MySQL查詢有兩個連接

Question

我有查詢我需要在三個鏈表上執行MySQL。 我可以用嵌套查詢做懶惰的方式，但我無法弄清楚如何使用單個查詢來完成它。

表是：

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

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

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

我需要遍歷所有區域（使用$area變量），以便列出所有區域和每個區域以指示“活動”顧問的數量...因此所有區域必須列在其旁邊的值（其中如果沒有相關的活躍顧問，則可以為零）

查詢的第一部分（無論顧問是否活躍）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

..但是當我想要包括顧問活躍的條件時，我無法確定正確的連接。 它只列出了> 0活躍顧問的區域，而我需要所有區域。

      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

有人幫嗎？

Answer 1

這是因為mysql的行為。 左連接后的內連接使左連接成為內連接。

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

在這里你可以看到我只使用左連接，更重要的是直接從左邊連接ON子句過濾consultants.active狀態。

Answer 2

你想要的是LEFT JOIN （又名左外連接）。

JOIN （實際上是一個內連接）只有在兩個表中都有相應的行時才會選擇JOIN產生的行。 如果左表只有匹配的行，則左連接將選擇行，無論右表是否匹配。

所以在你加入顧問表時：

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

您需要執行LEFT JOIN並更正GROUP BY 。 嘗試這個：

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

這將返回一個表格，其中包含區域名稱和活躍顧問的數量

Answer 4

嘗試這個：

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