如何在bigquery中的结构数组内排列数组中的元素

Question

如何在结构体数组中排列数组中的元素？

with races as(
select "800m" race, "Rudisha" name,23.4 s1, 26.3 s2, 26.4 s3, 26.1 s4 union all
select "800m" race, "Makhloufi" name,24.5 , 25.4, 26.6, 26.1 union all
select "800m" race, "Murphy" name,23.9, 26.0, 27.0, 26.0 union all
select "800m" race, "Bosse" name,23.6, 26.2, 26.5, 27.1 union all
select "800m" race, "Rotich" name,24.7, 25.6, 26.9, 26.4 union all
select "800m" race, "Lewandowski" name,25.0, 25.7, 26.3, 27.2 union all 
select "800m" race, "Kipketer" name,23.2, 26.1, 27.3, 29.4 union all
select "800m" race, "Berian" name,23.7, 26.1, 27.0, 29.3
)

我尝试了以下查询，但不起作用。

select race,array(select x splits from unnest(arr.splits)x order by x )from
(select race,[struct(name as name, [s1,s2,s3,s4] as splits)]arr from races)

Answer 1

那么，您只需要排列 s1、s2、s3、s4 列中的值，以便它们在相同的 s1、s2、s3、s4 中排序？- 是的，你是对的

以下是 BigQuery 标准 SQL

#standardSQL
SELECT
  race,
  name,
  splits[OFFSET(0)] AS s1,
  splits[OFFSET(1)] AS s2,
  splits[OFFSET(2)] AS s3,
  splits[OFFSET(3)] AS s4
FROM (
  SELECT 
    race,
    name,
    ARRAY(SELECT s FROM UNNEST([s1, s2, s3, s4]) s ORDER BY s) splits
  FROM `project.dataset.races`
)

您可以使用问题中的虚拟数据来测试、玩上面的内容，如下所示

#standardSQL
WITH `project.dataset.races` AS (
  SELECT "800m" race, "Rudisha" name,23.4 s1, 26.3 s2, 26.4 s3, 26.1 s4 UNION ALL
  SELECT "800m" race, "Makhloufi" name,24.5 , 25.4, 26.6, 26.1 UNION ALL
  SELECT "800m" race, "Murphy" name,23.9, 26.0, 27.0, 26.0 UNION ALL
  SELECT "800m" race, "Bosse" name,23.6, 26.2, 26.5, 27.1 UNION ALL
  SELECT "800m" race, "Rotich" name,24.7, 25.6, 26.9, 26.4 UNION ALL
  SELECT "800m" race, "Lewandowski" name,25.0, 25.7, 26.3, 27.2 UNION ALL 
  SELECT "800m" race, "Kipketer" name,23.2, 26.1, 27.3, 29.4 UNION ALL
  SELECT "800m" race, "Berian" name,23.7, 26.1, 27.0, 29.3
)
SELECT
  race,
  name,
  splits[OFFSET(0)] AS s1,
  splits[OFFSET(1)] AS s2,
  splits[OFFSET(2)] AS s3,
  splits[OFFSET(3)] AS s4
FROM (
  SELECT 
    race,
    name,
    ARRAY(SELECT s FROM UNNEST([s1, s2, s3, s4]) s ORDER BY s) splits
  FROM `project.dataset.races`
)

结果/输出为

Row race    name        s1      s2      s3      s4   
1   800m    Murphy      23.9    26.0    26.0    27.0     
2   800m    Bosse       23.6    26.2    26.5    27.1     
3   800m    Lewandowski 25.0    25.7    26.3    27.2     
4   800m    Makhloufi   24.5    25.4    26.1    26.6     
5   800m    Rudisha     23.4    26.1    26.3    26.4     
6   800m    Kipketer    23.2    26.1    27.3    29.4     
7   800m    Berian      23.7    26.1    27.0    29.3     
8   800m    Rotich      24.7    25.6    26.4    26.9