如何避免列中重复出现

Question

我有一张表，其中描述了为不同客户预订机票的代理。 以下数据描述了一个客户数据。

从上面的数据我期待的是

输出的意思是，我想把队列分组，他先订了一些去新加坡的票，然后是奥斯汀，再是新加坡和德里

我们如何在 SQL 中实现这一点，请帮助我

如果输出如下也有帮助

Answer 1

这是一个差距和孤岛问题。要解决它，您需要生成相邻记录组。这通常通过比较两个不同分区的行号来完成。

考虑：

select 
    agent_id,   
    travel_destination,
    min(date_of_booking) first_date_of_booking,
    max(date_of_booking) max_date_of_booking
from (
    select 
        t.*,
        row_number() 
            over(partition by agent_id order by date_of_booking) rn1,
        row_number() 
            over(partition by agent_id, travel_destination order by date_of_booking) rn2
    from mytable t
) t 
group by 
    agent_id, 
    rn1 - rn2,
    travel_destination
order by first_date_of_booking

请注意，我在答案中添加了每个组的开始和结束日期，因为我发现它使答案更有意义。

另外备注：根据您的样本数据，不清楚是否要将customerid放入组中；我假设不是（如果是，您需要将该列添加到两个分区）。

Demo on DB Fiddle：

鉴于这个（简化的）数据集：

agent_id | travel_destination |客户 ID | date_of_booking :------- | :----------------- | :------------ | :-------------- A1001 |新加坡 | C1001 | 2019-06-10 A1001 |新加坡 | C1001 | 2019-06-11 A1001 |奥斯汀 | C1001 | 2019-06-12 A1001 |新加坡 | C1001 | 2019-06-13 A1001 |新加坡 | C1001 | 2019-06-14 A1001 |新德里 | C1001 | 2019-06-15

查询返回：

agent_id | travel_destination | first_date_of_booking | max_date_of_booking :------- | :----------------- | :-------------------- | :----------------- A1001 |新加坡 | 2019-06-10 | 2019-06-11 A1001 |奥斯汀 | 2019-06-12 | 2019-06-12 A1001 |新加坡 | 2019-06-13 | 2019-06-14 A1001 |新德里 | 2019-06-15 | 2019-06-15

---

要实现您演示的第二个输出，您可以添加另一个级别的聚合并使用string_agg()

select 
    agent_id,
    string_agg(travel_destination order by first_date_of_booking) travel_destination
from (
  -- above query
) t
group by agent_id

Answer 2

试试这个 - 至少如果你的数据库有像 LISTAGG 这样的功能，就像在 Vertica 中一样......

WITH
-- this is your input - next time put it in so it can be 
-- copy-pasted and formatted to the below ....                                                                                                                                                    
input(agent_id,travel_dest,cust_id,bookdt) AS (
          SELECT 'A1001','Singapore','C1001',DATE '2109-06-10'
UNION ALL SELECT 'A1001','Singapore','C1001',DATE '2019-06-11'
UNION ALL SELECT 'A1001','Austin'   ,'C1001',DATE '2019-06-19'
UNION ALL SELECT 'A1001','Austin'   ,'C1001',DATE '2019-06-19'
UNION ALL SELECT 'A1001','Austin'   ,'C1001',DATE '2019-06-20'
UNION ALL SELECT 'A1001','Singapore','C1001',DATE '2019-07-30'
UNION ALL SELECT 'A1001','Singapore','C1001',DATE '2019-07-31'
UNION ALL SELECT 'A1001','Delhi'    ,'C1001',DATE '2019-08-01'
UNION ALL SELECT 'A1001','Delhi'    ,'C1001',DATE '2019-08-10'
UNION ALL SELECT 'A1001','Delhi'    ,'C1001',DATE '2019-08-10'
UNION ALL SELECT 'A1001','Delhi'    ,'C1001',DATE '2019-08-10'
UNION ALL SELECT 'A1001','Delhi'    ,'C1001',DATE '2019-08-10'
UNION ALL SELECT 'A1001','Delhi'    ,'C1001',DATE '2019-08-25'
)
-- real WITH clause starts here - substitute comma below with "WITH" ...
,
with_prev AS (
  SELECT
    agent_id
  , travel_dest
  , LAG(travel_dest,1,'') OVER (PARTITION BY agent_id ORDER BY bookdt) AS prev_dest
  FROM input
)
,
de_duped AS (
  SELECT
    agent_id
  , travel_dest
   FROM with_prev
   WHERE travel_dest <> prev_dest
)
SELECT
  agent_id
, LISTAGG(travel_dest) AS travel_dest
FROM de_duped
GROUP BY 1
;

你得到：

 agent_id |                travel_dest                 
----------+--------------------------------------------
 A1001    | Singapore,Austin,Singapore,Delhi,Singapore                                                                                                  

Answer 3

我只会使用lag():

SELECT t.agent_id, t.travel_dest
FROM (SELECT t.*,
             LAG(travel_dest) OVER (PARTITION BY agent_id ORDER BY bookdt) as prev_travel_dest
      FROM t
     ) t
WHERE prev_travel_dest IS NULL OR prev_travel_dest <> travel_dest
ORDER BY agent_id, bookdt;

我想不出更简单的解决方案。

Answer 4

以下是 BigQuery 标准 SQL

#standardSQL
SELECT agent_id, 
  STRING_AGG(DISTINCT travel_destination) AS travel_destination
FROM `project.dataset.table`
GROUP BY agent_id    

它将产生以下输出

Row agent_id    travel_destination   
1   A1001       Singapore,Austin,Delhi      

看起来预期的输出是Singapore,Austin,Singapore,Delhi - 下面是另一个选项

#standardSQL
CREATE TEMP FUNCTION DedupConsecutive(line STRING) RETURNS STRING LANGUAGE js AS """
  return line.split(",").filter(function(value,index,arr){return value != arr[index+1];}).join(",");
""";
SELECT agent_id, 
  DedupConsecutive(STRING_AGG(travel_destination ORDER BY date_of_booking)) destinations
FROM `project.dataset.table`
GROUP BY agent_id   

与 Gordon 的观点相同 - I cannot think of a simpler solution. :o)