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  • Is there a way to accept '%' as part of input that works both in python 2.6 & 3.0?

    - by bug11
    In 2.6, if I needed to accept input that allowed a percent sign (such as "foo % bar"), I used raw_input() which worked as expected. In 3.0, input() accomplishes that same (with raw_input() having left the building). As an exercise, I'm hoping that I can have a backward-compatible version that will work with both 2.6 and 3.0. When I use input() in 2.6 and enter "foo % bar", the following error is returned: File "<string>", line 1, in <module> NameError: name "foo" is not defined ...which is expected. Anyway to to accomplish acceptance of input containing a percent sign that works in both 2.6 and 3.0? Thx.

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