Question

在BigQuery中，我有一个带有path列的表格，如下所示：

ID .     | Path
---------+----------------------------------------
1        | foo/bar/baz
2        | foo/bar/quux/blat

我希望能够在正斜杠（/）上分割路径，并选择一个或多个路径部分，然后重新加入它们。

在PostgreSQL中，这很容易：

select array_to_string((regexp_split_to_array(path, '/'))[1:3], '/')

但是BigQuery似乎没有任何类型的范围偏移量或数组切片功能。

Answer 1

以下是用于BigQuery标准SQL

#standardSQL
SELECT id, path,
  (
    SELECT STRING_AGG(part, '/' ORDER BY index) 
    FROM UNNEST(SPLIT(path, '/')) part WITH OFFSET index 
    WHERE index BETWEEN 1 AND 3
  ) adjusted_path
FROM `project.dataset.table`

您可以使用问题中的示例数据来进行测试，如上示例所示

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 1 id, 'foo/bar/baz/foo1/bar1/baz1/' path UNION ALL
  SELECT 2, 'foo/bar/quux/blat/foo2/bar2/quux2/blat2' 
)
SELECT id, path,
  (
    SELECT STRING_AGG(part, '/' ORDER BY index) 
    FROM UNNEST(SPLIT(path, '/')) part WITH OFFSET index 
    WHERE index BETWEEN 1 AND 3
  ) adjusted_path
FROM `project.dataset.table`

有结果

Row     id      path                                        adjusted_path    
1       1       foo/bar/baz/foo1/bar1/baz1/                 bar/baz/foo1     
2       2       foo/bar/quux/blat/foo2/bar2/quux2/blat2     bar/quux/blat

如果出于某种原因您希望保持查询与PostgreSQL中使用的查询“内联/相似”（array_to_string（（regexp_split_to_array（path，'/'））[1：3]，'/'））-您可以像下面的示例一样引入SQL UDF（将其命名为ARRAY_SLICE）

#standardSQL
CREATE temp  FUNCTION ARRAY_SLICE(arr ARRAY<STRING>, start INT64, finish INT64) 
RETURNS ARRAY<STRING> AS (
  ARRAY(
    SELECT part FROM UNNEST(arr) part WITH OFFSET index 
    WHERE index BETWEEN start AND finish ORDER BY index
  )
);
SELECT id, path, 
  ARRAY_TO_STRING(ARRAY_SLICE(SPLIT(path, '/'), 1, 3), '/') adjusted_path
FROM `project.dataset.table`

很显然，如果要应用于相同的样本数据-您将获得相同的结果

#standardSQL
CREATE temp  FUNCTION ARRAY_SLICE(arr ARRAY<STRING>, start INT64, finish INT64) 
RETURNS ARRAY<STRING> AS (
  ARRAY(
    SELECT part FROM UNNEST(arr) part WITH OFFSET index 
    WHERE index BETWEEN start AND finish ORDER BY index
  )
);
WITH `project.dataset.table` AS (
  SELECT 1 id, 'foo/bar/baz/foo1/bar1/baz1/' path UNION ALL
  SELECT 2, 'foo/bar/quux/blat/foo2/bar2/quux2/blat2' 
)
SELECT id, path, 
  ARRAY_TO_STRING(ARRAY_SLICE(SPLIT(path, '/'), 1, 3), '/') adjusted_path
FROM `project.dataset.table`   

Row     id      path                                        adjusted_path    
1       1       foo/bar/baz/foo1/bar1/baz1/                 bar/baz/foo1     
2       2       foo/bar/quux/blat/foo2/bar2/quux2/blat2     bar/quux/blat

如何在BigQuery Standard SQL中获取数组的一部分？

1 个答案: