如何有效地在同一表的两个实例之间交换值

Question

我有两个Book实例，我想在两个sequence之间交换books列值

## Original values
# book_1.sequence = 10  
# book_2.sequence = 15

book_id = [1,2]

book_1 = Book.find(book_id.first)
book_2 = Book.find(book_id.second)

book_1_sequence = book_1.sequence
book_2_sequence = book_2.sequence

book_1.update_attributes!(sequence: book_2_sequence)
book_2.update_attributes!(sequence: book_1_sequence)

程序执行后：

book_1.sequence = 15
            book_2.sequence = 10

这可以解决问题，但这很简单，还有更好的方法吗？整洁高效

Answer 1

至少可以减少一行代码：

book_1.sequence = 10  
book_2.sequence = 15

book_id = [1,2]

book_1, book_2 = Book.find(book_ids)

book_1_sequence = book_1.sequence

book_1.update_attributes!(sequence: book_2.sequence)
book_2.update_attributes!(sequence: book_1_sequence)

根据Robbert van ELk的建议进行了更新。

Answer 2

另一种方法：

# Original values
book_1.sequence
 => 10

book_2.sequence
 => 15

book_ids = [1,2]

# Fetch all the books in single query
book_1, book_2 = Book.find(book_ids)

book_1.sequence, book_2.sequence = book_2.sequence, book_1.sequence
[book_1, book_2].each(&:save!)

Answer 3

用红宝石换行

# Original values
# book_1.sequence = 10  
# book_2.sequence = 15

book_id = [1,2]

book_1, book_2 = Book.find(book_ids)

book_1.sequence, book_2.sequence = book_2.sequence, book_1.sequence
book_1.save; book_2.save

甚至更短（不推荐，因为它很混乱，绝对不是整洁和不直观的）：

Book.find(book_ids).tap { |book_1, book_2| book_1.sequence, book_2.sequence = book_2.sequence, book_1.sequence }.map &:save

#tap 产生 self in块，它映射到book_1和book_2，交换序列值并返回 books 的顺序相同，然后save保存它。