Write a query to display the patron ID and the average number of days that patron keeps books during a checkout. Limit the results to only patrons who have at least three checkouts. Sort the results in descending order by the average days the book is kept.

Sure, here's the SQL query:

SELECT patron_id, AVG(DATEDIFF(checkin_date, checkout_date)) AS avg_days
FROM checkouts
GROUP BY patron_id
HAVING COUNT(*) >= 3
ORDER BY avg_days DESC;

This query selects the patron ID and the average number of days that patron keeps books during a checkout from the `checkouts` table. It then groups the results by patron ID and only includes patrons who have at least three checkouts (specified by the `HAVING` clause). Finally, it sorts the results in descending order by the average number of days the book is kept (specified by the `ORDER BY` clause).

相关问题

Write a query to display the patron ID, last name, number of times that patron has ever checked out a book, and the number of different books the patron has ever checked out. For example, if a given patron has checked out the same book twice, that would count as two checkouts but only one book. Limit the results to only patrons who have made at least three checkouts. Sort the results in descending order by number of books, then in descending order by number of checkouts, and then in ascending order by patron ID.

Sure, here's the SQL query that will display the required information:

SELECT p.patron_id, p.last_name, COUNT(*) AS num_checkouts, COUNT(DISTINCT b.book_id) AS num_books
FROM patrons p
JOIN checkouts c ON p.patron_id = c.patron_id
JOIN books b ON c.book_id = b.book_id
GROUP BY p.patron_id
HAVING num_checkouts >= 3
ORDER BY num_books DESC, num_checkouts DESC, p.patron_id ASC;

This query joins the `patrons`, `checkouts`, and `books` tables together, and groups the results by patron ID. The `COUNT()` function is used to count the number of checkouts and the number of distinct books that each patron has checked out. The `HAVING` clause filters out patrons who have made less than three checkouts, and the `ORDER BY` clause sorts the results as required.

Jupyter语法：提示用户输入他们的patron type：science rookie或science enthusiast

### 回答1：
你可以使用Python的input函数提示用户输入patron type，代码如下：

patron_type = input("请输入您的patron type（science rookie或science enthusiast）：")

这行代码将会在Jupyter Notebook中输出提示语句“请输入您的patron type（science rookie或science enthusiast）：”，并等待用户输入patron type。用户可以在Jupyter Notebook中的输入框中输入patron type，然后按下回车键来提交他们的输入。程序将会把用户输入的值存储在变量patron_type中，你可以在程序中使用这个变量来做出相应的决策。

### 回答2：
Jupyter是一个交互式的编程环境，常用于Python语言的开发和数据分析。在使用Jupyter时，可以使用语法来提示用户输入特定的信息。

如果我们想要提示用户输入他们的patron type，即科学爱好者的类型，可以使用`input()`函数来实现。下面是一个简单的例子：

patron_type = input("请输入您的patron type（可选择'science rookie'或'science enthusiast'）：")

上述代码中，`input("请输入您的patron type（可选择'science rookie'或'science enthusiast'）：")`会输出一个提示信息，要求用户输入他们的patron type。用户将会看到类似于“请输入您的patron type（可选择'science rookie'或'science enthusiast'）：”的提示，并且可以在提示后输入相应的选项。

用户输入的内容将被存储在变量`patron_type`中，以便后续使用。

在这个例子中，我们限定了用户可以输入的选项为'science rookie'或'science enthusiast'，用户输入其他选项时，可能需要添加相应的错误处理逻辑。

以上就是使用Jupyter的简单语法来提示用户输入patron type的方法。当然，这只是一个简单的示例，实际使用中可能需要更复杂的逻辑和代码。

### 回答3：
在Jupyter中，我们可以使用`input()`函数来提示用户输入信息。下面是一个简单的例子，用于提示用户输入他们的patron type：science rookie或science enthusiast：

patron_type = input("请输入您的patron type（science rookie或science enthusiast）：")

if patron_type == "science rookie":
    print("您的patron type是science rookie。")
elif patron_type == "science enthusiast":
    print("您的patron type是science enthusiast。")
else:
    print("您输入的patron type不是有效选项。请重新输入。")

在这个例子中，`input()`函数用于提示用户输入信息，并将输入的结果存储在`patron_type`变量中。然后，我们使用条件语句来判断`patron_type`的值，并打印出相应的提示信息。如果用户输入的不是有效选项，则打印出相应的错误提示。

注意：这只是一个简单的示例，仅用于演示如何在Jupyter中提示用户输入patron type并根据输入做出相应的反应。您可以根据实际场景进行相应的调整和扩展。