mysql查询获取2日期之间的第一个特定日期

Question

如何获取2个日期之间的第一个星期日？ 我只有2 myTable的字段（dt_from和dt_to）

dt_from = '2016-08-6'; 
dt_to = '2016-08-19'; 

SELECT firstsunday FROM myTable BETWEEN dt_from AND dt_to; 

Answer 1

大约从日期本身获得的第一个星期日刚过怎么过吗？

SELECT DATE_ADD(A.date_from, INTERVAL (6 - WEEKDAY(A.date_from)) DAY) from myTable as A; 

http://sqlfiddle.com/#!9/7fce5

如果星期天没有日期间：

SELECT DATE_ADD(A.date_from, INTERVAL (6 - WEEKDAY(A.date_from)) DAY) as firstsunday from myTable as A where DATE_ADD(A.date_from, INTERVAL (6 - WEEKDAY(A.date_from)) DAY) between A.date_from and A.date_to; 

http://sqlfiddle.com/#!9/83d0f0/4

一点点短：

SELECT firstsunday from (SELECT A.date_from, A.date_to, DATE_ADD(A.date_from, INTERVAL (6 - WEEKDAY(A.date_from)) DAY) as firstsunday from myTable as A) as B where B.firstsunday between B.date_from and B.date_to; 

http://sqlfiddle.com/#!9/6adab3/1

Answer 2

试试这个：

SELECT dt, DAYNAME(dt) as day 
FROM myTable WHERE dt BETWEEN date1 AND date2 
ORDER BY dt 
WHERE day = "Sunday" 
LIMIT 0,1 

Answer 3

假设你只需要起诉分贝来获得日期范围，你不希望任何国王查询。

<?php 

echo date('m/d/Y', strtotime('next Sunday', strtotime('2016-08-6'))); 

你可以添加一些验证，以确保它的结束日期