用DISTINCT因此，如果他们有多个共同的标签，我们就不会为同一位医生获取多行。

如果您只是查找与特定医生相关的医生，而不是所有医生，则查询将变为：

SELECT doctor_name
FROM doctor_tags AS dt1
JOIN doctor_tags AS dt2 ON dt1.tag_name = dt2.tag_name AND dt1.doctor_id != dt2.doctor_id
JOIN doctors AS d2 ON dt2.doctor_id = d2.doctor_id
WHERE dt1.doctor_id = :doctor_id

其中:doctor_id是给定医生的占位符。您不需要第一个doctor表格，因为您没有使用该表格中的任何信息。

如果您想在一行中获取所有相关医生姓名，请使用GROUP_CONCAT：

SELECT d1.doctor_id, GROUP_CONCAT(DISTINCT d2.doctor_name) AS related_doctors
FROM doctors AS d1
JOIN doctor_tags AS dt1 ON d1.doctor_id = dt1.doctor_id
JOIN doctor_tags AS dt2 ON dt1.tag_name = dt2.tag_name AND dt1.doctor_id != dt2.doctor_id
JOIN doctors AS d2 ON dt2.doctor_id = d2.doctor_id
GROUP BY d1.doctor_id

对于第二个问题，查询是相同的，除了您使用doctor_tags_join而不是doctor_tags。由于您未在结果中显示标记名称，因此doctor_tags表格无关紧要。

SELECT DISTINCT d1.doctor_id, d2.doctor_name
FROM doctors AS d1
JOIN doctor_tags_join AS dt1 ON d1.doctor_id = dt1.doctor_id
JOIN doctor_tags_join AS dt2 ON dt1.tag_id = dt2.tag_id 
    AND dt1.doctor_id != dt2.doctor_id
JOIN doctors AS d2 ON dt2.doctor_id = d2.doctor_id
ORDER BY d1.doctor_id, d2.doctor_name

Answer 2

SELECT DISTINCT r.doctor_id, r.doctor_name
FROM doctors d
    INNER JOIN doctor_tags dt_d
        ON d.doctor_id = dt_d.doctor_id
    INNER JOIN doctor_tags dt_r
        ON dt_d.tag_name = dt_r.tag_name
    INNER JOIN doctors r
        ON dt_r.doctor_id = r.doctor_id
WHERE d.doctor_id = 12345

Answer 3

您可以使用子查询来实现此目的。子查询只是另一个查询中的查询。这是如何逐步构建的

首先，您需要指定的doctor_id具有的tag_name，对吗？那就是：

select tag_name from doctors_tags where doctor_id='(your doctor_id)'

让所有医生使用相同的tag_name

SELECT doctor_id 
FROM doctors_tags 
WHERE tag_name in(
 SELECT tag_name FROM doctors_tags 
 WHERE doctor_id='(your doctor_id)' 
)

和最终查询，获取名称将是：

select doctors.doctor_name, doctors.doctor_id 
from doctors, doctors_tags 
where doctors_tags.tag_name in(
  SELECT tag_name FROM doctors_tags 
  WHERE doctor_id='(your doctor_id)' 
)
and doctors.doctor_id = doctors_tags.doctor_id;

希望有所帮助

修改

包含您的doctor_tags_joins表后，事情看起来像这样

select doctors.doctor_name, doctors.doctor_id
from doctors, doctors_tags, doctors_tags_joins
where doctors_tags.tag_name in
(
    select tag_name 
    from doctors_tags, doctors_tags_joins
    where doctors_tags_joins.doctor_id=(your id)
    and doctors_tags.tag_id=doctors_tags_joins.tag_id
)
and doctor_tags_joins.tag_id=doctors_tags.tag_id
and doctors.doctor_id=doctors_tags.doctor_id;

在查询MySQL中进行多个连接

3 个答案:

修改