如何使用Django模型字段定义保持DRY

Question

在定义Django模型字段时观察DRY原则的最佳实践是什么.

场景1:

file_one = models.FilePathField(path=FIELD_PATH, allow_files=True, allow_folders=True, recursive=True)
file_two = models.FilePathField()
file_three = models.FilePathField()

我可以这样做:

file_one = models.FilePathField(path=FIELD_PATH, allow_files=True, allow_folders=True, recursive=True)
file_two = file_one
...

场景2:

base = models.FilePathField(allow_files=True, allow_folders=True, recursive=True)
file_one = models.FilePathField(path=FIELD_PATH1)
file_two = models.FilePathField(path=FIELD_PATH2)
file_three = models.FilePathField(path=FIELD_PATH3)

如何让file_one,_two和_three继承/扩展规则,base = models...同时能够为每个规则分配不同的path=...

我觉得像Django:动态模型字段定义很接近,但不是我想要的!

保持棒极了Stack Overflow!

Answer 1

老实说,DRY代码很重要,应该努力,但有限制:)在这种情况下,你在DRY和python的第二行之间存在分歧Explicit is better than implicit.如果我维护你的代码,我宁愿进来看看:

file_one = models.FilePathField(path=FIELD_PATH1, allow_files=True, allow_folders=True, recursive=True)
file_two = models.FilePathField(path=FIELD_PATH2, allow_files=True, allow_folders=True, recursive=True)
file_three = models.FilePathField(path=FIELD_PATH3, allow_files=True, allow_folders=True, recursive=True)

因为虽然不是"干",但是很明显发生了什么,我不必浪费时间去"等,什么？"

(实际上严格来说,我希望看到的是:

# Useful comments
file_one = models.FilePathField(
    path=FIELD_PATH1,
    allow_files=True,
    allow_folders=True,
    recursive=True
)

# Useful comments
file_two = models.FilePathField(
    path=FIELD_PATH2,
    allow_files=True,
    allow_folders=True,
    recursive=True
)

..但那是因为我是PEP8的坚持者!):)