在 mysql 中不使用连接合并两个表？

Question

我尝试合并来自两个不同 tables.The 的值，但没有成功，第一个显示如我所愿，但第二个每次只显示第一行。

Select * From (Select date as Date, Sum(qty) as qtySum, Sum(weight) 
as weightSum From stock_list Group by date) as A,
(Select Sum(weight) as weightSum,Count(barcode)
as barcodeCount From selected_items Group by date) as B Group by date;

这是我得到的输出。

These is my selected_items table.

这是我的stock_list.

我的两个查询都是单独工作的，只有当我尝试运行它们在一起时我才能得到正确的输出它给第二个问题 query.Can 有人指出我的错误或告诉我更好的方法来做到这一点。这就是我的最终 objective 是

Answer 1

您可以使用 join。但是，如果两个表中的日期集不相同，那么您可能需要 full outer join。但这在 MySQL 中不可用。相反：

select date, sum(qtySum), sum(weightsum1), sum(weightsum2), sum(barcodeCount)
from ((Select date as Date, Sum(qty) as qtySum, Sum(weight) as weightSum1,
              NULL as weightsum2, NULL as barcodeCount
        From stock_list
        Group by date
       ) union all
       (Select date, null, null, Sum(weight), Count(barcode) as barcodeCount 
        From selected_items
        Group by date
       )
      ) t
Group by date;

我不确定您想要的输出如何与您提供的查询相对应。但这应该按日期聚合和合并两个表中的数据，以便您可以完成查询。

Answer 2

如果某些日期可能不存在于两个表中，您可以使用 FULL JOIN 运算符来解决您的任务。

Select ISNULL(A.date, B.date) AS date,
       A.qtySum,
       A.weightSum,
       B.weightSum,
       B.barcodeCount
From 
    (Select date as Date, 
            Sum(qty) as qtySum, 
            Sum(weight) as weightSum 
     From stock_list 
     Group by date) as A
FULL JOIN
(Select date,
        Sum(weight) as weightSum,
        Count(barcode) as barcodeCount 
 From selected_items 
 Group by date) as B ON A.date = B.date

Answer 3

第一个问题是您在子查询 B 中按日期分组，但您没有 select 它，因此您的结果集可能类似于：

weightSum       barcodeCount
---------------------------
26              8
9               14
4               7

这是 3 个日期的结果，但您不知道哪一行指的是哪个日期。

你的下一个问题是你正在使用交叉连接，因为你的两个查询之间没有 link，这意味着如果你的第一个查询 returns:

Date            qtySum      weightSum
----------------------------------------
2016-01-20      1           1
2016-01-21      2           2

完成此交叉连接后，您将得到：

Date            qtySum      a.weightSum     b.weightSum     barcodeCount
--------------------------------------------------------------------------
2016-01-20      1           1               26              8
2016-01-20      1           1               9               14
2016-01-20      1           1               4               7
2016-01-21      2           2               26              8
2016-01-21      2           2               9               14
2016-01-21      2           2               4               7

所以 A 中的每一行都与 B 中的每一行相匹配，总共有 6 行。

你的第三个问题是你然后按日期分组，但不执行任何聚合，没有深入研究 SQL 标准的细则、按子句分组和函数依赖，让我们将其简化为 MySQL 允许这样做，但除非您了解限制（这在 this answer 中对此有更详细的介绍），否则您不应该这样做。 select 中不在 group by 子句中的任何内容都应该在聚合中。

因此，由于 MySQL 的 GROUP BY 扩展 select 对所有内容进行分组并仅按日期分组，您实际上是说每个日期取 1 行，但您无法控制在哪一行，它可能是上面显示的每个组的第一行，所以你会得到的结果是：

Date            qtySum      a.weightSum     b.weightSum     barcodeCount
--------------------------------------------------------------------------
2016-01-20      1           1               26              8
2016-01-21      2           2               26              8

我认为这就是为什么您最终会重复子查询 B 中的所有相同值。

所以这涵盖了错误的地方，现在开始解决方案，假设 stock_list 中的日期在 selected_items 中不存在，反之亦然，您将需要一个完整的连接, 但由于 MySQL 不支持，您必须使用 UNION，最简单的方法是：

SELECT  t.Date,
        SUM(t.StockQuantity) AS StockQuantity, 
        SUM(t.StockWeight) AS StockWeight, 
        SUM(t.SelectedWeight) AS SelectedWeight, 
        SUM(t.BarcodeCount) AS BarcodeCount
FROM    (   SELECT  date, 
                    SUM(qty) AS StockQuantity, 
                    SUM(weight) AS StockWeight, 
                    0 AS SelectedWeight, 
                    0 AS BarcodeCount
            FROM    stock_list
            GROUP BY Date
            UNION ALL
            SELECT  date, 
                    0 AS StockQuantity, 
                    0 AS StockWeight, 
                    SUM(weight) AS SelectedWeight, 
                    COUNT(BarCode) AS BarcodeCount
            FROM    selected_items
            GROUP BY Date
        ) AS t
GROUP BY t.Date;

编辑

我无法对此进行测试，也不确定您的确切逻辑，但您可以使用 variables to calculate a running total in MySQL。这应该给出如何做的想法：

SELECT  Date,
        StockQuantity,
        StockWeight,
        SelectedWeight,
        BarcodeCount,
        (@w := @w + StockWeight -  SelectedWeight) AS TotalWeight,
        (@q := @q + StockQuantity - BarcodeCount) AS TotalQuantity
FROM    (   SELECT  t.Date,
                    SUM(t.StockQuantity) AS StockQuantity, 
                    SUM(t.StockWeight) AS StockWeight, 
                    SUM(t.SelectedWeight) AS SelectedWeight, 
                    SUM(t.BarcodeCount) AS BarcodeCount
            FROM    (   SELECT  date, 
                                SUM(qty) AS StockQuantity, 
                                SUM(weight) AS StockWeight, 
                                0 AS SelectedWeight, 
                                0 AS BarcodeCount
                        FROM    stock_list
                        GROUP BY Date
                        UNION ALL
                        SELECT  date, 
                                0 AS StockQuantity, 
                                0 AS StockWeight, 
                                SUM(weight) AS SelectedWeight, 
                                COUNT(BarCode) AS BarcodeCount
                        FROM    selected_items
                        GROUP BY Date
                    ) AS t
            GROUP BY t.Date
        ) AS t
        CROSS JOIN (SELECT @w := 0, @q := 0) AS v
GROUP BY t.Date;

在 mysql 中不使用连接合并两个表？

Merging two tables without using join in mysql?

mysql

sql

union

join

date