以八进制打印文字字符串python

Question

Hi I am having trouble trying to print a literal string in a proper format. For starters i have an object with a string parameter which is used for metadata such that it looks like:

obj {
    metadata: <str>
}

The object is being returned as a protocol response and we have the object to use as such.

print obj gives:

    metadata: "\n)\n\022foobar"

When I print the obj.metadata python treats the value as a string and converts the escapes to linebreaks and the corresponding ascii values as expected.

When i tried print repr(obj.metadata)

"\n)\n\x12foobar"

Unfortunately python prints the literal but converts the escaped characters from octal to hex. Is there a way i can print the python literal with the escaped characters in octal or convert the string such that I can have the values printed as it is in the object?

Thanks for the help

The extremely bad solution I have so far is

print str(obj).rstrip('\"').lstrip('metadata: \"')

to get the correct answer, but i am assuming there must be a smarter way

TLDR:

x = "\n)\n\022foobar"
print x

)
foobar

print repr(x)
'\n)\n\x12foobar'

how do i get x to print the way it was assigned

Answer 1

Please try this:

print('\\\n)\\\n\\\022foobar') 

or

print(r'\n)\n\022foobar')

The escape character '\\' interprets the character following it differently, for example \\n is used for new line.

The double escape character '\\\\' or letter 'r' nullifies the interpretation of the escape character. This is similar in C language.