MySQL查询有两个连接

Question

我有查询我需要在三个链表上执行MySQL。 我可以用嵌套查询做懒惰的方式，但我无法弄清楚如何使用单个查询来完成它。

表是：

Area:-
 : id     (int)
 : name   (string)

Consultant:-
 :id      (int)
 :active  (1/0)

ConsArea:-
 : areaID        (int)
 : consultantID  (int)

我需要遍历所有区域（使用$area变量），以便列出所有区域和每个区域以指示“活动”顾问的数量...因此所有区域必须列在其旁边的值（其中如果没有相关的活跃顾问，则可以为零）

查询的第一部分（无论顾问是否活跃）我可以做：

      SELECT areas.name AS aname, COUNT(consAreas.areaID) AS cct 
        FROM areas LEFT OUTER JOIN consAreas 
          ON consAreas.areaID = areas.id 
       WHERE areas.areaID = $area 
    GROUP BY areas.id 
    ORDER BY areas.name

..但是当我想要包括顾问活跃的条件时，我无法确定正确的连接。 它只列出了> 0活跃顾问的区域，而我需要所有区域。

      SELECT areas.name AS aname, COUNT(consAreas.area) AS cct 
        FROM areas LEFT OUTER JOIN consAreas 
          ON consAreas.area = areas.id 
        **JOIN consultants ON consultants.id = consAreas.cons**
       WHERE areas.areaID = $area 
         **AND consultants.active = 1**
    GROUP BY areas.id 
    ORDER BY areas.name

有人帮吗？

Answer 1

这是因为mysql的行为。 左连接后的内连接使左连接成为内连接。

SELECT areas.name AS aname, COUNT(consultants.id) AS cct 
FROM areas
    LEFT JOIN consAreas ON consAreas.area = areas.id 
    LEFT JOIN consultants ON consultants.id = consAreas.cons AND consultants.active = 1
WHERE 
    areas.areaID = $area 
GROUP BY areas.id 
ORDER BY areas.name

在这里你可以看到我只使用左连接，更重要的是直接从左边连接ON子句过滤consultants.active状态。

Answer 2

你想要的是LEFT JOIN （又名左外连接）。

JOIN （实际上是一个内连接）只有在两个表中都有相应的行时才会选择JOIN产生的行。 如果左表只有匹配的行，则左连接将选择行，无论右表是否匹配。

所以在你加入顾问表时：

SELECT areas.name AS aname, COUNT(consAreas.area) AS cct 
FROM areas LEFT OUTER JOIN consAreas 
ON consAreas.area = areas.id 
LEFT JOIN consultants ON consultants.id = consAreas.cons

Answer 3

您需要执行LEFT JOIN并更正GROUP BY 。 尝试这个：

SELECT areas.name AS aname, COUNT(consultants.active) AS cct 
    FROM areas 
    LEFT JOIN consAreas 
      ON consAreas.area = areas.id 
    LEFT JOIN consultants 
      ON consultants.id = consAreas.cons
   WHERE areas.areaID = $area 
     AND consultants.active = 1
GROUP BY areas.name
ORDER BY areas.name

这将返回一个表格，其中包含区域名称和活跃顾问的数量

Answer 4

尝试这个：

SELECT areas.name AS aname, COUNT(consultants.id) AS cct 
    FROM areas LEFT OUTER JOIN consAreas 
      ON consAreas.areaID = areas.id 
    LEFT OUTER JOIN consultants ON consAreas.consultantID=consultants.id AND consultants.active = 1
   WHERE areas.areaID = $area 
     GROUP BY areas.id 
ORDER BY areas.name