mysql中的复杂查询

Question

我有两个表reports和holidays。

reports：(username varchar(30),activity varchar(30),hours int(3),report_date date)

holidays：(holiday_name varchar(30), holiday_date date)

select * from reports给

+----------+-----------+---------+------------+ 
| username | activity | hours | date  | 
+----------+-----------+---------+------------+ 
| prasoon | testing | 3 | 2009-01-01 | 
| prasoon | coding | 4 | 2009-01-03 | 
| gautam | coding | 1 | 2009-01-05 | 
| prasoon | coding | 4 | 2009-01-06 | 
| prasoon | coding | 4 | 2009-01-10 | 
| gautam | coding | 4 | 2009-01-10 | 
+----------+-----------+---------+------------+ 

select * from holidays给

+--------------+---------------+ 
| holiday_name | holiday_date | 
+--------------+---------------+ 
| Diwali  | 2009-01-02 | 
| Holi   | 2009-01-05 | 
+--------------+---------------+ 

当我用下面的查询

SELECT dates.date AS date, 
    CASE 
    WHEN holiday_name IS NULL THEN COALESCE(reports.activity, 'Absent') 
    WHEN holiday_name IS NOT NULL and reports.activity IS NOT NULL THEN reports.activity 
    ELSE '' 
    END 
    AS activity, 
    CASE WHEN holiday_name IS NULL THEN COALESCE(reports.hours, 'Absent') 
    WHEN holiday_name IS NOT NULL and reports.hours IS NOT NULL THEN reports.hours 
    ELSE '' 
    END 
    AS hours, 
    CASE 
    WHEN holiday_name IS NULL THEN COALESCE(holidays.holiday_name, '') 
    ELSE holidays.holiday_name 
    END 
    AS holiday_name 
    FROM dates 
    LEFT OUTER JOIN reports ON dates.date = reports.date 
    LEFT OUTER JOIN holidays ON dates.date = holidays.holiday_date 
    where reports.username='gautam' and dates.date>='2009-01-01' and dates.date<='2009-01-10'; 

我得到了以下输出

+----------+-----------+---------+------------+ 
    | date | activity | hours | holiday | 
    +----------+-----------+---------+------------+ 
    |2009-01-05| coding | 1 | Holi  | 
    +----------+-----------+---------+------------+ 
    |2009-01-10| coding | 4 |   | 
    +----------+-----------+---------+------------+ 

，但我预计今年

+----------+-----------+---------+------------+ 
    | date | activity | hours | holiday | 
    +----------+-----------+---------+------------+ 
    |2009-01-01| Absent | Absent |   | 
    +----------+-----------+---------+------------+ 
    |2009-01-02|   |   | Diwali  | 
    +----------+-----------+---------+------------+ 
    |2009-01-03| Absent | Absent |   | 
    +----------+-----------+---------+------------+ 
    |2009-01-04| Absent | Absent |   | 
    +----------+-----------+---------+------------+ 
    |2009-01-05| Coding | 1  | Holi  | 
    +----------+-----------+---------+------------+ 
    |2009-01-06| Absent | Absent |   | 
    +----------+-----------+---------+------------+ 
    |2009-01-07| Absent | Absent |   | 
    +----------+-----------+---------+------------+ 
    |2009-01-08| Absent | Absent |   | 
    +----------+-----------+---------+------------+ 
    |2009-01-09| Absent | Absent |   | 
    +----------+-----------+---------+------------+ 
    |2009-01-10| Coding | 4  |   | 
    +----------+-----------+---------+------------+ 

如何修改上面的查询以获得所需的输出（用于特定用户（高塔姆在这种情况下））？

编辑

我也有一个表dates(date date)包含所有2009-01-01之间的日期2020-12-31

Answer 1

移动JOIN子句中的WHERE子句。我唯一改变的是最后几行：

FROM dates 
LEFT JOIN reports ON dates.date = reports.date 
AND reports.username='gautam' 
LEFT JOIN holidays ON dates.date = holidays.holiday_date 
AND dates.date >= '2009-01-01' 
AND dates.date <= '2009-01-10' 

你同意这个答案的上方，但进一步的测试表明，虽然它为你的作品，它没有为我工作。下面是一些对我的作品：

FROM dates 
LEFT JOIN reports ON dates.date = reports.date 
AND reports.username='gautam' 
LEFT JOIN holidays ON dates.date = holidays.holiday_date 
WHERE dates.date >= '2009-01-01' 
AND dates.date <= '2009-01-10' 

Answer 2

马克的回答会是不错的，但你正在做的东西的做法很可怜 - 你使用SQL格式化您的信息。这不好！您应该只使用它来检索您的信息，然后使用HTML格式或任何将数据拉入的格式。你的选择应该是一个简单：

SELECT * FROM reports WHERE username='guatam' 
AND date>='2009-01-01' AND date<='2009-01-9' 

和一个单独的一个假期，如果你需要它：

SELECT * from holidays 

然后利用这些信息，因为你需要。