Google bigquery BigQuery将新列添加到子查询中的嵌套结构中

下表包含嵌套的

STRUCT

s，在子查询中，我尝试在STRUCT级别添加其他列。到目前为止，我已经创建了一个可复制的示例：

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[18:70]第18行（使用安全除法）抛出一个错误

语法错误：意外“.”。如果我在第17行/第18行中切换，代码可以工作，但是fg_pct不在t1.o结构中，我希望它在那里

有没有办法像这样将列添加到子查询中的嵌套结构中
 下面是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    

下面是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    

有帮助的回答，但当我尝试使用
percent\u rank（）时，我收到此错误
SELECT without FROM子句无法在[92:6]使用分析函数
使用
SELECT AS STRUCT
使用此方法在您的问题中看不到任何分析函数-您可能希望发布包含所有详细信息的新问题，以便我们可以提供帮助：o）有用的答案，但我收到此错误
SELECT without FROM子句无法在[92:6]使用分析函数
使用
选择为结构
使用此方法在您的问题中看不到任何分析函数-您可能希望发布包含所有详细信息的新问题，以便我们提供帮助：o）