我有一张桌子（名字vcar）
id | createddatetime |
--------------------------
1 |2016-08-20 17:23:53 |
2 |2016-08-20 17:23:53 |
3 |2016-08-20 17:23:53 |
4 |2016-08-20 17:23:53 |
4 |2016-08-20 17:23:53 |
4 |2016-08-21 17:23:53 |
3 |2016-08-21 17:23:53 |
1 |2016-08-21 17:23:53 |
1 |2016-08-21 17:23:53 |
2 |2016-08-24 17:23:53 |
3 |2016-08-24 17:23:53 |
4 |2016-08-24 17:23:53 |
2 |2016-08-29 17:23:53 |
2 |2016-08-29 17:23:53 |
3 |2016-08-29 17:23:53 |
4 |2016-08-29 17:23:53 |
4 |2016-08-30 17:23:53 |
2 |2016-08-30 17:23:53 |
3 |2016-08-30 17:23:53 |
4 |2016-08-30 17:23:53 |
我尝试使用
获取特定日期范围中id的计数SELECT
count(CAST(createddatetime AS DATE)),
CAST(createddatetime AS DATE),
DAYNAME(CAST(createddatetime AS DATE))
FROM vcar
WHERE
CAST(createddatetime AS DATE)>=date('2016-08-20') AND
CAST(createddatetime AS DATE)<=date('2016-08-35')
GROUP BY CAST(createddatetime AS DATE);
这给了我o / p，但我还需要显示在某一天没有点击的日期。如何做到这一点？
O / p我正在
-------------------------------------- ------------------------------- ----------------------------------------
count(CAST(createddatetime AS DATE)) | CAST(createddatetime AS DATE) | DAYNAME(CAST(createddatetime AS DATE)) |
-------------------------------------- ------------------------------- ----------------------------------------
9 | 2016-08-20 | Saturday |
36 | 2016-08-23 | Tuesday |
4 | 2016-08-24 | Wednesday |
-------------------------------------- ------------------------------- ----------------------------------------
我需要什么
-------------------------------------- ------------------------------- ----------------------------------------
count(CAST(createddatetime AS DATE)) | CAST(createddatetime AS DATE) | DAYNAME(CAST(createddatetime AS DATE)) |
-------------------------------------- ------------------------------- ----------------------------------------
9 | 2016-08-20 | Saturday |
0 | 2016-08-21 | Sunday |
0 | 2016-08-22 | Monday |
36 | 2016-08-23 | Tuesday |
4 | 2016-08-24 | Wednesday |
9 | 2016-08-25 | Thursday |
0 | 2016-08-26 | Friday |
-------------------------------------- ------------------------------- ----------------------------------------
答案 0 :(得分：2)
日历表会让这更容易。由于您没有，因此以下查询将完成此任务。
SELECT
dateTable.day,
DAYNAME(dateTable.day) AS dayName,
COALESCE(yourQuery.total,0) AS cnt
FROM
(
SELECT ADDDATE('2016-08-20', INTERVAL @i:=@i+1 DAY) AS DAY
FROM (
SELECT a.a
FROM (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS a
CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS b
CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS c
) a
JOIN (SELECT @i := -1) r1
WHERE
@i < DATEDIFF('2016-08-31', '2016-08-20')
) AS dateTable
LEFT JOIN
(
SELECT
DATE(createddatetime) date,
count(id) As total
FROM vcar
WHERE DATE(createddatetime) BETWEEN '2016-08-20' AND '2016-08-31'
GROUP BY date
) AS yourQuery
ON dateTable.day = yourQuery.date
ORDER BY dateTable.day
注意：
dateTable别名中的查询将生成从2016-08-20到2016-08-31的日期。