BigQuery 将新列添加到子查询中的嵌套 STRUCT

Posted 2023-03-24

【中文标题】BigQuery 将新列添加到子查询中的嵌套 STRUCT

【英文标题】：BigQuery add new columns into nested STRUCT in subquery

【发布时间】：2019-11-10 04:03:00

【问题描述】：

我的下表包含嵌套的STRUCTs，并且在子查询中，我试图在结构级别添加其他列。到目前为止，我已经创建了一个可重复的示例：

WITH wide_stats AS (
  (
    SELECT 
      'joe' name, 'bills' team,
      struct(struct(7 as fga, 5 as fgm) as o, struct(8 as fga, 3 as fgm) as d) as t1,
      struct(struct(3 as fga, 4 as fgm) as o, struct(9 as fga, 2 as fgm) as d) as t2
  ) UNION ALL (
    SELECT 'nick' name, 'jets' team,
      struct(struct(12 as fga, 7 as fgm) as o, struct(13 as fga, 7 as fgm) as d) as t1,
      struct(struct(15 as fga, 7 as fgm) as o, struct(22 as fga, 7 as fgm) as d) as t2
  )
)

SELECT 
  *,
--   safe_divide(wide_stats.t1.o.fgm, wide_stats.t1.o.fga) as fg_pct,
  safe_divide(wide_stats.t1.o.fgm, wide_stats.t1.o.fga) as wide_stats.t1.o.fg_pct  

FROM wide_stats

当前代码在第 18 行（使用 safe_divide）抛出错误 Syntax error: Unexpected "." at [18:70]。如果我在第 17 行/第 18 行切换，代码可以工作，但是 fg_pct 不在 t1.o 结构中，我希望它在哪里。

有没有办法在这样的子查询中将列添加到嵌套结构中？

【参考方案1】：

以下是 BigQuery 标准 SQL

#standardSQL
WITH wide_stats AS (
  SELECT 'joe' name, 'bills' team,
    STRUCT(STRUCT(7 AS fga, 5 AS fgm) AS o, STRUCT(8 AS fga, 3 AS fgm) AS d) AS t1,
    STRUCT(STRUCT(3 AS fga, 4 AS fgm) AS o, STRUCT(9 AS fga, 2 AS fgm) AS d) AS t2 UNION ALL 
  SELECT 'nick' name, 'jets' team,
    STRUCT(STRUCT(12 AS fga, 7 AS fgm) AS o, STRUCT(13 AS fga, 7 AS fgm) AS d) AS t1,
    STRUCT(STRUCT(15 AS fga, 7 AS fgm) AS o, STRUCT(22 AS fga, 7 AS fgm) AS d) AS t2
)
SELECT * REPLACE (
  (SELECT AS STRUCT t1.* REPLACE (
    (SELECT AS STRUCT t1.o.*, SAFE_DIVIDE(wide_stats.t1.o.fgm, wide_stats.t1.o.fga) AS fg_pct ) 
    AS o))
  AS t1)
FROM wide_stats   

结果

Row name    team    t1.o.fga    t1.o.fgm    t1.o.fg_pct         t1.d.fga    t1.d.fgm    t2.o.fga    t2.o.fgm    t2.d.fga    t2.d.fgm     
1   joe     bills   7           5           0.7142857142857143  8           3           3           4           9           2    
2   nick    jets    12          7           0.5833333333333334  13          7           15          7           22          7    

【讨论】：

有用的答案，但是当我尝试使用这种方法将percent_rank() 与SELECT AS STRUCT 一起使用时，我收到了这个错误SELECT without FROM clause cannot use analytic functions at [92:6]

在您的问题中看不到任何分析函数 - 您可能想发布包含所有详细信息的新问题，以便我们提供帮助：o)