Question

我试图弄清楚如何根据一组值计算匹配值。下面是表结构的示例。我正在使用12c

Table 1                                        Table 2

id  term    course  course_num              course course_num equ_course equ_num       
1   201710  MATH    1020                    MATH   1020       MATH       2015                                            
1   201510  MATH    2015                    MATH   1020       MATH       101
1   201680  MATH    101                     MATH   101        MATH       1020
2   201380  ENGL    1010

我需要的是学生重复课程的数量。在示例中，id为1的学生同时参加MATH 1020,101和2015.由于在表2（等值表）中，MATH 2015和MATH 101计为MATH 1020，结果应如下所示。重复列是他们重复课程的次数。

id  term   course  course_num  repeated
1   201710 MATH    1020        3  
1   201510 MATH    2015        3 
1   201680 MATH    101         3

Answer 1

让我们改变样本数据，使其更复杂;而不是MATH/1020等同于MATH/2015和MATH/101我们可以将等效项链接起来，以便MATH/1020等同于MATH/2015，而MATH/101相当于Table2 {1}}。这使得简单地将CREATE TABLE Table1 ( id, term, course, course_num ) AS SELECT 1, 201710, 'MATH', 1020 FROM DUAL UNION ALL SELECT 1, 201510, 'MATH', 2015 FROM DUAL UNION ALL SELECT 1, 201680, 'MATH', 101 FROM DUAL UNION ALL SELECT 2, 201380, 'ENGL', 1010 FROM DUAL; CREATE TABLE Table2 ( course, course_num, equ_course, equ_num) AS SELECT 'MATH', 1020, 'MATH', 2015 FROM DUAL UNION ALL SELECT 'MATH', 2015, 'MATH', 101 FROM DUAL;加入到自身以合并等价物是不够的，因为可以存在多层深度的树。而是使用分层查询：

SQL Fiddle

Oracle 11g R2架构设置：

WITH equivalent_courses ( course, course_num, equ_course, equ_num ) AS (
  SELECT CONNECT_BY_ROOT( course ),
         CONNECT_BY_ROOT( course_num ),
         equ_course,
         equ_num
  FROM   Table2
  START WITH ( course, course_num )
             NOT IN ( SELECT equ_course, equ_num FROM TABLE2 )
  CONNECT BY PRIOR equ_course = course
  AND        PRIOR equ_num    = course_num
)
SELECT t.*,
       COUNT(*) OVER (
         PARTITION BY COALESCE( e.course, t.course ),
                       COALESCE( e.course_num, t.course_num )
       ) AS repeated
FROM   Table1 t
       LEFT OUTER JOIN equivalent_courses e
       ON ( t.course = e.equ_course AND t.course_num = e.equ_num )

查询1 ：

| ID |   TERM | COURSE | COURSE_NUM | REPEATED |
|----|--------|--------|------------|----------|
|  2 | 201380 |   ENGL |       1010 |        1 |
|  1 | 201710 |   MATH |       1020 |        3 |
|  1 | 201510 |   MATH |       2015 |        3 |
|  1 | 201680 |   MATH |        101 |        3 |

<强> Results ：

const fetch = require('jest-fetch-mock');

jest.setMock('portable-fetch', fetch);
const factory = require('./current-weather');

describe('current-weather', () => {
  const config = { appid: 'appid' };
  const currentWeather = factory(config);

  test('currentWeather()', async () => {
    const weather = { temperature: '25.17 °C' };
    fetch.mockResponse(JSON.stringify(weather));

    const current = await currentWeather({ city: 'London' });

    expect(current).toEqual(weather);
  });
});

Oracle根据值组计算匹配值

1 个答案: