Question

我现在无法找到 6 小时的确切答案...... !!

我想按顺序显示生日，首先显示即将到来的生日。

$today_date = date('Y-m-d');

$s = "select * from BDays order by TIMEDIFF(STR_TO_DATE(date,'%Y-%m-%d') , $today_date)";

包含日期的列在标记为“BDays”的表格中标记为“日期”。

在查询中，我试图从今天的日期中减去出生日期以获得即将到来的生日。

Answer 1

您希望仅将每个日期/时间的月份和日期与当前月份日期进行比较。有一个功能EXTRACT，你可以在这里做到这一点。像这样：

$s = "select * from BDays
      order by EXTRACT(MONTH_DAY, date) - EXTRACT(MONTH_DAY, $today_date)";

这将返回所有生日，即使已经过去的生日。要只有即将到来的生日，请尝试：

$s = "select * from BDays
       where (EXTRACT(MONTH_DAY, date) - EXTRACT(MONTH_DAY, $today_date)) > 0
       order by EXTRACT(MONTH_DAY, date) - EXTRACT(MONTH_DAY, $today_date)";

Answer 2

SELECT
    name,
    birthday,
    birthday + INTERVAL(YEAR(CURRENT_TIMESTAMP) - YEAR(birthday)) + 0 YEAR AS currbirthday,
    birthday + INTERVAL(YEAR(CURRENT_TIMESTAMP) - YEAR(birthday)) + 1 YEAR AS nextbirthday
FROM birthdays
ORDER BY CASE
    WHEN currbirthday >= CURRENT_TIMESTAMP THEN currbirthday
    ELSE nextbirthday
END

http://sqlfiddle.com/#!9/9a5c0/1

Answer 3

你可以试试这个：

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结果：

create table BDays (`date` date);
insert into BDays values ('1993-04-01'), ('2015-10-26'), ('1995-08-10'), ('1995-06-05'), ('1994-04-04');

select * from BDays
order by
case 
when `date` > current_date then `date`
else cast(concat(year(`date`)+1, right(`date`, 6)) as date)
end

示例：http://sqlfiddle.com/#!9/d60e2/2

我们在这里按日期订购。如果出生日期尚未到达，请使用该日期。否则采取明年，并将其与月和日结合起来。然后对它进行排序。

在您的日期显示为DD-MM-YYYY，您可以在这些行上使用| date | |---------------------------| | April, 01 1993 00:00:00 | | April, 04 1994 00:00:00 | | June, 05 1995 00:00:00 | | August, 10 1995 00:00:00 | | October, 26 2015 00:00:00 |或其他内容。

编辑：

cast( concat(left(date, 6), year(date)+1) as date)

结果：

-- added 1994-11-27 to the table
create table BDays (`date` date);
insert into BDays values ('1993-04-01'), ('2015-10-26'), ('1995-08-10'), ('1995-06-05'), ('1994-04-04'), ('1994-11-27');

-- updated query
select * from BDays
order by
case 
when `date` > current_date then `date`
else cast(concat(year(current_date)+1, right(`date`, 6)) as date)
end

SQLFiddle示例：http://sqlfiddle.com/#!9/1f448/13

通过mysql获得两个日期的差异

3 个答案: