Question

我正在尝试创建一个查询最新日期的查询并返回相应的值。仅当我具有max和hose part no列且没有其他内容时，才能使用max inspection函数。一旦我添加了scrap amount列，它就会复制hose part no列。我只希望每个软管零件号有一个记录。以下是发生情况的示例。

Hose Part No Max Inspection Date scrapamt1 scrapamt2 scrapamt3 scrapamt4

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Answer 1

如果我理解您说的没错，这就是您想要的：

df['group'] = pd.cut(df.value, range(0, highest_size), right=False, labels=labels)

子查询（名为Select * From Table1 Inner Join (Select Table1.[Hose Part No], Max(Table1.[Max Inspection Date]) As [MaxOfMax Inspection Date] From Table1 Group By Table1.[Hose Part No]) As MaxValues On Table1.[Hose Part No] = MaxValues.[Hose Part No] And Table1.[Max Inspection Date] = MaxValues.[MaxOfMax Inspection Date]）用于确定每个MaxValues的最大Max Inspection Date，并将其连接到同一表。

因为您也没有提及表格名称，所以我选择了Hose Part No。

Answer 2

这似乎可行-

tbl发布编号：

| Release No | Hose Part No | Release Date | Due Date   |
|------------|--------------|--------------|------------|
| 1          | Hose 1       | 01/01/2018   | 01/01/2018 |
| 2          | Hose 1       | 01/01/2018   | 01/01/2018 |
| 3          | Hose 2       | 01/01/2018   | 01/01/2018 |
| 4          | Hose 2       | 01/01/2018   | 01/01/2018 |
| 5          | Hose 3       | 01/01/2018   | 01/01/2018 |

NewScrapTble：

| Release No | Inspect Date | scrapamt1 | reason1 | scrapamt2 |
|------------|--------------|-----------|---------|-----------|
| 1          | 01/12/2018   | 10        |         | 15        |
| 2          | 12/12/2018   | 15        |         | 18        |
| 3          | 01/07/2018   | 12        |         | 12        |
| 4          | 01/08/2018   | 14        |         | 200       |
| 5          | 01/03/2017   | 22        |         | 20        |

SQL：

SELECT  [Hose Part No]
        , MAX([Inspect Date]) AS Inspect
        , LAST(scrapamt1) AS Amount1
        , LAST(scrapamt2) AS Amount2
FROM    (
        SELECT  [Hose Part No]
                , [Inspect Date]
                , scrapamt1
                , scrapamt2
        FROM    tblReleaseNo INNER JOIN NewScrapTble ON 
                    tblReleaseNo.[Release No] = NewScrapTble.[Release No]
        )
GROUP BY [Hose Part No]

最终表：

| Hose Part No | Inspect    | Amount1 | Amount2 |
|--------------|------------|---------|---------|
| Hose 1       | 12/12/2018 | 15      | 18      |
| Hose 2       | 01/08/2018 | 14      | 200     |
| Hose 3       | 01/03/2017 | 22      | 20      |

注意：我不是100％地确定LAST是正确的，它提取正确的金额，但不确定是否会在某些记录上跳闸？

使用相应记录访问选择最大日期

2 个答案: