Mysql union所有选择问题

我想弄清楚我在哪里出错了。我正在使用不同的Where语句从同一个表中加入多个select语句，其中一个选择也是一个连接。我试图使用工会，但我似乎失去了一些东西，我不知道它是什么。任何帮助都会很棒。下面是我的代码：Mysql union所有选择问题

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;

我要寻找从该查询使用导致while循环，像这样的最终结果是：

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>'; 
}

来源

2017-01-10 Action Coding

工会的所有集应该有相同的不列。在你的情况下，第一个查询也有fromName。要么删除或把占位符其他查询以及

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（还没有执行此）

(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)

来源

2017-01-10 21:51:01 uSeruSher

使用上面的这个问题是当循环遍历结果时，它将每个sel ect语句作为while循环中的新行。我认为使用Union将会返回每行所有三个选择的结果。 –

请参阅我对原始问题的编辑 –

所以我有可能会帮助其他人在这种情况下，这是什么给了我正在寻找的结果：

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

来源

2017-01-11 00:19:17

Mysql union所有选择问题

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