Question

create table db.temp
location '/user/temp' as
SELECT t1.mobile_no
FROM db.temp t1
WHERE NOT EXISTS ( SELECT NULL
                   FROM db.temp t2
                   WHERE t1.mobile_no = t2.mobile_no
                     AND t1.cell != t2.cell
                     AND t2.access_time BETWEEN t1.access_time
                                            AND t1.access_time_5);

我需要从 access_time_5 获取在时间间隔 (access time) 的 5 小时内使用同一单元格的所有用户。这段代码对impala 来说完全没问题。但不适用于 Hive。

出现错误

<块引用>

"编译语句时出错：失败： SemanticException [错误 10249]：第 23:25 行不受支持的子查询表达式"

我查看了与此错误相关的类似问题。想不出解决办法。任何帮助将不胜感激！

Answer 1

在 Hive 和非对等连接中不支持相关 BETWEEN。尝试使用 LEFT JOIN 重写，使用您的条件和过滤器计算行数：

select mobile_no from
(
SELECT t1.mobile_no, 
       sum(case when t1.cell != t2.cell
                 and t2.access_time between t1.access_time and t1.access_time_5
               then 1 else 0
           end) as cnt_exclude
  FROM db.temp t1
       LEFT JOIN db.temp t2 on t1.mobile_no = t2.mobile_no
 GROUP BY t1.mobile_no
)s
where cnt_exclude=0

这种解决方案的问题是LEFT JOIN可能会产生大量重复并且会影响性能，但如果数据不是太大，它可能会起作用。

Answer 2

在我看来，窗口函数对两个数据库都更好。让我假设 access_time 是 Unix 时间（即以秒为单位）。您可以轻松地将值转换为这样的时间：

SELECT t1.mobile_no
FROM (SELECT t1.*,
             MIN(t1.cell) OVER (PARTITION BY mobile_no
                                ORDER BY access_time
                                RANGE BETWEEN 17999 preceding AND CURRENT ROW
                               ) as min_cell,
             MAX(t1.cell) OVER (PARTITION BY mobile_no
                                ORDER BY access_time
                                RANGE BETWEEN 17999 preceding AND CURRENT ROW
                               ) as max_cell
      FROM db.temp t1
     ) t1
WHERE min_cell = max_cell;

Hive 子查询错误 - 不支持的子查询表达式

2 个答案: