2 COUNT內側留下了一個選擇加入

用戶：

ID | NAME 
1 | caio 
2 | mike 
3 | peter

發送：

ID | ID_SEND | ID_RECEIVE 
1 | 1 |  2 
2 | 2 |  3 
3 | 3 |  2 
4 | 2 |  1

用戶每次發送卡（肚裏通過電子郵件）給另一個用戶，提供表格。

問題：我想寫一個SELECT知道用戶有多少次是ID_SEND多少次用戶在ID_RECEIVE。我想是這樣的：

SELECT 
    user.email, 
    COUNT(sends.id_receive) AS numberReceive, 
    COUNT(sends.id_send) AS numberSend 
FROM 
    user 
LEFT JOIN 
    sends ON user.id = sends.id_send OR user.id = sends.id_receive 
GROUP BY 
    user.email

的問題是，無論是ID_SEND和ID_RECEIVE正在返回的值相同，則兩者之和......我要去的地方錯了嗎？

來源

2016-03-09 caiocafardo

LEFT JOIN兩次，一次用於發送，一旦收到

SELECT u.email, 
     COUNT(r.id_receive) AS numberReceive, 
     COUNT(s.id_send) AS numberSend 
FROM user u 
LEFT JOIN sends r ON u.id = r.id_receive 
LEFT JOIN sends s ON u.id = s.id_send 
GROUP BY u.email

另外，單LEFT JOIN，使用CASE表達式條件計數：

SELECT u.email, 
     SUM(case when u.id = s.id_receive then 1 else 0 end) AS numberReceive, 
     SUM(case when u.id = s.id_send then 1 else 0 end) AS numberSend 
FROM user u 
LEFT JOIN sends s ON u.id IN (s.id_receive, s.id_send) 
GROUP BY u.email

來源

2016-03-09 20:39:32 jarlh

2 LEFT JOIN我有一個錯誤：mysql_fetch_array（）預計參數1是資源 – caiocafardo

這是非常奇怪的......應該執行罰款。我會添加一個替代解決方案。請檢查5. – jarlh

謝謝它的工作原理！我真的不知道wy與2左加入dindt作品.... – caiocafardo

試試這個。

SELECT S.*, 
    R.received 
FROM (SELECT u.NAME, 
      Count(snd.id) sends 
    FROM USER u 
      JOIN sends snd 
      ON snd.id_send = u.id 
    GROUP BY u.NAME) S 
    LEFT JOIN (SELECT u.NAME, 
        Count(rec.id) received 
       FROM USER u 
        JOIN sends rec 
         ON rec.id_receive = u.id 
       GROUP BY u.NAME) R 
      ON S.NAME = R.NAME

來源

2016-03-09 21:08:29

2 COUNT內側留下了一個選擇加入

回答

相關問題