在python中打开模拟文件 [英] mock file open in python

问题描述

我正在尝试模拟打开文件，所有示例都表明我需要

I'm trying to mock file open, and all of the examples show that I need to

@patch('open', create=True) 

但我不断得到

Need a valid target to patch. You supplied: 'open'

我知道补丁需要 open 的全点划线路径，但我不知道它是什么.事实上，我什至不确定那是问题所在.

I know patch needs the full dotted path of open, but I have no idea what it is. As a matter of fact, I'm not even sure that's the problem.

推荐答案

您需要添加模块名称；如果在脚本中进行测试，则模块的名称为__main__:

You need to include a module name; if you are testing in a script, the name of the module is __main__:

@patch('__main__.open')

否则，请使用包含要测试的代码的模块的名称:

otherwise use the name of the module that contains the code you are testing:

@patch('module_under_test.open')

，以便任何使用内置open()的代码都将找到修补的全局代码.

so that any code that uses the open() built-in will find the patched global instead.

请注意，mock模块随附 实用程序，可让您使用文件数据构建合适的open()调用:

Note that the mock module comes with a mock_open() utility that'll let you build a suitable open() call with file data:

@patch('__main__.open', mock_open(read_data='foo\nbar\nbaz\n'))

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