Question

我有一个应用，我希望用户输入工作历史记录。我使用DATE格式（每个作业的start_date和end_date）将结果存储在mysql表中。因此，例如，在几个条目之后它可能看起来像这样（job_history表：

Job             Start              End
Dishwasher      2013-07-10         2014-03-28
Banker          2014-04-02         2014-10-10
Cahsier         2014-10-11         2014-12-20

我正在尝试设置一个mysql查询，该查询将检查特定年份是否有任何丢失的天数。对于我们的示例，请使用2014. 2014-03-29至2014-04-01期间缺少天数。

通过搜索，我找不到任何好的例子。我认为有一个更好的方法来做这个比做一个PHP循环和做365个查询（哎哟！）。感谢

编辑：更新了包含年末和年末的示例：

Job             Start              End
Dishwasher      2013-07-10         2013-12-01
Banker          2014-01-10         2014-10-10
Cahsier         2014-10-11         2014-12-20

在这个例子中，我们的输入年份将是2014年，我们将在年初的9天内被发现，而在年底则有11天。

Answer 1

对于后来提出这个问题的人来说，最初的答案就是这个

SELECT jh.id,
    CASE
        WHEN datediff(jh1.start, jh.end) > 1
        THEN datediff(jh1.start, jh.end)
        ELSE NULL
    END as DifferenceInDays
FROM job_history jh
LEFT JOIN job_history jh1 ON jh1.id = (jh.id + 1)

这样做是假设自连接的顺序ID，这样您可以将最后一行日期与当前行日期进行比较。

在对要求进行了大量补充之后，这将满足您的需求。

SELECT position, id, DifferenceInDays
FROM
(       SELECT 'BEGINNING' as position, id,
            @a := if(@a = 'BAD', 'BAD', if(DATE(concat(year(now()), '-01-01')) BETWEEN start AND end, 'BAD', '')),
            case
                WHEN @a = 'BAD'
                THEN NULL
                ELSE datediff(jstart, DATE(concat(year(now()), '-01-01')))
            end AS DifferenceInDays
        FROM
        (   SELECT jh.*, jh3.start as jstart
            FROM job_history jh
            CROSS JOIN(SELECT @a:='')t
            LEFT JOIN (SELECT min(start) as start FROM job_history where userid = 123456 AND YEAR(start) = YEAR(NOW())) jh3 ON jh3.start = jh.start
            WHERE jh.userid = 123456
            ORDER BY id
        )t
    UNION ALL
        SELECT 'MIDDLE', jh.id, '',
            case
                WHEN datediff(jh1.start, jh.end) > 1
                THEN datediff(jh1.start, jh.end) - 1
                ELSE NULL
            end as DifferenceInDays
        FROM
        (   SELECT *, @new_id := @new_id + 1 as new_id
            FROM job_history, (select @new_id := 0) temp
            WHERE userid = 123456
        ) jh
        LEFT JOIN 
        (   SELECT *, @new_id1 := @new_id1 + 1 as new_id
            FROM job_history, (select @new_id1 := 0) temp
            WHERE userid = 123456
        ) jh1 ON jh1.new_id = (jh.new_id + 1)

    UNION ALL
        SELECT 'ENDING', id, 
            @b := if(@b = 'BAD', 'BAD', if(DATE(concat(year(now()), '-12-31')) BETWEEN start AND end, 'BAD', '')),
            case
                WHEN @b = 'BAD'
                THEN NULL
                ELSE datediff(DATE(concat(year(now()), '-12-31')), jend)
            end
        FROM
        (   SELECT jh.*, jh2.end as jend
            FROM job_history jh
            CROSS JOIN(SELECT @b:='')t
            LEFT JOIN (SELECT max(end) as end FROM job_history WHERE userid = 123456) jh2 ON jh2.end = jh.end
            WHERE jh.userid = 123456
            ORDER BY id
        )t
    ORDER BY case position WHEN 'BEGINNING' THEN 1 WHEN 'MIDDLE' THEN 2 ELSE 3 END ASC, id ASC
) t
WHERE DifferenceInDays IS NOT NULL

FIDDLE WITH MULTIPLE ROWS

FIDDLE WITH ONLY MIDDLE AND END

FIDDLE WITH NO MIDDLE ONE ROW

FIDDLE WITH ALL THREE

FIDDLE WITH NO END

Answer 2

您可date_diff用于查找两个日期之间的差异

$interval = date_diff($datetime1, $datetime2);

PHP / mysql检查所有日期都被考虑在内

2 个答案: