0

数据如下所示——

category_id   category_name    associated_keys
    111          Books         CC34DE5W|SQA7ZZ87|LM24NO3P|SQA7ZZ87
    222          Office        LM24NO3P|AAB12B34
    444         Furniture      X34YY78Z|LM24NO3P|SQA7ZZ87|SEF5C6T4|CC34DE5W|AAB12B34
    222          Office        X34YY78Z|X34YY78Z

我想从 associated_keys 列中删除不同 category_id 的重复项。输出应如下所示 -

category_id   category_name    associated_keys
    111          Books         CC34DE5W|SQA7ZZ87|LM24NO3P
    222          Office        LM24NO3P|AAB12B34
    444         Furniture      X34YY78Z|LM24NO3P|SQA7ZZ87|SEF5C6T4|CC34DE5W|AAB12B34
    222          Office        X34YY78Z
4

2 回答 2

2

以下是 BigQuery 标准 SQL

#standardSQL
SELECT category_id, category_name, 
  (SELECT STRING_AGG(DISTINCT key, '|') FROM UNNEST(SPLIT(associated_keys, '|')) key) associated_keys
FROM (
  SELECT category_id, category_name, STRING_AGG(associated_keys, '|') AS associated_keys
  FROM `project.dataset.data` 
  GROUP BY category_id, category_name  
)

如果适用于您的示例中的示例数据 - 输出是

Row category_id category_name   associated_keys  
1   111         Books           CC34DE5W|SQA7ZZ87|LM24NO3P   
2   222         Office          LM24NO3P|AAB12B34|X34YY78Z   
3   444         Furniture       X34YY78Z|LM24NO3P|SQA7ZZ87|SEF5C6T4|CC34DE5W|AAB12B34

如果您不想按 category_id 分组（如最后一个问题） - 请在下面使用

#standardSQL
SELECT category_id, category_name, 
  (SELECT STRING_AGG(DISTINCT key, '|') FROM UNNEST(SPLIT(associated_keys, '|')) key) associated_keys
FROM `project.dataset.data`

带输出

Row category_id category_name   associated_keys  
1   111         Books           CC34DE5W|SQA7ZZ87|LM24NO3P   
2   222         Office          LM24NO3P|AAB12B34    
3   444         Furniture       X34YY78Z|LM24NO3P|SQA7ZZ87|SEF5C6T4|CC34DE5W|AAB12B34    
4   222         Office          X34YY78Z
于 2020-08-31T23:40:51.680 回答
0

不要将此类值存储为字符串！BigQuery 提供数组。因此，我将向您展示如何将结果从字符串转换为数组：

with t as (
      select 111 as category_id, 'Books' as category_name, 'CC34DE5W|SQA7ZZ87|LM24NO3P|SQA7ZZ87' as associated_keys union all
      select 222, 'Office', 'LM24NO3P|AAB12B34' union all
      select 444, 'Furniture', 'X34YY78Z|LM24NO3P|SQA7ZZ87|SEF5C6T4|CC34DE5W|AAB12B34' union all
      select 222, 'Office', 'X34YY78Z|X34YY78Z'
     )
select t.* except (associated_keys),
       (select array_agg(distinct key)
        from unnest(split(t.associated_keys, '|')) key
       ) as associated_keys
from t;

如果你真的想重建一个字符串，你可以使用string_agg()，但我不建议这样做。

于 2020-08-31T23:38:57.600 回答