Question

使用＆＃34; *＆＃34;

检查文件夹路径

这是我尝试过的。

microsecond

假

这将始终打印False，还有其他方式可以将其打印为true吗？我知道当我把静态路径，它可以

import os

    checkdir = "/lib/modules/*/kernel/drivers/char"
    path = os.path.exists(checkIPMI_dir)
    print path

我不能放静态，因为我有一些linux操作系统，它使用不同的内核版本，那里的号码将不一样。

Answer 1

这可能无法解决您的问题，但前段时间我需要能够通过目录结构递归未知次数来查找文件，所以我写了这个函数：

import os
import fnmatch
def get_files(root, patterns='*', depth=1, yield_folders=False):
    """
    Return all of the files matching patterns, in pattern.
    Only search to teh specified depth
    Arguments:
        root - Top level directory for search
        patterns - comma separated list of Unix style
                   wildcards to match NOTE: This is not
                   a regular expression.
        depth - level of subdirectory search, default
                1 level down
        yield_folders - True folder names should be returned
                        default False
    Returns:
        generator of files meeting search criteria for all
        patterns in argument patterns
    """
    # Determine the depth of the root directory
    root_depth = len(root.split(os.sep))
    # Figure out what patterns we are matching
    patterns = patterns.split(';')

    for path, subdirs, files in os.walk(root):
        # Are we including directories in search?
        if yield_folders:
            files.extend(subdirs)
        files.sort()

        for name in files:
            for pattern in patterns:
                # Determine if we've exceeded the depth of the
                # search?
                cur_depth = len(path.split(os.sep))
                if (cur_depth - root_depth) > depth:
                    break

                if fnmatch.fnmatch(name, pattern):
                    yield os.path.join(path, name)
                    break

使用此功能，您可以检查文件是否存在以下内容：

checkdir = "/lib/modules/*/kernel/drivers/char"
matches = get_files(checkdir, depth=100, yield_folders=True)
found = True if matches else False

这一切可能都有点过头了，但它应该有效！

python检查文件夹路径*

1 个答案: