[英]Count MySQL-Entries per day, inclusive days without any entries (Date Range)
我有一个MySQL表
id mydate content
----------------------------------
1 2015-06-20 some content
2 2015-06-20 some content
3 2015-06-22 some content
现在,我要计算每天的条目:
SELECT DATE(mydate) Date, COUNT(DISTINCT id) dayCount FROM mytable
GROUP BY DATE(mydate) HAVING dayCount > -1 ORDER BY DATE(mydate) DESC
这对我有用,结果:
2015-06-20 = 2
2015-06-22 = 1
我如何获取没有任何输入的日子? 在我的示例中,结果应为:
2015-06-19 = 0
2015-06-20 = 2
2015-06-21 = 0
2015-06-22 = 1
2015-06-23 = 0
基于此:
<?php
$today = date("Y-m-d");
$mystartdate = date_create($today);
date_sub($mystartdate, date_interval_create_from_date_string('14 days'));
$mystartdate = date_format($mystartdate, 'Y-m-d');
?>
最后,我想输出最近14天的计数,也要输出“ 0天”。 希望你理解我的问题。
为此,您可以创建一个包含增量数字的新表,但这不是一个好主意。 但是,如果采用这种方式,请使用此表通过DATE_ADD构造日期列表。
根据时间部分将其左联接到数据表上,以获取日期列表
有关更多信息,请通过链接
尝试以下-
SELECT a.date_field, COUNT(DISTINCT b.id) dayCount 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
) AAA
WHERE MONTH(date_field) = MONTH(NOW()) ) a
LEFT JOIN mytable b ON a.date_field=DATE(b.mydate)
GROUP BY a.date_field HAVING dayCount > -1 ORDER BY a.date_field DESC;
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