ind*_*ago 7 mysql datetime date
我在 MySQL 中有一个查询,它通过获取当月内的所有记录来很好地为我服务;
SELECT date_field,val FROM MY_TABLE WHERE date_field>=(CURDATE()-INTERVAL 1 MONTH);
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上面的查询运行良好。所以如果这个月我们只有两个记录和 28 天,它只会带来两个记录。
date_field | val
========================
2015-02-08 | 567
2015-02-09 | 345
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但我希望返回的记录数与当月的天数完全相同。如果当月有 28 天并且只有两个记录,它应该带来;
date_field | val
========================
2015-02-01 | 0
2015-02-02 | 0
2015-02-03 | 0
2015-02-04 | 0
2015-02-05 | 0
2015-02-06 | 0
2015-02-07 | 0
2015-02-08 | 567
2015-02-09 | 345
2015-02-10 | 0
2015-02-11 | 0
2015-02-12 | 0
2015-02-13 | 0
2015-02-14 | 0
2015-02-15 | 0
2015-02-16 | 0
2015-02-17 | 0
2015-02-18 | 0
2015-02-19 | 0
2015-02-20 | 0
2015-02-21 | 0
2015-02-22 | 0
2015-02-23 | 0
2015-02-24 | 0
2015-02-25 | 0
2015-02-26 | 0
2015-02-27 | 0
2015-02-28 | 0
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如何修改我的查询以获得上述结果?
您可以创建当月的动态日期表
SELECT date_field
FROM
(
SELECT
MAKEDATE(YEAR(NOW()),1) +
INTERVAL (MONTH(NOW())-1) MONTH +
INTERVAL daynum DAY date_field
FROM
(
SELECT t*10+u daynum
FROM
(SELECT 0 t UNION SELECT 1 UNION SELECT 2 UNION SELECT 3) A,
(SELECT 0 u UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) B
ORDER BY daynum
) AA
) AAA
WHERE MONTH(date_field) = MONTH(NOW());
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注意:如果您按原样剪切并粘贴上述查询,它将为您生成整个月
然后,您 LEFT JOIN 到您的原始查询
SELECT
AAA.date_field,
IFNULL(BBB.val,0) val
FROM
(
SELECT date_field
FROM
(
SELECT MAKEDATE(YEAR(NOW()),1) +
INTERVAL (MONTH(NOW())-1) MONTH +
INTERVAL daynum DAY date_field
FROM
(
SELECT t*10+u daynum FROM
(SELECT 0 t UNION SELECT 1 UNION SELECT 2 UNION SELECT 3) A,
(SELECT 0 u UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) B ORDER BY daynum
) AA
) AA WHERE MONTH(date_field) = MONTH(NOW())
) AAA LEFT JOIN (SELECT date_field,val FROM MY_TABLE) BBB
USING (date_field);
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首先,条件WHERE date_field >= (CURDATE()-INTERVAL 1 MONTH)
不会将您的结果限制为当月。它将获取从 30-31 天前到当前日期(以及未来,如果表中有未来日期的行)的所有日期。
它应该是:
WHERE date_field >= LAST_DAY(CURRENT_DATE) + INTERVAL 1 DAY - INTERVAL 1 MONTH
AND date_field < LAST_DAY(CURRENT_DATE) + INTERVAL 1 DAY
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现在,对于主要问题,要创建 28-31 个日期,即使表中没有包含所有日期的行,您也可以使用Calendar
表(包含所有日期,例如 1900 年到 2200 年)或动态创建它们,使用这样的东西(该days
表可以是临时表,也可以将其设为派生表,查询比这更复杂一些):
CREATE TABLE days
( d INT NOT NULL PRIMARY KEY ) ;
INSERT INTO days
VALUES (0), (1), (2), ....
..., (28), (29), (30) ;
SELECT
cal.my_date AS date_field,
COALESCE(t.val, 0) AS val
FROM
( SELECT
s.start_date + INTERVAL (days.d) DAY AS my_date
FROM
( SELECT LAST_DAY(CURRENT_DATE) + INTERVAL 1 DAY - INTERVAL 1 MONTH
AS start_date,
LAST_DAY(CURRENT_DATE)
AS end_date
) AS s
JOIN days
ON days.d <= DATEDIFF(s.end_date, s.start_date)
) AS cal
LEFT JOIN my_table AS t
ON t.date_field >= cal.my_date
AND t.date_field < cal.my_date + INTERVAL 1 DAY ;
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以上应该适用于任何类型的date_field
列(日期、日期时间、时间戳)。如果date_field
列的类型为DATE
,则最后一个连接可以简化为:
LEFT JOIN my_table AS t
ON t.date_field = cal.my_date ;
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