Question

我看到了这段代码，我想问你们是否可以向我们解释这段代码是如何运作的。

select SUM(IF(encodedRestDay & (1 << (DAYOFWEEK(selected_date)-1)) > 0, 1, NULL)) as numberOfRestDay from (
    select *, if ( encodedRestDay & (1 << (DAYOFWEEK(selected_date)-1)) > 0,
    'restDay', 'workDay'
    ) as dayStatus, dayName(selected_date) as `dayName` from (
        select *,  (sat << 6 | fri << 5 | thu << 4 | wed << 3 | tue << 2 | mon << 1 | sun) as encodedRestDay from (
            select adddate('2014-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) as selected_date,
                sat, fri, thu, wed, tue, mon, sun
            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,
            restday
            where id = 6
            ) v

            where selected_date between '2014-02' and '2014-04'
        )a
    )b

Answer 1

此查询正在执行复杂的日期算术，以计算两个日期之间的“休息”天数。（这可能是非工作日的数量。）数据结构足够灵活，可以处理工作日的不同“文化”。

这个子查询：

        select adddate('2014-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) as selected_date,
            sat, fri, thu, wed, tue, mon, sun
        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,
        restday
        where id = 6

正在创建从2014-01-01开始的接下来的10,000天的日历。对restday id = 6中的select *, (sat << 6 | fri << 5 | thu << 4 | wed << 3 | tue << 2 | mon << 1 | sun) as encodedRestDay from (行中的每一行执行此操作。如果没有表格的布局，很难理解这意味着什么，或者星期几列真正意味着什么。我推测他们会告诉你某一天是特定文化的工作日还是休息日。

下一个级别：

正在做一些转移。这样，像restday这样的东西代表星期六，星期三，星期一和星期日。

最外面的查询然后进行某种比较以获得“休息”日（如果我理解正确的话）。

这是查询的要点。我不完全理解{{1}}正在做什么;我怀疑它针对不同的文化或转变有不同的工作日模式。

有人可以向我解释这段代码吗？

1 个答案: