python regex dynamic string as input

Question

I am trying to get the string dynamically and use regex over it but i get the error:

coding partial snippet:

import re
def file_save():
    f = th1.asksaveasfile(mode='w', defaultextension=".txt")
    re.search(r"<[^>]*\bname='([^']*)", f).group(1)       
    f.close()

Error

    re.search(r"<[^>]*\bname='([^']*)", f).group(1)
  File "C:\Python34\lib\re.py", line 166, in search
    return _compile(pattern, flags).search(string)
TypeError: expected string or buffer

Here,the input string has to be attained from a 'filename to be saved' which produces the text 'f' as:

<_io.TextIOWrapper name='C:/Python34/abcd.txt' mode='w' encoding='cp1252'>

the regex works perfect,but icannot get the string and get the ouput as:

C:/Python34/abcd.txt

Please help me to fix my problem.Answers will be appreciated!

falsetru · Accepted Answer · 2014-08-07 14:07:35Z

1

I think you're using wrong tool.

If you want to get filename, just use asksaveasfilename instead of asksaveasfile.

def file_save():
    filename = th1.asksaveasfilename(defaultextension=".txt")
    # do something with the filename

Or, if you need to operate on the file, use asksaveasfile and use name attribute of the file returned.

def file_save():
    f = th1.asksaveasfile(mode='w', defaultextension=".txt")
    filename = f.name
    # do something with the filename
    f.close()

edited Aug 7, 2014 at 14:07

answered Aug 7, 2014 at 14:02

falsetru

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4 Comments

user3914506 Over a year ago

but,how do i get the input for the regex from filename saved?

falsetru Over a year ago

@user3914506, What do you mean? You don't need to use regular expression to get filename.

user3914506 Over a year ago

then how could I get my Filename stored and save in a variable say x?

falsetru Over a year ago

@user3914506, I updated the answer. Please check it out.

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