我有下表,其中包含嵌套的STRUCT
,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)引发错误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 不在在t1.o结构中,我希望它在其中。
有没有办法像这样在子查询中将列添加到嵌套结构中?
答案 0 :(得分:2)
以下是用于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
有结果
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