Question

我正在将一个巨大的py模块重构为包 - 为了不破坏现有代码我将其内容移动到package/__init__.py模块（Adding code to __init__.py）并继续从那里拆分。我注意到，在我的追溯中，我得到了：

Traceback (most recent call last):
      File "<string>", line 656, in DoItemMenu
      File "bash\balt.py", line 2109, in PopupMenu
        link.AppendToMenu(menu,parent,*args)
      File "bash\balt.py", line 2225, in AppendToMenu
        for link in self.links: link.AppendToMenu(subMenu,window,data)
    ...

其中File "<string>"中的行对应于特定的package/__init__.py模块。此外，PyCharm的调试器显示“框架不可用”行，并且不会进入__init__.py中的行。为什么？它与导入模式有关吗？

代码由launcher class导入：

class UnicodeImporter(object):
    def find_module(self,fullname,path=None):
        if isinstance(fullname,unicode):
            fullname = fullname.replace(u'.',u'\\')
            exts = (u'.pyc',u'.pyo',u'.py')
        else:
            fullname = fullname.replace('.','\\')
            exts = ('.pyc','.pyo','.py')
        if os.path.exists(fullname) and os.path.isdir(fullname):
            return self
        for ext in exts:
            if os.path.exists(fullname+ext):
                return self

    def load_module(self,fullname):
        if fullname in sys.modules:
            return sys.modules[fullname]
        else:
            sys.modules[fullname] = imp.new_module(fullname)
        if isinstance(fullname,unicode):
            filename = fullname.replace(u'.',u'\\')
            ext = u'.py'
            initfile = u'__init__'
        else:
            filename = fullname.replace('.','\\')
            ext = '.py'
            initfile = '__init__'
        if os.path.exists(filename+ext):
            try:
                with open(filename+ext,'U') as fp:
                    mod = imp.load_source(fullname,filename+ext,fp)
                    sys.modules[fullname] = mod
                    mod.__loader__ = self
                    return mod
            except:
                print 'fail', filename+ext
                raise
        mod = sys.modules[fullname]
        mod.__loader__ = self
        mod.__file__ = os.path.join(os.getcwd(),filename)
        mod.__path__ = [filename]
        #init file
        initfile = os.path.join(filename,initfile+ext)
        if os.path.exists(initfile):
            with open(initfile,'U') as fp:
                code = fp.read()
            exec code in mod.__dict__
        return mod

Answer 1

代码不是以传统方式导入的;相反，launcher code使用exec statement来加载__init__.py个文件。

削减包的启动器load_module()函数中的流程（因此不是模块的路径），你得到这个：

# the fullname module isn't yet loaded
sys.modules[fullname] = imp.new_module(fullname)
initfile = '__init__'  # or u'__init__' if a unicode path was used

# if no .py file was found, so not a module
mod = sys.modules[fullname]
mod.__loader__ = self
mod.__file__ = os.path.join(os.getcwd(),filename)
mod.__path__ = [filename]
#init file
initfile = os.path.join(filename,initfile+ext)
if os.path.exists(initfile):
    with open(initfile,'U') as fp:
        code = fp.read()
    exec code in mod.__dict__
return mod

这将创建一个空模块对象，手动加载源并将其作为字符串执行，将模块命名空间作为执行代码的全局变量传递。生成的代码对象始终会在回溯中列出<string>：

>>> import imp
>>> mod = imp.new_module('foo.bar')
>>> mod.__file__ = r'C:\some\location\foo\bar'
>>> mod.__path__ = [r'foo\bar']
>>> exec 'raise ValueError("oops")' in mod.__dict__
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<string>", line 1, in <module>
ValueError: oops

因为没有与代码关联的文件名，PyCharm也找不到原始源。

解决方法是使用compile() function创建代码对象首先，并将文件名附加到：

>>> exec compile('raise ValueError("oops")', r'C:\some\location\foo\bar\__init__.py', 'exec') in mod.__dict__
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\some\location\foo\bar\__init__.py", line 1, in <module>
ValueError: oops

请注意，我在文件名中包含了__init__.py;将其翻译回您使用的启动器：

if os.path.exists(initfile):
    with open(initfile,'U') as fp:
        code = fp.read()
    exec compile(code, initfile, 'exec') in mod.__dict__

文件＆＃34; <string>＆＃34;在python traceback中</string>

1 个答案: