Question

I am trying to use “With open()” with python 2.6 and it is giving error(Syntax error) while it works fine with python 2.7.3
Am I missing something or some import to make my program work!

Any help would be appreciated.

Br

My code is here:

def compare_some_text_of_a_file(self, exportfileTransferFolder, exportfileCheckFilesFolder) :
    flag = 0
    error = ""
    with open("check_files/"+exportfileCheckFilesFolder+".txt") as f1,open("transfer-out/"+exportfileTransferFolder) as f2:

        if f1.read().strip() in f2.read():
            print ""
        else:
            flag = 1
            error = exportfileCheckFilesFolder
            error = "Data of file " + error + " do not match with exported data\n"
        if flag == 1:   
            raise AssertionError(error)

Answer 1

The with open() statement is supported in Python 2.6, you must have a different error.

See PEP 343 and the python File Objects documentation for the details.

Quick demo:

Python 2.6.8 (unknown, Apr 19 2012, 01:24:00) 
[GCC 4.2.1 (Based on Apple Inc. build 5658) (LLVM build 2335.15.00)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> with open('/tmp/test/a.txt') as f:
...     print f.readline()
... 
foo

>>> 

You are trying to use the with statement with multiple context managers though, which was only added in Python 2.7:

Changed in version 2.7: Support for multiple context expressions.

Use nested statements instead in 2.6:

with open("check_files/"+exportfileCheckFilesFolder+".txt") as f1:
    with open("transfer-out/"+exportfileTransferFolder) as f2:
        # f1 and f2 are now both open.