Question

这是让我睡了2天的问题。

我有2张桌子

views 

id | postid |   date     | count
=================================  
13 |   8    | 2016-07-16 |  38
16 |   8    | 2016-07-17 |  35
15 |   9    | 2016-07-16 |  7
17 |   9    | 2016-07-17 |  32
14 |   12   | 2016-07-16 |  17
18 |   12   | 2016-07-17 |  13


visitors 

id | postid |   date     | ip
=================================  
13 |   8    | 2016-07-16 |  127.0.0.1
17 |   8    | 2016-07-17 |  127.0.0.1
18 |   8    | 2016-07-17 |  127.0.0.1
16 |   9    | 2016-07-16 |  127.0.0.1
19 |   9    | 2016-07-17 |  127.0.0.1
14 |   12   | 2016-07-16 |  127.0.0.1
15 |   12   | 2016-07-16 |  127.0.0.1
20 |   12   | 2016-07-17 |  127.0.0.1
21 |   12   | 2016-07-17 |  127.0.0.1

以下查询

$query = $wpdb->get_results("
    SELECT
        SUM(a.count) AS countviews, 
        COUNT(b.ip) AS countvisitors,
        a.postid
    FROM views a
    RIGHT JOIN visitors b
        ON a.postid=b.postid
        AND a.date=b.date
    WHERE 
        a.date
    BETWEEN 
        DATE_SUB('2016-07-17', INTERVAL 3 DAY) 
    AND 
        '2016-07-17' 
    GROUP BY 
        a.postid
    ORDER BY 
        countviews DESC
");

当我打印输出时，我会看到以下结果

Array
(
    [0] => stdClass Object
        (
            [countviews] => 108
            [countvisitors] => 3
            [postid] => 8
        )

    [1] => stdClass Object
        (
            [countviews] => 60
            [countvisitors] => 4
            [postid] => 12
        )

    [2] => stdClass Object
        (
            [countviews] => 39
            [countvisitors] => 2
            [postid] => 9
        )

 )

只有[countviews]结果高于已经过期的结果。我会计算并看到来自postid 8的countviews不能是＆＃39; 108＆＃39;但是＆＃39; 73＆＃39;。关于它的陌生事情是postid 8的最后一次计数是35＆＃39;。＆＃39; 108＆＃39;减去＆＃39; 35＆＃39; =＆＃39; 73＆＃39;。所以视图表计数加倍？

RIGHT JOIN，LEFT JOIN和INNER JOIN给出了相同的结果。

Answer 1

如果你想数数，你不能在这里加入。您所建立的关系是创建视图表的倍数，以防您的搜索参数中存在多天相同的postid。

您可以使用子查询来避免这种情况：

SELECT
    SUM(a.count) AS countviews,
    (SELECT COUNT(b.ip) FROM visitors i WHERE b.date BETWEEN DATE_SUB("2016-07-17", INTERVAL 3 DAY) AND "2016-07-17" AND i.postid = a.postid) AS countvisitors,
    a.postid
FROM views a
WHERE 
    a.date
BETWEEN 
    DATE_SUB('2016-07-17', INTERVAL 3 DAY) 
AND 
    '2016-07-17' 
GROUP BY 
    a.postid
ORDER BY 
    countviews DESC

希望我做对了。如果这有帮助，请告诉我:)）

MySQL Join的回报超过预期

1 个答案: