Does, With open() not works with python 2.6

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 m

Answer 1

It is the "extended" with statement with multiple context expressions which causes your trouble.

In 2.6, instead of

with open(...) as f1, open(...) as f2:
    do_stuff()

you should add a nesting level and write

with open(...) as f1:
    with open(...) as f2:
        do.stuff()

The docu says

Changed in version 2.7: Support for multiple context expressions.

Answer 2

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.

Answer 3

The with open() syntax is supported by Python 2.6. On Python 2.4 it is not supported and gives a syntax error. If you need to support PYthon 2.4, I would suggest something like:

def readfile(filename, mode='r'):
    f = open(filename, mode)
    try:
        for line in f:
            yield f
    except e:
        f.close()
        raise e
    f.close()

for line in readfile(myfile):
    print line