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How to define a string in python with double and single quotes

I am using python to communicate with the OS.

I need to create a string of the following form:

string = "done('1') && done('2')"

Note that my string MUST have the double quotes in it, but I am not sure how to do that since the double quotes are used in python for defining a string.

Then I do something like:

os.system(string)

But the system would only read the string with the double and single quotes in it.

I tried:

>>> s = '"done('1') && done('2')"'
  File "<stdin>", line 1
    s = '"done('1') && done('2')"'
                ^
SyntaxError: invalid syntax

I also tried the triple quotes suggested here but i get an error:

>>> s = """"done('1') && done('2')""""
  File "<stdin>", line 1
    s = """"done('1') && done('2')""""
                                     ^
SyntaxError: EOL while scanning string literal

How to store a string containing both single quote( ' ) and double quote( " ) in python

When you use a triply quoted string you need to remember that the string ends when Python finds a closing set of three quotes - and it is not greedy about it. So you can:

Change to wrapping in triple single quotes:

my_command = '''"done('1') && done('2')"'''

Escape the ending quote:

my_command = """"done('1') && done('2')\""""

or add space around your quotes and call strip on the resulting string:

my_command = """
"done('1') && done('2')"
""".strip()
# Blank lines are for illustrative purposes only
# You can do it all on one line as well (but then it looks like you have
# 4 quotes (which can be confusing)

你可以逃避两种报价:

s = '"done(\'1\') && done(\'2\')"'

All four flavors of quotes:

print('''"done('1') && done('2')"''')  # No escaping required here.
print(""""done('1') && done('2')\"""")
print("\"done('1') && done('2')\"")
print('"done(\'1\') && done(\'2\')"')

Output:

"done('1') && done('2')"
"done('1') && done('2')"
"done('1') && done('2')"
"done('1') && done('2')"

I think this is what you are expecting: string = "\\"done('1') && done('2')\\""

Ignore if this does not answer your question.

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