路径无效,不允许使用反斜杠

时间:2013-12-21 07:16:37

标签: python dropbox dropbox-api

我正在尝试实现一个从用户的Dropbox帐户下载文件的Dropbox应用程序。在用户的本地目录中创建目标路径时,它会崩溃说

  

发生错误[400] {u'path:u''invalid path / New folder \\ img1.jpg:不允许索引11反斜杠的字符}

我认为dropbox的文件夹层次结构使用正斜杠来表示dorectories的嵌套,而windows使用反斜杠,因此它们可能存在冲突。然后我使用python的BIF replace(),如下所示,用于不同的路径

  

sample_path.replace(“\\”,“/”)

但仍然

  

complete_path

我的代码中的

变量给出包含反斜杠的路径,之后程序崩溃。 我的保管箱帐户中的文件夹层次结构为:

New Folder :
            Img1.jpg
dtu.jpg
img.jpg

代码是:

def download_file(self,source_path,target_path):
    print 'Downloading %s' % source_path
    file_path = os.path.expanduser(target_path)
    (dir_path,tail) = os.path.split(target_path)
    self.check_dir(dir_path)
    to_file = open(file_path,"wb")
    print source_path+"!!!!!!!!!!!!!!!!!!!!!!!!!!"
    source_path.replace("\\","/")        
    f= self.mClient.get_file(source_path) #  request to server !
    to_file.write(f.read())
    return
def download_folder(self, folderPath):


    # try to download 5 times to handle http 5xx errors from dropbox

    try:
        response = self.mClient.metadata(folderPath)
            # also ensure that response includes content
        if 'contents' in response:
            for f in response['contents']:
                name = os.path.basename(f['path'])
                complete_path = os.path.join(folderPath, name)
                if f['is_dir']:            
                    # do recursion to also download this folder
                    self.download_folder(complete_path)
                else:
                    # download the file
                    self.download_file(complete_path, os.path.join(self._target_folder, complete_path))
        else:
            raise ValueError
    except (rest.ErrorResponse, rest.RESTSocketError, ValueError) as error:
            print 'An error occured while listing a directory. Will try again in some seconds.'
            print "Error occured "+ str(error)

1 个答案:

答案 0 :(得分:3)

在Python控制台中尝试使用此功能来查看问题:

>>> x = "hello"
>>> x.replace("hello", "goodbye")
'goodbye'
>>> x
'hello'

在字符串上调用replace实际上并不会修改字符串。它返回一个带有替换的新字符串。所以你可能想要这样做:

source_path = source_path.replace("\\", "/")