Question

如何在sqlite中获得相同的结果？

在excel中，我有这个功能，它每天根据分配的字母代码计算员工人数：

这是我的Excel公式：=SUMIF(G$2:G$5,E9,$E$2:$E$5)

在我的Sqlite数据库中，我有3个表：

TABLE 1
+----------------+---------------+--------+
| ID | status_name | status_code | status |
+----+-------------+-------------+--------+
| 1  | Available   | A           | true   |
+------------------+-------------+--------+
| 2  | HalfDay     | H           | true   |
+------------------+-------------+--------+
| 3  | On Leave    | OL          | true   |            
+------------------+-------------+--------+
| 4  | Restday     | R           | true   |            
+------------------+-------------+--------+
| 5  | Vacation    | V           | true   |            
+------------------+-------------+--------+

TABLE 2
+--------------+-------+-------+------+----------+
| EmployeeName | Site  | Shift | Team | JobTitle |
+--------------+-------+-------+------+----------+
| Steve        | Bldg1 | Night | N1   | Doctor   |
+--------------+-------+-------+------+----------+
| Dave         | Bldg1 | Night | N2   | Nurse    | 
+--------------+-------+-------+------+----------+
| Jack         | Bldg1 | Night | N2   | Nurse    |
+--------------+-------+-------+------+----------+
| Jacob        | Bldg2 | Day   | D1   | Doctor   |
+--------------+-------+-------+------+----------+
| Noah         | Bldg2 | Day   | D2   | Nurse    | 
+--------------+-------+-------+------+----------+
| MAX          | Bldg2 | Day   | D2   | Nurse    | 
+--------------+-------+-------+------+----------+

TABLE 3
+----------+-------+-------+------+-----+-----+-----+-----+-----+-----+-----+
| JobsType | Site  | Shift | Team | SUN | MON | TUE | WED | THU | FRI | SAT |
+----------+-------+-------+------+-----+-----+-----+-----+-----+-----+-----+
| Doctor   | Bldg1 | Night | N1   | A   | H   | A   | A   | OL  | A   | A   |
+----------+-------+-------+------+-----+-----+-----+-----+-----+-----+-----+
| Nurse    | Bldg1 | Night | N2   | A   | H   | H   | A   | A   | A   | A   |
+----------+-------+-------+------+-----+-----+-----+-----+-----+-----+-----+
| Doctor   | Bldg2 | Day   | D1   | H   | A   | H   | H   | A   | A   | OL  |
+----------+-------+-------+------+-----+-----+-----+-----+-----+-----+-----+
| Nurse    | Bldg1 | Night | N2   | A   | H   | H   | A   | A   | A   | A   |
+----------+-------+-------+------+-----+-----+-----+-----+-----+-----+-----+

通过使用上面的3个表，如何在查询中获得此结果？

+--------------+-----+-----+-----+-----+-----+-----+-----+
| STATUS TYPES | SUN | MON | TUE | WED | THU | FRI | SAT |
+--------------+-----+-----+-----+-----+-----+-----+-----+
| Available    | 5   | 4   | 4   | 5   | 5   | 6   | 5   |
+--------------+-----+-----+-----+-----+-----+-----+-----+
| HalfDay      | 1   | 5   | 5   | 1   | 0   | 0   | 0   |
+--------------+-----+-----+-----+-----+-----+-----+-----+
| On Leave     | 0   | 0   | 0   | 0   | 1   | 0   | 1   |
+--------------+-----+-----+-----+-----+-----+-----+-----+
| Restday      | 0   | 0   | 0   | 0   | 0   | 0   | 0   |
+--------------+-----+-----+-----+-----+-----+-----+-----+
| Vacation     | 0   | 0   | 0   | 0   | 0   | 0   | 0   |
+--------------+-----+-----+-----+-----+-----+-----+-----+

到目前为止，我已经来过这个


    SELECT DISTINCT M.status_name, M.status_code, A.SUN, count(*),count(*) FROM TABLE3 M LEFT join TABLE2 A ON M.status_code= A.SUN LEFT join TABLE2 B ON (A.Team = B.Team AND A.Shift = B.Shift AND A.Site = B.Site)  
WHERE M.product_status = 1  
GROUP BY M.status_name;

Answer 1

第一步，生成一个测试数据库（带有几个有用的索引）：

CREATE TABLE status(id INTEGER PRIMARY KEY, status_name TEXT, status_code TEXT, status TEXT);
INSERT INTO status VALUES(1,'Available','A','true');
INSERT INTO status VALUES(2,'HalfDay','H','true');
INSERT INTO status VALUES(3,'On Leave','OL','true');
INSERT INTO status VALUES(4,'Restday','R','true');
INSERT INTO status VALUES(5,'Vacation','V','true');
CREATE TABLE employees(EmployeeName TEXT, Site TEXT, Shift TEXT, Team TEXT, JobTitle TEXT);
INSERT INTO employees VALUES('Steve','Bldg1','Night','N1','Doctor');
INSERT INTO employees VALUES('Dave','Bldg1','Night','N2','Nurse');
INSERT INTO employees VALUES('Jack','Bldg1','Night','N2','Nurse');
INSERT INTO employees VALUES('Jacob','Bldg2','Day','D1','Doctor');
INSERT INTO employees VALUES('Noah','Bldg2','Day','D2','Nurse');
INSERT INTO employees VALUES('MAX','Bldg2','Day','D2','Nurse');
CREATE TABLE schedule(JobsType TEXT, Site TEXT, Shift TEXT, Team TEXT, SUN TEXT, MON TEXT, TUE TEXT, WED TEXT, THU TEXT, FRI TEXT, SAT TEXT);
INSERT INTO schedule VALUES('Doctor','Bldg1','Night','N1','A','H','A','A','OL','A','A');
INSERT INTO schedule VALUES('Nurse','Bldg1','Night','N2','A','H','H','A','A','A','A');
INSERT INTO schedule VALUES('Doctor','Bldg2','Day','D1','H','A','H','H','A','A','OL');
INSERT INTO schedule VALUES('Nurse','Bldg1','Night','N2','A','H','H','A','A','A','A');
CREATE INDEX employees_idx ON employees(JobTitle, Site, Shift, Team);
CREATE INDEX status_idx_code ON status(status_code);

此查询：

SELECT st.status_name, st.status_code
     , sum(sc.SUN = st.status_code) AS SUN
     , sum(sc.MON = st.status_code) AS MON
     , sum(sc.TUE = st.status_code) AS TUE
     , sum(sc.WED = st.status_code) AS WED
     , sum(sc.THU = st.status_code) AS THU
     , sum(sc.FRI = st.status_code) AS FRI
     , sum(sc.SAT = st.status_code) AS SAT
FROM status AS st
JOIN schedule AS sc ON st.status_code IN (sc.SUN, sc.MON, sc.TUE, sc.WED
                                        , sc.THU, sc.FRI, sc.SAT)
JOIN employees AS e ON sc.JobsType = e.JobTitle AND sc.Site = e.Site
                   AND sc.Shift = e.Shift AND sc.Team = e.Team
GROUP BY st.status_name, st.status_code
ORDER BY st.status_name, st.status_code;

会产生

status_name  status_code  SUN         MON         TUE         WED         THU         FRI         SAT       
-----------  -----------  ----------  ----------  ----------  ----------  ----------  ----------  ----------
Available    A            5           1           1           5           5           6           5         
HalfDay      H            1           5           5           1           0           0           0         
On Leave     OL           0           0           0           0           1           0           1

工作原理：

对于状态表中的每一行，将具有该状态的计划表中的每一行联接至少一天，然后将雇员表中与每个计划行的工作类型匹配的所有行联接起来（使用此示例数据），则总共生成了14行）。然后将状态中的所有行分组，并针对每一天，将当天的计划代码与状态匹配的行中的行数相加。

如何简化从Excel到sqlite的查询？

1 个答案: