mysql计算两个时间戳/日期间的工作日，兼容节假日
设置起止时间参数
set @date1='2019/01/01';
set @date2='2019/12/31';
建表
DROP TABLE IF EXISTS `calendar`;
CREATE TABLE `calendar` (
`day` date DEFAULT NULL,
`holiday` int(11) DEFAULT '0',
`dow` int(11) DEFAULT NULL
) ENGINE=MyISAM DEFAULT CHARSET=utf8mb4;
写入时间函数
delimiter $$
drop procedure if exists test;
create procedure test()
begin
declare tday date;
set tday=@date1;
while (tday <= @date2) do
insert into calendar(day) values(tday);
set tday=date_add(tday,interval 1 day);
end while;
end$$
调用存储过程
call test();
更新星期数据
update calendar set dow=dayofweek(day)-1;
update calendar set dow=7 where dow=0;
使用
SELECT COUNT(*) FROM calendar
WHERE day BETWEEN @date1 AND @date2
AND ((DAYOFWEEK(day) NOT IN(1,7) AND holiday=0) or holiday=2);
排除周末
delimiter //
drop function if exists getworkdays;
create function getworkdays(s varchar(16),e varchar(16))
returns varchar(16)
begin
DECLARE result VARCHAR(16) default '';
set e = FROM_UNIXTIME(e,'%Y-%m-%d');
set s = FROM_UNIXTIME(s,'%Y-%m-%d');
select (floor(days/7)*5
+ days%7
- case when 6 between wd and wd+days%7-1 then 1 else 0 end
- case when 7 between wd and wd+days%7-1 then 1 else 0 end) into result
from
(select datediff(e,s)+1 as days, weekday(s)+1 as wd ) as a;
RETURN result;
end //
delimiter ;
## 函数分析
1、编写存储过程，便于测试。存储过程命名为getWorkDays(d1,d2)
2、先求出日期d1到d2之间的天数days【使用datediff()函数】，再求出d2所在的日期为星期几【使用weekday()函数】
3、使用floor()函数求出days中所包含的星期数为(floor(days/7))，那么工作日数就是floor(days/7)*5。
4、最后一步就是求出零散的工作日：days%7表示的是，当前日期被7整除后的余数，这个余数就是代表零散的天数。但是我们不能将这个零散的天数全部加上，因为这个零散的天数中可能包含着周六，周末。这就是为什么我们前面计算d2是星期几的原因。
针对上述问题，后面再接上case when()语句，其中的between wd and wd+days%7-1意思就是指周六或者周末是不是在多出来的这个零散日期里。【需要注意这里的wd是weekday(s)即开始时间的星期，笔者在这个位置坑了好久！！】
---------------------
以上来自：https://blog.csdn.net/liu16659/article/details/82584807
排除节假日
delimiter //
drop function if exists getworkdays;
create function getworkdays(s varchar(16),e varchar(16))
returns varchar(16)
begin
DECLARE result VARCHAR(16) default '';
set e = FROM_UNIXTIME(e,'%Y-%m-%d');
set s = FROM_UNIXTIME(s,'%Y-%m-%d');
select count(*) into result from calendar where holiday = 0 and day between s and e
RETURN result;
end //
delimiter ;
注：节假日需要维护calendar表
calendar数据录入
<?php
// 工作日对应结果为 0, 休息日对应结果为 1, 节假日对应的结果为 2；
date_default_timezone_set('PRC');
$day='2019';
$url='http://tool.bitefu.net/jiari/vip.php?type=0&apikey=123456&d='.$day;
$result=curlget($url);
function curlget($url){
$ch = curl_init();
curl_setopt ($ch, CURLOPT_URL, $url);
curl_setopt ($ch, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt ($ch, CURLOPT_SSL_VERIFYHOST, false);
curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt ($ch, CURLOPT_CONNECTTIMEOUT,10);
$dxycontent = curl_exec($ch);
curl_close($ch);
return $dxycontent;
}
$conn = mysqli_connect('localhost','root','');
if(!$conn){
var_dump(mysqli_error());
}
$jsonDay = json_decode($result,true);
$dayData = $jsonDay['data']['2019'];
mysqli_select_db($conn,'efuzhou3');
$sql = "select * from calendar";
$res = mysqli_query($conn,$sql);
while($row=mysqli_fetch_assoc($res)){
$timestamp = strtotime($row['day']);
$mdData = date('md',$timestamp);
if(isset($dayData[$mdData])){
$row['holiday'] = $dayData[$mdData];
}
$sql = "update calendar set holiday = $row[holiday] where `day` = '$row[day]'";
mysqli_query($conn,$sql);
$data[] = $row;
}
var_dump($data);
?>
