Question

拥有一个事件表，每个用户有多种类型的UUID，我们希望找到一种方法将所有这些UUID拼接在一起，以获得单个用户的最高可能定义。

例如：

   UUID1 | UUID2
   1      a
   1      a
   2      a
   2      b
   3      c
   4      c

这里有2位用户，第一位是uuid1 = {1,2}，uuid2 = {a，b}，第二位是uuid1 = {3,4}，uuid2 = {c}。这些链可能很长。没有交叉点（即1c不存在），所有行都按时间戳排序。

红移是否有办法产生这些独特的＆＃34;客人＆＃34;标识符没有创建具有多个连接的巨大查询？

提前致谢！

Answer 1

创建测试数据表

-- DROP TABLE uuid_test;
CREATE TEMP TABLE uuid_test AS
          SELECT   1 row_id, 1::int uuid1, 'a'::char(1) uuid2
UNION ALL SELECT   2 row_id, 1::int uuid1, 'a'::char(1) uuid2
UNION ALL SELECT   3 row_id, 2::int uuid1, 'a'::char(1) uuid2
UNION ALL SELECT   4 row_id, 2::int uuid1, 'b'::char(1) uuid2
UNION ALL SELECT   5 row_id, 3::int uuid1, 'c'::char(1) uuid2
UNION ALL SELECT   6 row_id, 4::int uuid1, 'c'::char(1) uuid2
UNION ALL SELECT   7 row_id, 4::int uuid1, 'd'::char(1) uuid2
UNION ALL SELECT   8 row_id, 5::int uuid1, 'e'::char(1) uuid2
UNION ALL SELECT   9 row_id, 6::int uuid1, 'e'::char(1) uuid2
UNION ALL SELECT  10 row_id, 6::int uuid1, 'f'::char(1) uuid2
UNION ALL SELECT  11 row_id, 7::int uuid1, 'f'::char(1) uuid2
UNION ALL SELECT  12 row_id, 8::int uuid1, 'g'::char(1) uuid2
UNION ALL SELECT  13 row_id, 8::int uuid1, 'h'::char(1) uuid2
;

通过使用严格排序来查找唯一用户更改的每个位置，将其捕获为查找表，然后将其应用于原始数据，解决了实际问题。

-- Create lookup table with a from-to range of IDs for each unique user
WITH unique_user AS (

-- Calculate the end of the id range using LEAD() to look ahead
-- Use an inline MAX() to find the ending ID for the last entry

SELECT row_id AS from_id
     , NVL(LEAD(row_id,1) OVER (ORDER BY row_id)-1, (SELECT MAX(row_id) FROM uuid_test) ) AS to_id
     , unique_uuid

-- Mark unique user change when there is discontinuity in either UUID
FROM (SELECT row_id
            ,CASE WHEN NVL(LAG(uuid1,1) OVER (ORDER BY row_id),  0) <> uuid1
                   AND NVL(LAG(uuid2,1) OVER (ORDER BY row_id), '') <> uuid2
             THEN MD5(uuid1||uuid2)
             ELSE NULL END unique_uuid
      FROM uuid_test) t
WHERE unique_uuid IS NOT NULL
ORDER BY row_id
)

-- Apply the unique user value to each row using a range join to the lookup table
SELECT a.row_id, a.uuid1, a.uuid2, b.unique_uuid
FROM uuid_test   AS a
JOIN unique_user AS b
  ON a.row_id BETWEEN b.from_id AND b.to_id
ORDER BY a.row_id
;

这是输出

 row_id | uuid1 | uuid2 |           unique_uuid
--------+-------+-------+----------------------------------
      1 |     1 | a     | efaa153b0f682ae5170a3184fa0df28c
      2 |     1 | a     | efaa153b0f682ae5170a3184fa0df28c
      3 |     2 | a     | efaa153b0f682ae5170a3184fa0df28c
      4 |     2 | b     | efaa153b0f682ae5170a3184fa0df28c
      5 |     3 | c     | 5fcfcb7df376059d0075cb892b2cc37f
      6 |     4 | c     | 5fcfcb7df376059d0075cb892b2cc37f
      7 |     4 | d     | 5fcfcb7df376059d0075cb892b2cc37f
      8 |     5 | e     | 18a368e1052b5aa0388ef020dd9a1e20
      9 |     6 | e     | 18a368e1052b5aa0388ef020dd9a1e20
     10 |     6 | f     | 18a368e1052b5aa0388ef020dd9a1e20
     11 |     7 | f     | 18a368e1052b5aa0388ef020dd9a1e20
     12 |     8 | g     | 321fcc2447163a81d470b9353e394121
     13 |     8 | h     | 321fcc2447163a81d470b9353e394121

在redshift中查找具有多个UUID标识符的唯一实体

1 个答案: