Mysql union all select问题

Posted 2023-03-24

【中文标题】Mysql union all select问题

【英文标题】：Mysql union all select issue

【发布时间】：2017-01-10 21:49:06

【问题描述】：

我正试图找出我哪里出错了。我正在使用不同的 Where 语句从同一个表中加入多个选择语句，其中一个选择也是一个连接。我正在尝试使用 union all 但我似乎遗漏了一些东西，我无法弄清楚它是什么。任何帮助都会很棒。以下是我的代码：

SET @LASTQTR:=IF((QUARTER(CURDATE( ))-1) = 0, 4, QUARTER(CURDATE( ))-1);
SET @YR:=IF(@LASTQTR = 4, YEAR(NOW( ))-1, YEAR(NOW( )));

(SELECT COUNT(r.id) AS total, CONCAT(m1.first_name,' ', m1.last_name) AS fromName  FROM Referrals AS r JOIN Members AS m1 ON m1.id=r.from_id WHERE QUARTER(rdate) = @LASTQTR AND YEAR(rdate) = @YR  GROUP BY r.from_id)
    UNION ALL
    (SELECT COUNT(id) AS external FROM Referrals WHERE QUARTER(rdate) = @LASTQTR AND YEAR(rdate) = @YR AND rtype=1 GROUP BY from_id)
    UNION ALL
    (SELECT COUNT(id) AS internal FROM Referrals WHERE QUARTER(rdate) = @LASTQTR AND YEAR(rdate) = @YR AND rtype=2  GROUP BY from_id)
    ORDER BY total DESC LIMIT 10;

我从这个查询中寻找的最终结果是在 while 循环中使用结果，如下所示：

while ( $re = $q8->fetch(PDO::FETCH_ASSOC)) 
    echo'<tr>
    <td style="width:40%;">'.$re['fromName'].'</td>
    <td class="text-right">'.$re['external'].'</td>
    <td class="text-right">'.$re['internal'].'</td>
    <td class="text-right">'.$re['total'].'</td>
    </tr>';

【参考方案1】：

所以我可能会在这种情况下帮助其他人，这就是我正在寻找的结果：

SELECT COUNT(r.id) AS total, CONCAT(m1.first_name,' ', m1.last_name) AS fromName, 
(SELECT COUNT(id) FROM Referrals WHERE rtype=1 AND from_id=r.from_id GROUP BY fromName) AS external,
(SELECT COUNT(id) FROM Referrals WHERE rtype=2 AND from_id=r.from_id GROUP BY fromName) AS internal
FROM Referrals AS r JOIN Members AS m1 ON m1.id=r.from_id WHERE QUARTER(rdate) = @LASTQTR AND YEAR(rdate) = @YR  GROUP BY fromName ORDER BY total DESC LIMIT 10

【参考方案2】：

对于联合，所有集合都应该具有相同的列。在您的情况下，第一个查询也有 fromName 。要么删除它，要么在其他查询中放置占位符

SET @LASTQTR:=IF((QUARTER(CURDATE( ))-1) = 0, 4, QUARTER(CURDATE( ))-1);
SET @YR:=IF(@LASTQTR = 4, YEAR(NOW( ))-1, YEAR(NOW( )));

(SELECT COUNT(r.id) AS total, CONCAT(m1.first_name,' ', m1.last_name) AS fromName  FROM Referrals AS r JOIN Members AS m1 ON m1.id=r.from_id WHERE QUARTER(rdate) = @LASTQTR AND YEAR(rdate) = @YR  GROUP BY r.from_id)
    UNION ALL
    (SELECT COUNT(id) AS external,'' FROM Referrals WHERE QUARTER(rdate) = @LASTQTR AND YEAR(rdate) = @YR AND rtype=1 GROUP BY from_id)
    UNION ALL
    (SELECT COUNT(id) AS internal,'' FROM Referrals WHERE QUARTER(rdate) = @LASTQTR AND YEAR(rdate) = @YR AND rtype=2  GROUP BY from_id)
    ORDER BY total DESC LIMIT 10;

---

根据您的评论（尚未执行此操作）

(SELECT COUNT(r.id) AS total, CONCAT(m1.first_name,' ', m1.last_name) AS fromName,
(SELECT COUNT(id)  FROM Referrals WHERE QUARTER(rdate) = @LASTQTR AND YEAR(rdate) = @YR AND rtype=1 GROUP BY from_id) AS external,
(SELECT COUNT(id)  FROM Referrals WHERE QUARTER(rdate) = @LASTQTR AND YEAR(rdate) = @YR AND rtype=2  GROUP BY from_id) AS internal
  FROM Referrals AS r JOIN Members AS m1 ON m1.id=r.from_id WHERE QUARTER(rdate) = @LASTQTR AND YEAR(rdate) = @YR  GROUP BY r.from_id)

【讨论】：

使用上述方法的这个问题是，当循环遍历结果时，它将每个 select 语句作为一个新行放入 while 循环中。我认为使用 Union all 会返回每行所有三个选择的结果。