如何上传长度超过2147483647字节的字符串？

Question

我试图上传约5GB大小的文件,如下所示,但它会抛出错误string longer than 2147483647 bytes.听起来上传的限制为2 GB.有没有办法以块的形式上传数据？有人可以提供指导吗？

logger.debug(attachment_path)
currdir = os.path.abspath(os.getcwd())
os.chdir(os.path.dirname(attachment_path))
headers = self._headers
headers['Content-Type'] = content_type
headers['X-Override-File'] = 'true'
if not os.path.exists(attachment_path):
    raise Exception, "File path was invalid, no file found at the path %s" % attachment_path
filesize = os.path.getsize(attachment_path) 
fileToUpload = open(attachment_path, 'rb').read()
logger.info(filesize)
logger.debug(headers)
r = requests.put(self._baseurl + 'problems/' + problemID + "/" + attachment_type + "/" + urllib.quote(os.path.basename(attachment_path)), 
                 headers=headers,data=fileToUpload,timeout=300)

错误:

string longer than 2147483647 bytes

更新:

def read_in_chunks(file_object,chunk_size=30720*30720):
    """Lazy function (generator) to read a file piece by piece.
    Default chunk size: 1k."""
    while True:
        data = file_object.read(chunk_size)
        if not data:
            break
        yield data
        f = open(attachment_path)

for piece in read_in_chunks(f):
      r = requests.put(self._baseurl + 'problems/' + problemID + "/" + attachment_type + "/" + urllib.quote(os.path.basename(attachment_path)), 
                        headers=headers,data=piece,timeout=300)

Answer 1

您的问题已在requestsbug跟踪器上询问; 他们的建议是使用流媒体上传.如果这不起作用,您可能会看到块编码的请求是否有效.

[编辑]

基于原始代码的示例:

# Using `with` here will handle closing the file implicitly
with open(attachment_path, 'rb') as file_to_upload:
    r = requests.put(
        "{base}problems/{pid}/{atype}/{path}".format(
            base=self._baseurl,
            # It's better to use consistent naming; search PEP-8 for standard Python conventions.
            pid=problem_id,
            atype=attachment_type,
            path=urllib.quote(os.path.basename(attachment_path)),
        ),
        headers=headers,
        # Note that you're passing the file object, NOT the contents of the file:
        data=file_to_upload,
        # Hard to say whether this is a good idea with a large file upload
        timeout=300,
    )

我不能保证这会按原样运行,因为我无法实际测试它,但它应该是接近的.我链接的错误跟踪器注释也提到发送多个标头可能会导致问题,因此如果您指定的标头实际上是必需的,这可能无效.

关于块编码:这应该是你的第二选择.您的代码未指定'rb'为模式open(...),因此更改该代码可能会使上面的代码工作.如果没有,你可以试试这个.

def read_in_chunks():
    # If you're going to chunk anyway, doesn't it seem like smaller ones than this would be a good idea?
    chunk_size = 30720 * 30720

    # I don't know how correct this is; if it doesn't work as expected, you'll need to debug
    with open(attachment_path, 'rb') as file_object:
        while True:
            data = file_object.read(chunk_size)
            if not data:
                break
            yield data


# Same request as above, just using the function to chunk explicitly; see the `data` param
r = requests.put(
    "{base}problems/{pid}/{atype}/{path}".format(
        base=self._baseurl,
        pid=problem_id,
        atype=attachment_type,
        path=urllib.quote(os.path.basename(attachment_path)),
    ),
    headers=headers,
    # Call the chunk function here and the request will be chunked as you specify
    data=read_in_chunks(),
    timeout=300,
)