我将以下代码用于根据用户确定的道路开立档案,但正出现错误。 任何建议?
f = raw_input("
Hello, user. "
"
Please type in the path to your file and press Enter : ")
file = open( f , r )
它说,尽管我正在界定,但还是没有确定或不存在这样的东西。 查阅文件。
我将以下代码用于根据用户确定的道路开立档案,但正出现错误。 任何建议?
f = raw_input("
Hello, user. "
"
Please type in the path to your file and press Enter : ")
file = open( f , r )
它说,尽管我正在界定,但还是没有确定或不存在这样的东西。 查阅文件。
页: 1
myfile = open(f, r )
<代码> f/code>系指由字母 f组成的示意图,因此,您的代码正在寻找一个称为<代码>f的文档,而不是找到。 而是使用<代码>f,这意味着变量 f的价值。
另外,没有把变数称作存储你的文档file
。 这样做容易,但试图避免。 已经有了一个叫做file
的现成班级,其最佳做法是,不以你自己的姓名隐藏任何已建班或职能。 这是因为你看到的其他代码将预期<代码>>>>>><>代码>代表档案类别而不是您的变量。
采用<代码>求助<>/代码>功能的一种方法:
>>> help(file)
Help on class file in module __builtin__:
class file(object)
| file(name[, mode[, buffering]]) -> file object
|
| Open a file. The mode can be r , w or a for reading (default),
| writing or appending. The file will be created if it doesn t exist
而由于登革热在斯图尔德建议,在登上此刻的法典时,你完全有理由 in。
你试图打开这块地块。 为此:
file = open(f, r )
Don t putf
in quotes. <代码>f是一种可变数,可加以扼制,但在你开放时,你正在使用直观价值 f。
file = open(f, r )
公开(姓名、方式)归还档案标,最常见的是用两种论点:公开(档案名称、方式)。
>>> f = open(/tmp/workfile , w >/code>
http://docs.python.org/tutorial/inputoutput.html http://www.un.org。
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