以下是 BigQuery 标准 SQL - 希望您将其用于您的实际用例。
#standardSQL
WITH metals AS (
SELECT 'Gold' Metal, 1 RANK UNION ALL SELECT 'platinum', 2 UNION ALL
SELECT 'silver', 3 UNION ALL SELECT 'copper', 4 UNION ALL SELECT 'bronze', 5
)
SELECT Employee_number,
MAX(IF(pos=0, Metal, NULL)) Rank_1,
MAX(IF(pos=1, Metal, NULL)) Rank_2,
MAX(IF(pos=2, Metal, NULL)) Rank_3,
MAX(IF(pos=3, Metal, NULL)) Rank_4,
MAX(IF(pos=4, Metal, NULL)) Rank_5
FROM (
SELECT Employee_number,
ARRAY_CONCAT(
ARRAY(SELECT Metal FROM (
SELECT 1 a, Rank_1 Metal UNION ALL SELECT 2, Rank_2 UNION ALL
SELECT 3, Rank_3 UNION ALL SELECT 4, Rank_4 UNION ALL
SELECT 5, Rank_5 )
WHERE NOT Metal IS NULL
ORDER BY a
), ARRAY(SELECT Metal FROM metals m
WHERE NOT LOWER(Metal) IN (
SELECT x FROM UNNEST(ARRAY(
SELECT LOWER(b) FROM (
SELECT Rank_1 b UNION ALL SELECT Rank_2 UNION ALL
SELECT Rank_3 UNION ALL SELECT Rank_4 UNION ALL
SELECT Rank_5 )
WHERE NOT b IS NULL
)) x
) ORDER BY RANK
)) arr
FROM `project.dataset.employee`
), UNNEST(arr) Metal WITH OFFSET pos
GROUP BY Employee_number
ORDER BY Employee_number
您可以使用您问题中的虚拟数据进行测试,如下所示
#standardSQL
WITH `project.dataset.employee` AS (
SELECT 1 Employee_number, 'Gold' Rank_1, NULL Rank_2, NULL Rank_3, NULL Rank_4, NULL Rank_5 UNION ALL
SELECT 2, 'bronze', 'Gold', NULL, NULL, NULL UNION ALL
SELECT 3, 'Gold', 'platinum', NULL, NULL, NULL UNION ALL
SELECT 4, 'Gold', 'copper', NULL, NULL, NULL UNION ALL
SELECT 5, 'Gold', 'bronze', 'platinum', NULL, NULL UNION ALL
SELECT 6, 'Gold', 'bronze', 'platinum', NULL, NULL UNION ALL
SELECT 7, 'Gold', 'platinum', 'Silver', NULL, NULL UNION ALL
SELECT 8, 'Gold', 'platinum', 'Silver', NULL, NULL UNION ALL
SELECT 9, 'Gold', 'platinum', 'business', NULL, NULL UNION ALL
SELECT 10, NULL, NULL, NULL, NULL, NULL UNION ALL
SELECT 11, 'Silver', 'bronze', 'business', 'platinum', 'Gold'
), metals AS (
SELECT 'Gold' Metal, 1 RANK UNION ALL SELECT 'platinum', 2 UNION ALL
SELECT 'silver', 3 UNION ALL SELECT 'copper', 4 UNION ALL SELECT 'bronze', 5
)
SELECT Employee_number,
MAX(IF(pos=0, Metal, NULL)) Rank_1,
MAX(IF(pos=1, Metal, NULL)) Rank_2,
MAX(IF(pos=2, Metal, NULL)) Rank_3,
MAX(IF(pos=3, Metal, NULL)) Rank_4,
MAX(IF(pos=4, Metal, NULL)) Rank_5
FROM (
SELECT Employee_number,
ARRAY_CONCAT(
ARRAY(SELECT Metal FROM (
SELECT 1 a, Rank_1 Metal UNION ALL SELECT 2, Rank_2 UNION ALL
SELECT 3, Rank_3 UNION ALL SELECT 4, Rank_4 UNION ALL
SELECT 5, Rank_5 )
WHERE NOT Metal IS NULL
ORDER BY a
), ARRAY(SELECT Metal FROM metals m
WHERE NOT LOWER(Metal) IN (
SELECT x FROM UNNEST(ARRAY(
SELECT LOWER(b) FROM (
SELECT Rank_1 b UNION ALL SELECT Rank_2 UNION ALL
SELECT Rank_3 UNION ALL SELECT Rank_4 UNION ALL
SELECT Rank_5 )
WHERE NOT b IS NULL
)) x
) ORDER BY RANK
)) arr
FROM `project.dataset.employee`
), UNNEST(arr) Metal WITH OFFSET pos
GROUP BY Employee_number
ORDER BY Employee_number
结果
Row Employee_number Rank_1 Rank_2 Rank_3 Rank_4 Rank_5
1 1 Gold platinum silver copper bronze
2 2 bronze Gold platinum silver copper
3 3 Gold platinum silver copper bronze
4 4 Gold copper platinum silver bronze
5 5 Gold bronze platinum silver copper
6 6 Gold bronze platinum silver copper
7 7 Gold platinum Silver copper bronze
8 8 Gold platinum Silver copper bronze
9 9 Gold platinum business silver copper
10 10 Gold platinum silver copper bronze
11 11 Silver bronze business platinum Gold
注意:上述解决方案假设填充金属和空金属之间没有混合,这意味着三个选项:
1. all Rank fields filled already with Metal
2. all Rank fields are NULL
3. first 1 or more fields filled with Metal and rest are NULLs
话虽如此,第一个数组是由填充字段构建的;第二个数组由 Metals 表中的其余 Metal 字段构建;然后连接两个数组,前 5 个元素用于重新创建原始表
希望这不会太混乱
PS 上述解决方案可以相对容易地扩展到 NULL 和填充金属混合的情况 - 但看起来这超出了问题的范围:o)