是,使用open()不适用于python 2.6

Question

我正在尝试使用带有python 2.6的"with open()"并且它正在给出错误(语法错误),而它与python 2.7.3工作正常我错过了一些东西或一些导入使我的程序工作!

任何帮助,将不胜感激.

BR

我的代码在这里:

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

with open()Python 2.6支持该语句,您必须有不同的错误.

有关详细信息,请参阅PEP 343和python 文件对象文档.

快速演示:

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

>>> 

您正尝试将该with语句与多个上下文管理器一起使用,但这只是在Python 2.7中添加的:

在2.7版中更改:支持多个上下文表达式.

在2.6中使用嵌套语句:

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

Answer 2

它是with带有多个上下文表达式的"扩展" 语句,会导致您遇到麻烦.

在2.6中,而不是

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

你应该添加一个嵌套级别并写入

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

文件说

在2.7版中更改:支持多个上下文表达式.