Question

我有2个查询，我想将其合并为1.但是，我收到了错误：

//first query:
SELECT
  @rownum := @rownum +1 `index`
  , t. *
FROM ipAll t, ( SELECT @rownum :=0 ) r

//second query:

SELECT DISTINCT
  COUNT( ip )
  , SUM( views )
  , DATE
FROM ipAll
WHERE date LIKE '%-12-%'
GROUP BY DATE
ORDER BY DATE ASC
LIMIT 31

// combined query: (Totally wrong. How do I fix this?)

 SELECT
  @rownum := @rownum +1 `index`
  , DISTINCT COUNT( ip )
  , SUM( views )
  , DATE
FROM ipAll t, ( SELECT @rownum :=0 ) r
WHERE date LIKE '%-12-%'
GROUP BY DATE
ORDER BY DATE ASC
LIMIT 31

Answer 1

我建议删除distinct。

我们假设您的表格包含以下内容：

|                         dt |      ip | views |
|----------------------------|---------|-------|
| December, 01 2015 00:00:00 | 1.1.1.1 |   100 |
| December, 01 2015 00:00:00 | 2.2.2.2 |    10 |
| December, 02 2015 00:00:00 | 3.3.3.3 |     4 |
| December, 01 2014 00:00:00 | 1.1.1.1 |   100 |
| November, 01 2015 00:00:00 | 1.1.1.1 |     1 |
| November, 01 2015 00:00:00 | 2.2.2.2 |     2 |

查询：

SELECT 
@rownum := @rownum +1 `index`,
COUNT( ip ), SUM( views ) , dt 
FROM ipAll,  ( SELECT @rownum :=0 ) r 
WHERE dt LIKE '%-12-%' 
GROUP BY dt 
ORDER BY dt ASC 
LIMIT 31;

将导致：

| index | COUNT( ip ) | SUM( views ) |                         dt |
|-------|-------------|--------------|----------------------------|
|     3 |           1 |          100 | December, 01 2014 00:00:00 |
|     1 |           2 |          110 | December, 01 2015 00:00:00 |
|     2 |           1 |            4 | December, 02 2015 00:00:00 |

获得类似（不相同）结果的另一种方法是：

select @rownum := @rownum +1 `index`, main.*
from ( SELECT @rownum :=0 ) r, (
  SELECT COUNT( ip ), SUM( views ) , dt 
  FROM ipAll
  WHERE dt LIKE '%-12-%' 
  GROUP BY dt 
  LIMIT 31
) main
ORDER BY dt ASC

结果将是：

| index | COUNT( ip ) | SUM( views ) |                         dt |
|-------|-------------|--------------|----------------------------|
|     1 |           1 |          100 | December, 01 2014 00:00:00 |
|     2 |           2 |          110 | December, 01 2015 00:00:00 |
|     3 |           1 |            4 | December, 02 2015 00:00:00 |

SQLFiddle示例：http://sqlfiddle.com/#!9/beb98/3

将2个不同的查询合并为一个

1 个答案: