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主页 / user-181592

arilwan's questions

Martin Hope
arilwan
Asked: 2022-09-27 16:20:55 +0800 CST

我在这里的交叉表查询中缺少什么?

  • 6

我很难理解如何从我正在使用的数据库中正确转置该表。该表是由几年前设计数据库的人创建的,但我设法使用pg_dump.

这是带有示例条目的表:

CREATE TABLE response (
    session_id integer NOT NULL,
    seconds integer NOT NULL,
    question_id integer NOT NULL,
    response character varying(500),
    file bytea
);

INSERT INTO response(session_id, seconds, question_id, response, file)
VALUES (758,1459505869,31,'0',''),  (758,1459505869,32,'0',''), 
(758,1459505869,33,'0',''), (758,1459505869,34,'0',''), 
(758,1459505869,35,'1',''), (758,1459505869,36,'0',''), 
(758,1459505869,37,'0',''), (758,1459505869,38,'0',''), 
(758,1459506973,38,'0',''), (758,1459506973,37,'0',''), 
(758,1459506973,36,'0',''),(758,1459506973,35,'1',''),  
(758,1459506973,34,'0',''),(758,1459506973,33,'0',''),  
(758,1459506973,32,'0',''),(758,1459506973,31,'0',''),
(758,1459508676,31,'0',''),(758,1459508676,32,'0',''),  
(758,1459508676,33,'0',''),(758,1459508676,34,'0',''),  
(758,1459508676,35,'1',''),(758,1459508676,36,'0',''),  
(758,1459508676,37,'0', ''),    (758,1459508676,38,'0', '');

SELECT * FROM response LIMIT 5;
session_id  seconds   question_id   response    file
   758     1459505869     31           0         [null]
   758     1459505869     32           0         [null]
   758     1459505869     33           0         [null]
   758     1459505869     34           0         [null]
   758     1459505869     35           1         [null]

列中的问题 idquestion_id代表以下内容:

30  -- not_foot_count 
31  -- not_moving
32  -- foot
33  -- bicycle
34  -- motorcycle
35  -- car
36  -- bus
37  -- metro
38  -- other
39  -- train

响应可以是文本(错误的用户响应),但主要是 a1或 a 0(我感兴趣)。

所以我想把这个表转置成一个新表survey,这样返回的查询结果对于每一列都会有对应的响应码值作为列名(32 -> foot; 33 -> bike; 35 -> car.等)

我对所有这些回复都不感兴趣,但 5 : foot, bike(for bike), bus, car, and metro.

因为我在仅检索 5 个感兴趣的响应时遇到了很大的麻烦,所以我开始检索所有这些值以查看我是否正确地做事。事实证明我做错了。

这是我的尝试:

CREATE TABLE survey
AS
SELECT aresult.session_id,
   aresult.not_foot_count,
    aresult.not_moving,
    aresult.foot,
    aresult.bike,
    aresult.motor,
    aresult.car,
    aresult.bus,
    aresult.metro,
    aresult.train,
    aresult.other
   FROM crosstab('select session_id, question_id, response 
                from  response
                order by session_id,question_id'::text) 
  aresult(session_id integer, not_foot_count character varying(500), 
  not_moving character varying(500), foot character varying(500), 
  bike character varying(500), motor character varying(500), 
  car character varying(500), bus character varying(500), 
  metro character varying(500), train character varying(500), 
  other character varying(500));

这使:

SELECT * FROM survey;
session_id seconds not_foot_count not_moving foot bike motor car bus metro train other
758       1459505869       0          0        0   0    0     0   0   0      0     0
758       1459506973       0          0        0   0    0     0   0   0      0     0
758       1459508676       0          0        0   0    0     0   0   0      0     0

请注意,这不正确,因为列car应该是1.

此外,我并不是对所有的价值观都不感兴趣。相反,只希望是感兴趣的值。

预计出局

我希望将我的返回结果限制为以下(正确答案):

session_id  seconds    foot bike car bus metro 
    758   1459505869    0     0   1   0    0
    758   1459506973    0     0   1   0    0
    758   1459508676    0     0   1   0    0

注意:此dbfiddle说明了我的尝试。

编辑

就评论而言,已编辑问题以显示完整的预期输出。

postgresql pivot
  • 1 个回答
  • 208 Views
Martin Hope
arilwan
Asked: 2019-10-09 10:59:33 +0800 CST

使用 WITH 子句创建两个子查询

  • 0

我正在努力处理涉及两个使用此处WITH描述的子句的子查询的非常复杂的查询。

但在我看来,我不能在主查询中使用一个子查询。我这里是我的查询语句:

WITH trip_labelled AS
(
    SELECT DISTINCT ON (trips_1.trip_id) trips_1.trip_id,
            sum(segments.length::double precision) AS segments_length,
            sum(travelmode_profile.foot * segments.length)::real / GREATEST(sum(segments.length), 1::real) AS foot,
            sum(travelmode_profile.bike * segments.length)::real / GREATEST(sum(segments.length), 1::real) AS bike,
            sum(travelmode_profile.car * segments.length)::real / GREATEST(sum(segments.length), 1::real) AS car,
            sum(travelmode_profile.bus * segments.length)::real / GREATEST(sum(segments.length), 1::real) AS bus,
            sum(travelmode_profile.metro * segments.length)::real / GREATEST(sum(segments.length), 1::real) AS metro
           FROM trips trips_1
             JOIN segments ON segments.session_id = ANY (trips_1.session_ids)
             JOIN travelmode_profile USING (session_id, segment_id)
             LEFT JOIN session USING (session_id)
             LEFT JOIN ( SELECT DISTINCT ON (session_1.user_id) session_1.user_id,
                    response.response AS role
                   FROM session session_1
                     LEFT JOIN response USING (session_id)
                  WHERE response.question_id = 24) demography USING (user_id)
          WHERE segments.movement = true
          GROUP BY trips_1.trip_id
),
   trip_mode AS
(   SELECT trips.trip_id AS trip_id,
       trips.seconds_start,
       trips.seconds_end,
       trips.distance,
         CASE
            WHEN GREATEST(trip_labelled.car, trip_labelled.bike, trip_labelled.bus, trip_labelled.metro, trip_labelled.foot) = trip_labelled.car THEN 'car'::text
            WHEN GREATEST(trip_labelled.car, trip_labelled.bike, trip_labelled.bus, trip_labelled.metro, trip_labelled.foot) = trip_labelled.bike THEN 'bike'::text
            WHEN GREATEST(trip_labelled.car, trip_labelled.bike, trip_labelled.bus, trip_labelled.metro, trip_labelled.foot) = trip_labelled.bus THEN 'bus'::text
            WHEN GREATEST(trip_labelled.car, trip_labelled.bike, trip_labelled.bus, trip_labelled.metro, trip_labelled.foot) = trip_labelled.metro THEN 'metro'::text
            WHEN GREATEST(trip_labelled.car, trip_labelled.bike, trip_labelled.bus, trip_labelled.metro, trip_labelled.foot) = trip_labelled.foot THEN 'foot'::text
            ELSE NULL::text
        END AS travel_mode
   FROM trips
     JOIN trip_labelled USING (trip_id)
  )
  SELECT  session_id, 
         COALESCE(gps.gpstime, gslocation.gpstime) AS gpstime,
         COALESCE(gps.lat, gslocation.lat) AS lat,
        COALESCE(gps.lon, gslocation.lon) AS lon,
        COALESCE(gps.alt, gslocation.alt) AS alt,
        COALESCE(gps.speed, gslocation.speed) AS speed,
        COALESCE(gps.acc, gslocation.acc) AS acc
        trip_mode.travel_mode
FROM raw_data.gslocation
FULL JOIN raw_data.gps
    USING (session_id, seconds, millis);

错误:

ERROR:  syntax error at or near "trip_mode"
LINE 53:   trip_mode.travel_mode

我不能在主查询中使用travel_mode或trip_mode.travel_mode来自第二个子trip_mode查询。但致力于注释掉travel_mode/trip_mode.travel_mode。我哪里出错了?

postgresql
  • 2 个回答
  • 61 Views
Martin Hope
arilwan
Asked: 2019-06-04 07:45:08 +0800 CST

更改列中的值

  • 0

我手头有下表(用于数据迁移):

recvtimets     |         recvtime         | fiwareservicepath | entityid |   entitytype    |     attrname     |    attrtype     |               attrvalue               | attrmd
---------------+--------------------------+-------------------+----------+-----------------+------------------+-----------------+---------------------------------------+--------
 1559573267088 | 2019-06-03T14:47:47.88Z  | /basic            | 001      | WeatherObserved | location         | StructuredValue | {"coordinates":[-8.613502,41.143994]} | []
 1559573267088 | 2019-06-03T14:47:47.88Z  | /basic            | 001      | WeatherObserved | relativeHumidity | Number          | 59.1                                  | []
 1559573267088 | 2019-06-03T14:47:47.88Z  | /basic            | 001      | WeatherObserved | dateObserved     | Text            | 2015-09-01T10:15:57+01:00             | []
 1559573267088 | 2019-06-03T14:47:47.88Z  | /basic            | 001      | WeatherObserved | illuminance      | None            | null                                  | []
 1559573267088 | 2019-06-03T14:47:47.88Z  | /basic            | 001      | WeatherObserved | temperature      | Number          | 24.9                                  | []
 1559573267088 | 2019-06-03T14:47:47.88Z  | /basic            | 001      | WeatherObserved | name             | Text            | Rua das Flores                        | []
 1559573267782 | 2019-06-03T14:47:47.782Z | /basic            | 001      | WeatherObserved | name             | Text            | Praca do Marques - pole 1             | []
 1559573267782 | 2019-06-03T14:47:47.782Z | /basic            | 001      | WeatherObserved | illuminance      | None            | null                                  | []
 1559573267782 | 2019-06-03T14:47:47.782Z | /basic            | 001      | WeatherObserved | temperature      | Number          | 21.9                                  | []
 1559573267782 | 2019-06-03T14:47:47.782Z | /basic            | 001      | WeatherObserved | dateObserved     | Text            | 2015-09-01T10:16:24+01:00             | []
 1559573267782 | 2019-06-03T14:47:47.782Z | /basic            | 001      | WeatherObserved | location         | StructuredValue | {"coordinates":[-8.603802,41.161671]} | []
 1559573267782 | 2019-06-03T14:47:47.782Z | /basic            | 001      | WeatherObserved | relativeHumidity | Number          | 80.8                                  | []
 1559573267696 | 2019-06-03T14:47:47.696Z | /basic            | 001      | WeatherObserved | name             | Text            | Fundacao de Serralves                 | []
 1559573267696 | 2019-06-03T14:47:47.696Z | /basic            | 001      | WeatherObserved | illuminance      | None            | null                                  | []
 1559573267696 | 2019-06-03T14:47:47.696Z | /basic            | 001      | WeatherObserved | location         | StructuredValue | {"coordinates":[-8.65915,41.1591]}    | []
 1559573267696 | 2019-06-03T14:47:47.696Z | /basic            | 001      | WeatherObserved | relativeHumidity | Number          | 1                                     | []
 1559573267696 | 2019-06-03T14:47:47.696Z | /basic            | 001      | WeatherObserved | dateObserved     | Text            | 2015-09-01T10:16:22+01:00             | []

前两列描述了应用程序写入数据库的时间戳(从某个数据库读取)。然而,实际记录数据的时间是列中的dateObserved属性,attrname其对应的值是attrvalue列中显示的值。

这意味着前 6 行(2019-06-03T14:47:47.88Z在recvtime列中记录在2015-09-01T10:15:57+01:00(显示在 attrvalue列上),即dateObserved = 2015-09-01T10:15:57+01:00

2019-06-03T14:47:47.782Z列中的接下来的 6 行recvtime记录在 上2015-09-01T10:16:24+01:00,即dateObserved =2015-09-01T10:16:24+01:00

同样,最后 6 行有recvtimeas2019-06-03T14:47:47.696Z记录在2015-09-01T10:16:22+01:00iedateObserved = 2015-09-01T10:16:22+01:00

目标是更改recvtime列的内容,以便将这些值替换为记录为包含在attrvalue相应行中的日期。

因此,最终结果如下表所示(即产生此结果的最终表):

recvtimets     |           recvtime         | fiwareservicepath | entityid |   entitytype    |     attrname     |    attrtype     |               attrvalue               | attrmd
---------------+----------------------------+-------------------+----------+-----------------+------------------+-----------------+---------------------------------------+--------
 1559573267088 | 2015-09-01T10:15:57+01:00  | /basic            | 001      | WeatherObserved | location         | StructuredValue | {"coordinates":[-8.613502,41.143994]} | []
 1559573267088 | 2015-09-01T10:15:57+01:00  | /basic            | 001      | WeatherObserved | relativeHumidity | Number          | 59.1                                  | []
 1559573267088 | 2015-09-01T10:15:57+01:00  | /basic            | 001      | WeatherObserved | dateObserved     | Text            | 2015-09-01T10:15:57+01:00             | []
 1559573267088 | 2015-09-01T10:15:57+01:00  | /basic            | 001      | WeatherObserved | illuminance      | None            | null                                  | []
 1559573267088 | 2015-09-01T10:15:57+01:00  | /basic            | 001      | WeatherObserved | temperature      | Number          | 24.9                                  | []
 1559573267088 | 2015-09-01T10:15:57+01:00  | /basic            | 001      | WeatherObserved | name             | Text            | Rua das Flores                        | []
 1559573267782 | 2015-09-01T10:16:24+01:00  | /basic            | 001      | WeatherObserved | name             | Text            | Praca do Marques - pole 1             | []
 1559573267782 | 2015-09-01T10:16:24+01:00  | /basic            | 001      | WeatherObserved | illuminance      | None            | null                                  | []
 1559573267782 | 2015-09-01T10:16:24+01:00  | /basic            | 001      | WeatherObserved | temperature      | Number          | 21.9                                  | []
 1559573267782 | 2015-09-01T10:16:24+01:00  | /basic            | 001      | WeatherObserved | dateObserved     | Text            | 2015-09-01T10:16:24+01:00             | []
 1559573267782 | 2015-09-01T10:16:24+01:00  | /basic            | 001      | WeatherObserved | location         | StructuredValue | {"coordinates":[-8.603802,41.161671]} | []
 1559573267782 | 2015-09-01T10:16:24+01:00  | /basic            | 001      | WeatherObserved | relativeHumidity | Number          | 80.8                                  | []
 1559573267696 | 2015-09-01T10:16:22+01:00  | /basic            | 001      | WeatherObserved | name             | Text            | Fundacao de Serralves                 | []
 1559573267696 | 2015-09-01T10:16:22+01:00  | /basic            | 001      | WeatherObserved | illuminance      | None            | null                                  | []
 1559573267696 | 2015-09-01T10:16:22+01:00  | /basic            | 001      | WeatherObserved | location         | StructuredValue | {"coordinates":[-8.65915,41.1591]}    | []
 1559573267696 | 2015-09-01T10:16:22+01:00  | /basic            | 001      | WeatherObserved | relativeHumidity | Number          | 1                                     | []
 1559573267696 | 2015-09-01T10:16:22+01:00  | /basic            | 001      | WeatherObserved | dateObserved     | Text            | 2015-09-01T10:16:22+01:00             | []

我如何在 SQL 中实现这一点?

postgresql
  • 1 个回答
  • 48 Views

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