MySQL 使用时间戳按天分组结果 [英] MySQL Group Results by day using timestamp

问题描述

我需要进行以下查询并提取总订单数和按天分组的订单总和.我使用时间戳存储所有内容.

I need to take the following query and pull the total order counts and sum of the orders grouped by day. I'm storing everything using timestamps.

SELECT
    COUNT(id) as order_count,
    SUM(price + shipping_price) as order_sum,
    DAY(FROM_UNIXTIME(created)) as day
FROM `order`
WHERE '.implode(' AND ', $where).'

我需要按 DAY 分组，但是当我为上周末的销售进行分组时，它需要我的 order_count 并将其设为 1 而不是 3.如何提取按天分组的上述值?

I need to group by DAY but when I do for this past weekend's sales it takes my order_count and makes it 1 instead of 3. How can I pull the above values grouped by day?

注意:内爆仅用于定义时间段(创建位置 >= TIMESTAMP AND <= TIMESTAMP)

NOTE: The implode is used ONLY to define the time period (WHERE created >= TIMESTAMP AND <= TIMESTAMP)

更新

没有 GROUP BY day

Without GROUP BY day

Array ( 
    [order_count] => 3
    [order_sum] => 69.70
    [day] => 17
)

使用 GROUP BY day

With GROUP BY day

Array ( 
    [order_count] => 1
    [order_sum] => 24.90
    [day] => 17
)

我需要这个查询来返回每天的销售额、订单数量以及这些销售额的总和.我在这里的某个地方遗漏了一块拼图......

I need this query to return each day that had sales, how many orders, and the sum of those sales. I'm missing a piece of the puzzle here somewhere....

推荐答案

你是不是忘了在最后加上GROUP BY ...?

Are you just forgetting to add GROUP BY ... at the end?

SELECT
    COUNT(id) as order_count,
    SUM(price + shipping_price) as order_sum,
    DAY(FROM_UNIXTIME(created)) as order_day
FROM `order`
WHERE '.implode(' AND ', $where).'
GROUP BY order_day

注意:

您不能将 as day 用于您的 day 列，因为 day 是 MySQL 函数.使用 order_day 之类的内容.

NOTE:

You cannot use as day for your day column because day is a MySQL function. Use something like order_day.

根据@OMG Unicorn 的评论，您可以使用:

Per @OMG Unicorn's comment, you can use:

DAY(FROM_UNIXTIME(created)) as `day`

只要将 day 包裹在 ` 反引号中.

So long as wrap day in ` backticks.

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