Question

我有一个表，其中包含来自其他2个数据库（textfile和htmltables）的数据他们有一个共同点是列order_number，这就是为什么我把它们组合到同一个表中的原因。 dr_ *来自文本文件，来自htmltable

的oi_ *

Select * from data;
+--------------+-----------+----------+-----------+-------+
| order_number | dr_amount | dr_speed | oi_amount | oi_up |
+--------------+-----------+----------+-----------+-------+
|         9699 | 10000     | 26000    | NULL      | NULL  |
|         9699 | 20000     | 47619    | NULL      | NULL  |
|        10135 | 18000     | 12676    | NULL      | NULL  |
|         9979 | 25000     | 14286    | NULL      | NULL  |
|         9699 | NULL      | NULL     | 4800      | 4     |
|        10135 | NULL      | NULL     | 8700      | 2     |
|         9979 | NULL      | NULL     | 3000      | 8     |
+--------------+-----------+----------+-----------+-------+

首先，我必须使用

从表格中选出order_number（使用dr_amount）

select order_number, count(*) as c from data where oi_amount IS NOT NULL group by order_number having c<2;
+--------------+---+
| order_number | c |
+--------------+---+
|         9699 | 1 |
|         9979 | 1 |
+--------------+---+

这将删除拥有9699的order_number，因为它分为2行＆amp;＆amp; dr_amount = IS NOT NULL （从我后来的处理中，如果它是一个重复行，我将无法将特定行与另一行匹配，因此我将所有重复项排除在dr_amount = IS NOT NULL之外）

接下来我想通过将没有重复的order_number（对于相同的order_number的dr_amount）与具有oi_ *值的那些组合来产生这种输出。像这样：

+--------------+-----------+----------+-----------+-------+
| order_number | dr_amount | dr_speed | oi_amount | oi_up |
+--------------+-----------+----------+-----------+-------+
|        10135 | 18000     | 12676    | 8700      | 2     |
|         9979 | 25000     | 14286    | 3000      | 8     |
+--------------+-----------+----------+-----------+-------+

正如你所看到的，order_number 9699被整理出来，因为排序没有重复，第3和第6行被合并，以及第4和第7行。

我正在考虑使用第一个过滤来获取非重复的order_number并将其传递给第二个select-query作为where =的结果但是这给了我一个问题：

select order_number, dr_amount, dr_speed, oi_amount, oi_up from data where order_number=(select order_number, count(*) as c from data where oi_amount IS NOT NULL group by order_number having c<2);

Operand should contain 1 column(s)

我理解为什么会出现错误，而不是如何解决错误。当使用排序重复所需的嵌套选择时，它将返回2列（order_number和count（*）为c）。那么我如何使用嵌套选择，所以当我将它传递给真正的select时它只包含1列？

祝你好运尼克拉斯古斯塔夫森

Answer 1

该上下文中的子查询不能返回两列。您还需要in而不是=，因为它可能会返回多行。这可能就是你想要做的事情：

select order_number, dr_amount, dr_speed, oi_amount, oi_up
from data
where order_number in (select order_number
                       from data
                       where oi_amount IS NOT NULL
                       group by order_number
                       having count(*) < 2
                      ) and
      dr_amount is not null;

Answer 2

感谢提示将计数（*）移到最后。这帮助我把其余部分组合在一起！您建议的查询无法开箱即用。但这给了我想要的结果：

select order_number, max(dr_amount) as dr_amount, max(oi_amount) as oi_amount from data where order_number IN (select order_number from data where oi_amount IS NOT NULL group by order_number having count(*)<2) group by order_number having dr_amount IS NOT NULL && oi_amount IS NOT NULL;

使用此查询，行将合并，因为我希望它们是:) 我删除了dr_speed和oi_up，因为它们并不是真的需要。

/尼克拉斯

mySQL - 根据第一行的计数结合行的结果？

2 个答案: