根据另一列中的最高值从列中选择不同的值

I have a table in which I need to return a distinct value for each id based on the highest reference value. As an example, this table:

我有一个表，我需要根据最高参考值为每个id返回一个不同的值。作为一个例子，这个表：

my table
|--------|----------|-------------|
| ID     | Amount   | Ref_value   |
|---------------------------------|
| 1      | 200      | 5           |
| 1      | 120      | 8           |
| 2      | 150      | 3           |
| 3      | 200      | 4           |
|--------|----------|-------------|

I need to get one return per ID, but since ID=1 appears more than once, I need to select the one with the highest "Ref_value". So my result would be:

我需要为每个ID获得一次返回，但由于ID = 1出现不止一次，我需要选择具有最高“Ref_value”的那个。所以我的结果是：

result of the query over my table
|--------|----------|-------------|
| ID     | Amount   | Ref_value   |
|---------------------------------|
| 1      | 120      | 8           |
| 2      | 150      | 3           |
| 3      | 200      | 4           |
|--------|----------|-------------|

3 个解决方案

#1

you could use a inner join on the max value grouped by id

您可以在按ID分组的最大值上使用内部联接

select * from my_table 
inner join (
   select id, max(ref_value) max_value
   from my_table 
   group by id
)  t on t.id = my_table_id, t.max_vale = my_table_value

#2

If your database supports window functions, this solution will access the table only once:

如果您的数据库支持窗口函数，则此解决方案仅访问该表一次：

SELECT id, amount, ref_value
FROM (
  SELECT t.*, row_number() OVER (PARTITION BY id ORDER BY ref_value DESC) rn
  FROM t
) t
WHERE rn = 1

If you're using Oracle, you might use this solution that I've found to be a bit faster in most cases:

如果你正在使用Oracle，你可能会使用我发现在大多数情况下更快一点的解决方案：

SELECT
  max(id)        KEEP (DENSE_RANK FIRST ORDER BY ref_value DESC) id,
  max(amount)    KEEP (DENSE_RANK FIRST ORDER BY ref_value DESC) amount,
  max(ref_value) KEEP (DENSE_RANK FIRST ORDER BY ref_value DESC) ref_value,
FROM t
GROUP BY id

Both of the above solutions will return an arbitrary row if two ref_value values are tied.

如果两个ref_value值相关联，则上述两种解决方案都将返回任意行。

If none of the above apply, scaisEdge's solution works on all databases.

如果以上都不适用，scaisEdge的解决方案适用于所有数据库。

#3

Select * might throw an ambiguous column names error. So you might want to select the only required columns there.

选择*可能会抛出一个模糊的列名错误。因此，您可能希望在那里选择唯一需要的列。

SELECT A.ID, A.AMOUNT, A.REF_VALUE
FROM
MYTABLE A
INNER JOIN
(SELECT ID, MAX(REF_VALUE) AS MAX_REF FROM MYTABLE GROUP BY ID) B
ON A.ID = B.ID AND A.REF_VALUE = B.MAX_REF;

#1