错误：SELECT列表表达式引用了created_utc列，该列既未在[2:49]进行分组也未聚合

Posted 2021-04-16

我有一张桌子：t

我的目标：仅提取表中得分最高的“id”，并将其按week_number分组。

查询：

SELECT id, 
       CAST(EXTRACT(WEEK FROM TIMESTAMP_SECONDS(created_utc)) AS STRING) AS week_number, 
       MAX(score) AS highest_score
FROM t 
WHERE body='r/twinpeaks'
GROUP BY id;

我收到此错误：错误：SELECT列表表达引用列created_utc既未分组也未汇总在[2:49]

我尝试这样做：

SELECT id, 
       CAST(EXTRACT(WEEK FROM TIMESTAMP_SECONDS(created_utc)) AS STRING) AS week_number, 
       MAX(score) AS highest_score
FROM  t 
WHERE body='r/twinpeaks'
GROUP BY week_number, id;

但这就是我得到的：

Row  id            week_number  highest_score    
1    dmkb6sv      36            1    
2    dn1cd2s      37            2    
3    dn43h1k      38            16   
4    dn3xf18      38            1    
5    dn7i1ko      38            1
6    dnpr9b1      39            1

我要这个：

Row  id            week_number  highest_score    
1    dmkb6sv      36            1    
2    dn1cd2s      37            2    
3    dn43h1k      38            16   
6    dnpr9b1      39            1

答案

你可以尝试在这里使用ROW_NUMBER：

SELECT *
FROM
(
    SELECT id,
        CAST(EXTRACT(WEEK FROM TIMESTAMP_SECONDS(created_utc)) AS STRING) AS week_number,
        score,
        ROW_NUMBER() OVER (PARTITION BY
            CAST(EXTRACT(WEEK FROM TIMESTAMP_SECONDS(created_utc)) AS STRING)
            ORDER BY score DESC) rn
    FROM t 
    WHERE body = 'r/twinpeaks'
) t
WHERE rn = 1;

这将返回每个相应周数的最高分的记录。我在这里假设你要么不关心第一关系，要么不关系。如果您需要处理关系，则可以使用排名函数代替行号。

另一答案

以下是BigQuery Standard SQL

#standardSQL
SELECT 
  EXTRACT(WEEK FROM TIMESTAMP_SECONDS(created_utc)) AS week_number,
  ARRAY_AGG(id ORDER BY score DESC LIMIT 1)[OFFSET(0)] id,
  ARRAY_AGG(score ORDER BY score DESC LIMIT 1)[OFFSET(0)] highest_score    
FROM `project.dataset.table` 
WHERE body = 'r/twinpeaks'
GROUP BY week_number
ORDER BY week_number