1

我有以下查询

SELECT MONTH, COUNT(DISTINCT VISITS) AS BRAND_VISITS, NULL AS NONB_VISITS
FROM Table1
WHERE KEYWORD_TYPE = BRAND(
AND DATE >= '2013-01-01'
GROUP BY MONTH

UNION ALL

SELECT MONTH, NULL, COUNT(DISTINCT VISITS) AS NONB_VSTS 
FROM Table1
WHERE KEYWORD_TYPE = NON-BRAND
AND DATE >= '2013-01-01'
GROUP BY MONTH

我得到以下结果：

1   352540  NULL
2   309834  NULL
3   228764  NULL
4   236054  NULL
5   218096  NULL
6   172527  NULL
1   NULL    5337
2   NULL    14120
3   NULL    9954
4   NULL    23755
5   NULL    19771
6   NULL    30797

但是，我想要的是没有 NULLS 的内联结果

1   352540  5337
2   309834  14120
3   228764  9954
4   236054  23755
5   218096  19771
6   172527  30797
4

2 回答 2

0

您可以使用带有 CASE 的单个语句或使用 JOIN on month 而不是 UNION 来做到这一点。如果您采用连接方法，您可能需要考虑空值（一个月内没有访问关键字）。您将需要对它们进行概要分析，以查看您的数据和表结构哪个更快。这实际上与您需要聚合的索引和数据量有关。

假设您不必担心基于示例中的计数的空值，这就是您想要的。 

SELECT brand.month, brand.brand_visits,nonbrand.non_brand_visits
FROM (SELECT month, COUNT(visits) AS brand_visits
      FROM Table1
      WHERE keyword_type = 'BRAND'
      AND date >= '2013-01-01'
      GROUP BY month) brand
INNER JOIN
      (SELECT month, COUNT(visits) AS non_brand_visits 
       FROM Table1
       WHERE keyword_type = 'NON-BRAND'
       AND date >= '2013-01-01'
       GROUP BY month) nonbrand 
ON brand.month=nonbrand.month

这是 CASE 方法。您应该根据您正在聚合的实际数据和索引进行分析，以查看哪种方法更快。 

SELECT month, 
SUM(CASE WHEN keyword_type = 'BRAND' THEN 1 ELSE 0 END) AS brand_visits, 
SUM(CASE WHEN keyword_type = 'NON-BRAND' THEN 1 ELSE 0 END) AS non_brand_visits 
FROM Table1
WHERE date >= '2013-01-01'
GROUP BY month

最后，您没有提供表结构或示例数据，所以我在上面做了一些假设。我坚信您不需要COUNT(DISTINCT原始声明中的 。我已将其删除并验证了上面的两个语句产生了相同的结果。如果COUNT(DISTINCT需要，则 CASE 方法将不起作用，但 join 方法仍然可以正常工作。

于 2013-07-20T04:15:04.517 回答
0

使用您的列：

SELECT month, 
count(distinct CASE WHEN keyword_type = 'BRAND' THEN visits END) AS BRAND_VISITS, 
count(distinct CASE WHEN keyword_type = 'NON-BRAND' THEN visits END) AS NONB_VSTS  
FROM Table1
WHERE date >= '2013-01-01'
and keyword_type in ('BRAND','NON-BRAND')
GROUP BY month

我很想相信月份只是日期列中的月份，我更喜欢这个解决方案，它涵盖的年份多于 1 年，并且相同的查询在 2014 年仍然有效

SELECT cast(dateadd(month, datediff(month, 0, date), 0) as date) month, 
count(distinct CASE WHEN keyword_type = 'BRAND' THEN visits END) AS BRAND_VISITS, 
count(distinct CASE WHEN keyword_type = 'NON-BRAND' THEN visits END) AS NONB_VSTS 
FROM Table1
WHERE date >= '2013-01-01'
and keyword_type in ('BRAND','NON-BRAND')
GROUP BY datediff(month, 0, date)

如果您想坚持使用旧脚本，可以通过以下方式修复它：

SELECT MONTH, max(BRAND_VISITS) BRAND_VISITS, max(NONB_VISITS) NONB_VISITS 
FROM
(
SELECT MONTH, COUNT(DISTINCT VISITS) AS BRAND_VISITS, NULL AS NONB_VISITS
FROM Table1
WHERE KEYWORD_TYPE = 'BRAND'
AND DATE >= '2013-01-01'
GROUP BY MONTH
UNION ALL
SELECT MONTH, NULL, COUNT(DISTINCT VISITS) AS NONB_VSTS 
FROM Table1
WHERE KEYWORD_TYPE = 'NON-BRAND'
AND DATE >= '2013-01-01'
GROUP BY MONTH
) a
GROUP BY MONTH
于 2013-07-20T06:51:55.270 回答