Question

我有一张这样的表 -

CREATE TABLE sent_bulletin_list
(
    `id` int,
    `date` found_time,
    `value` int,
);

我希望每天能够找到最近7天的参赛作品。

我所做的是 -

SELECT
DATE(found_time) as date,
count( * ) as total
FROM
sent_bulletin_list
WHERE
found_time > ( UNIX_TIMESTAMP() - ( 7 * 24 * 60 * 60 ) )
GROUP BY
DATE(found_time);

找到这样的东西 -

+------------+-------+
| date       | total |
+------------+-------+
| 2016-07-01 |     8 |
+------------+-------+

但我喜欢这样的事情 -

+------------+-------+
| date       | total |
+------------+-------+
| 2016-06-25 |     0 |
| 2016-06-26 |     0 |
| 2016-06-27 |     0 |
| 2016-06-28 |     0 |
| 2016-06-29 |     0 |
| 2016-06-30 |     0 |
| 2016-07-01 |     8 |
+------------+-------+

更新 -

我尝试使用case像这样 -

count( CASE found_time IS NOT NULL 
      THEN 1
      ELSE 0
      END AS date )

但它也无效。

有人可以帮忙吗？

提前感谢您的帮助。

Answer 1

您需要一个表格，其中查询中date range中指定的所有日期都会驻留。

你可以尝试一下：

SELECT 
dateTable.day,
COALESCE(your_query.total,0) AS total
FROM 
(
    SELECT DATE(ADDDATE(FROM_UNIXTIME(UNIX_TIMESTAMP() - ( 7 * 24 * 60 * 60 )), 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(FROM_UNIXTIME(UNIX_TIMESTAMP()), FROM_UNIXTIME(UNIX_TIMESTAMP() - ( 7 * 24 * 60 * 60 )))
) dateTable

LEFT JOIN 
(
    SELECT
    DATE(found_time) as date,
    count( * ) as total
    FROM
    sent_bulletin_list
    WHERE
    found_time > ( UNIX_TIMESTAMP() - ( 7 * 24 * 60 * 60 ) )
    GROUP BY    DATE(found_time)
) your_query
ON dateTable.day = your_query.date
ORDER BY dateTable.day

WORKING DEMO

<强>测试

假设您的表格中包含以下数据：

DROP TABLE IF EXISTS `sent_bulletin_list`;
CREATE TABLE `sent_bulletin_list` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `found_time` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00' ON UPDATE CURRENT_TIMESTAMP,
  PRIMARY KEY (`id`)
);
INSERT INTO `sent_bulletin_list` VALUES ('1', '2016-07-17 00:00:00');
INSERT INTO `sent_bulletin_list` VALUES ('2', '2016-07-17 00:00:00');
INSERT INTO `sent_bulletin_list` VALUES ('3', '2016-07-17 00:00:00');
INSERT INTO `sent_bulletin_list` VALUES ('4', '2016-07-17 00:00:00');
INSERT INTO `sent_bulletin_list` VALUES ('5', '2016-07-17 00:00:00');

然后运行上面的查询将给出以下输出：

<强>输出：

day         total
2016-07-14  0
2016-07-15  0
2016-07-16  0
2016-07-17  5
2016-07-18  0
2016-07-19  0
2016-07-20  0
2016-07-21  0

注意：

您需要将end and start date范围放在此处：

@i < DATEDIFF(FROM_UNIXTIME(UNIX_TIMESTAMP()), FROM_UNIXTIME(UNIX_TIMESTAMP() - ( 7 * 24 * 60 * 60 )))

你需要把start date放在这里：

SELECT DATE(ADDDATE(FROM_UNIXTIME(UNIX_TIMESTAMP() - ( 7 * 24 * 60 * 60 )), INTERVAL @i:=@i+1 DAY)) AS DAY

Answer 2

试试这个

{
  "value": "43165913081459",
  "name": "AvailableStorage"
}

MySQL - 按组日期值获取计数值为0

更新 -

2 个答案: