Question

到目前为止，我有这个：

select created_at,
DATEDIFF(TO_DATE(current_date()), TO_DATE(sales_flat_order.created_at)) as delay,
count(*) over() as NumberOfOrders 
FROM
magentodb.sales_flat_order 
WHERE
status IN ( 'packed' , 'cod_confirmed' )
GROUP BY TO_DATE(created_at)

但这不起作用。

语法错误：

编译语句时出错：FAILED：SemanticException [错误10004]：第1行：7无效的表别名或列引用'created_at':(可能的列名是：（tok_function to_date（tok_table_or_col created_at）））
< / LI>
count(*) does not give sum for each grouped by date but instead all of the rows。

注意：我实际上使用的是hive，但它与查询

Answer 1

试试这个：

select created_at,
DATEDIFF(TO_DATE(current_date()), TO_DATE(sales_flat_order.created_at)) as delay,
count(*) as NumberOfOrders 
FROM
magentodb.sales_flat_order 
WHERE
status IN ( 'packed' , 'cod_confirmed' )
GROUP BY Date(created_at)

Answer 2

我的手机不允许我发表评论。但试试这个链接，它可能会以正确的方式引导你。

stackoverflow.com/questions/29704904/invalid-table-alias-or-column-reference-b

Answer 3

我认为您希望使用created_at的日期部分（包括年，月和日）进行分组。

select
date(created_at) as created_at_day,
datediff(curdate(), sales_flat_order.created_at) as delay,
count(*) as numberOfOrders
from magentodb.sales_flat_order
WHERE status IN ('packed', 'cod_confirmed' ) GROUP BY created_at_day

此查询仅显示当天创建的第一个订单。因为你按天分组。您可以使用average查找当天创建的订单的平均延迟。

按日期时间分组日期时间并获取每个

3 个答案: