相关疑难解决方法(0)

这就是我所拥有的:

glob(os.path.join('src','*.c'))

但我想搜索src的子文件夹.像这样的东西会起作用:

glob(os.path.join('src','*.c'))
glob(os.path.join('src','*','*.c'))
glob(os.path.join('src','*','*','*.c'))
glob(os.path.join('src','*','*','*','*.c'))

但这显然是有限和笨重的.

如何让这篇文章在python 2.6中遵循符号链接？

def load_recursive(self, path):
    for subdir, dirs, files in os.walk(path):
        for file in files:
            if file.endswith('.xml'):
                file_path = os.path.join(subdir, file)
                try:
                    do_stuff(file_path) 
                except:
                    continue