Question

我有这样的日子列表

CREATE TABLE `date` (
  `fecha` date NOT NULL,
  `feriado` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

-- 
-- Volcar la base de datos para la tabla `date`
-- 
INSERT INTO `date` VALUES ('2016-10-28', 0);
INSERT INTO `date` VALUES ('2016-10-29', 0);
INSERT INTO `date` VALUES ('2016-10-30', 1);
INSERT INTO `date` VALUES ('2016-10-31', 1);
INSERT INTO `date` VALUES ('2016-11-01', 0);
INSERT INTO `date` VALUES ('2016-11-02', 0);
INSERT INTO `date` VALUES ('2016-11-03', 0);

此查询添加日期

SELECT ADDDATE(CURDATE(),INTERVAL 2+(select COUNT(fecha) FROM date where fecha BETWEEN curdate() AND ADDDATE(CURDATE(),INTERVAL 2 DAY) AND feriado=1) DAY)

如果结果日期为feriado = 1，我怎么能这样做呢？它会在将来返回最接近feriado = 0的那个？谢谢！

Answer 1

如果你想在第二天用feriado = 0来获得第二天，那么另一种方法可能是排序和选择：

SELECT max(fecha) FROM (
  SELECT fecha FROM date
  WHERE feriado = 0
  AND fecha > curdate()
  ORDER by fecha
  LIMIT 2
) AS foo;

Answer 2

CREATE TABLE date (
  fecha date NOT NULL PRIMARY KEY,
  feriado tinyint NOT NULL
);

INSERT INTO `date` VALUES 
('2016-10-28', 0),
('2016-10-29', 0),
('2016-10-30', 1),
('2016-10-31', 1),
('2016-11-01', 0),
('2016-11-02', 0),
('2016-11-03', 0);

SELECT x.* 
     , MIN(y.fecha) nearest
  FROM date x
  LEFT
  JOIN date y
    ON y.fecha >= x.fecha
   AND y.feriado = 1
 GROUP 
    BY x.fecha;

+------------+---------+------------+
| fecha      | feriado | nearest    |
+------------+---------+------------+
| 2016-10-28 |       0 | 2016-10-30 |
| 2016-10-29 |       0 | 2016-10-30 |
| 2016-10-30 |       1 | 2016-10-30 |
| 2016-10-31 |       1 | 2016-10-31 |
| 2016-11-01 |       0 | NULL       |
| 2016-11-02 |       0 | NULL       |
| 2016-11-03 |       0 | NULL       |
+------------+---------+------------+

选择具有一个值的最近行

2 个答案: