获取两个日期之间的日期

Question

我有两个这样的表：

表格1

emp_leave_summary(id,emp_id,leave_from_date,leave_to_date,leave_type)

表2

emp_leave_daywise(id,emp_id,leave_date,leave_type)

我想emp_id, leave_type从 Table1中 选择并插入 Table2中 。

例如： 在表1中，我有这个

id,emp_id,leave_from_date,leave_to_date,leave_type
 1, 12345,2017-07-01     ,2017-07-03   ,Sick Leave

在表2中，我想要这个

id,emp_id,leave_date,leave_type
 1,12345,2017-07-01,Sick Leave
 2,12345,2017-07-02,Sick Leave
 3,12345,2017-07-03,Sick Leave

带有样本数据的表结构

    CREATE TABLE `emp_leave_summary` (
  `id` int(11) NOT NULL,
  `emp_id` int(11) NOT NULL,
  `leave_from_date` date NOT NULL,
  `leave_to_date` date NOT NULL,
  `leave_type` varchar(30) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


INSERT INTO `emp_leave_summary` (`id`, `emp_id`, `leave_from_date`, `leave_to_date`, `leave_type`) VALUES
(1, 123, '2017-02-01', '2017-02-15', 'Earned Vacation Leave'),
(2, 123, '2017-07-12', '2017-07-26', 'Earned Vacation Leave'),
(3, 456, '2017-03-20', '2017-04-20', 'Earned Vacation Leave'),
(4, 789, '2017-01-15', '2017-02-23', 'Earned Vacation Leave'),
(5, 789, '2017-02-26', '2017-02-27', 'Sick Leave');

CREATE TABLE `emp_leave_daywise` (
  `id` int(11) NOT NULL,
  `emp_id` int(11) NOT NULL,
  `leave_date` date NOT NULL,
  `leave_type` varchar(30) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


ALTER TABLE `emp_leave_daywise`
ADD PRIMARY KEY (`id`),
ADD KEY `emp_id` (`emp_id`),
ADD KEY `leave_date` (`leave_date`),
ADD KEY `leave_type` (`leave_type`);


ALTER TABLE `emp_leave_summary`
ADD PRIMARY KEY (`id`),
ADD KEY `emp_id` (`emp_id`),
ADD KEY `leave_type` (`leave_type`),
ADD KEY `leave_from_date` (`leave_from_date`),
ADD KEY `leave_to_date` (`leave_to_date`);

Answer 1

感谢您的架构。它使处理您的问题变得容易。我对您的架构进行了一些更改以利用auto_increment

CREATE TABLE `emp_leave_summary` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `emp_id` int(11) NOT NULL,
  `leave_from_date` date NOT NULL,
  `leave_to_date` date NOT NULL,
  `leave_type` varchar(30) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


INSERT INTO `emp_leave_summary` (`emp_id`, `leave_from_date`, `leave_to_date`, `leave_type`) VALUES
( 123, '2017-02-01', '2017-02-15', 'Earned Vacation Leave'),
( 123, '2017-07-12', '2017-07-26', 'Earned Vacation Leave'),
( 456, '2017-03-20', '2017-04-20', 'Earned Vacation Leave'),
( 789, '2017-01-15', '2017-02-23', 'Earned Vacation Leave'),
( 789, '2017-02-26', '2017-02-27', 'Sick Leave');

CREATE TABLE `emp_leave_daywise` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `emp_id` int(11) NOT NULL,
  `leave_date` date NOT NULL,
  `leave_type` varchar(30) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

在这里，我在emp_leave_daywise表上添加了唯一约束，因为id整数上的主键不能确保记录不重复。

ALTER TABLE `emp_leave_daywise`
ADD UNIQUE KEY `emp_leave_daywise_unique_key` (`emp_id`,`leave_date`,`leave_type`),
ADD KEY `emp_id` (`emp_id`),
ADD KEY `leave_date` (`leave_date`),
ADD KEY `leave_type` (`leave_type`);

emp_leave_summary的唯一键需要一些思考。例如…您是否允许摘要涵盖重叠的日期范围？…

ALTER TABLE `emp_leave_summary`
ADD UNIQUE KEY `emp_leave_summary_unique_key` (`emp_id`,`leave_from_date`),
ADD KEY `emp_id` (`emp_id`),
ADD KEY `leave_type` (`leave_type`),
ADD KEY `leave_from_date` (`leave_from_date`),
ADD KEY `leave_to_date` (`leave_to_date`);

现在使用左连接对现有数据进行数据提取。

/*
    insert any missing records using a left join on existing records
    */
insert into emp_leave_daywise ( emp_id, leave_date, leave_type )
select `new`.* from 
( 
    select summary.emp_id, dates.date_ leave_date, summary.leave_type 
    from emp_leave_summary summary
    inner join (
    /* 
        get dates to match against
        https://stackoverflow.com/questions/9295616/how-to-get-list-of-dates-between-two-dates-in-mysql-select-query
        */
        select adddate('1970-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) date_ from
        ( select 0 i 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) t0,
        ( select 0 i 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) t1,
        ( select 0 i 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) t2,
        ( select 0 i 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) t3,
        ( select 0 i 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) t4) dates
 on dates.date_ >= summary.leave_from_date 
 and dates.date_ <= summary.leave_to_date
 ) `new`
 left join emp_leave_daywise old
 on `new`.emp_id = old.emp_id
 and `new`.leave_date = old.leave_date 
 and `new`.leave_type = old.leave_type 
 where old.id is null ;

 select * from emp_leave_daywise order by leave_date, emp_id;

根据给定的数据返回104行

 id emp_id  leave_date  leave_type
 1  789 2017-01-15  Earned Vacation Leave
 2  789 2017-01-16  Earned Vacation Leave
 3  789 2017-01-17  Earned Vacation Leave
 4  789 2017-01-18  Earned Vacation Leave
 5  789 2017-01-19  Earned Vacation Leave
 6  789 2017-01-20  Earned Vacation Leave
 ...
 102    123 2017-07-24  Earned Vacation Leave
 103    123 2017-07-25  Earned Vacation Leave
 104    123 2017-07-26  Earned Vacation Leave

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