如何上传超过2147483647字节的字符串块？

问题内容：

我正在尝试上传一个大小约为5GB的文件，如下所示，但是会引发错误string longer than 2147483647 bytes。听起来上传的文件限制为2 GB。有没有一种方法可以分块上传数据？谁能提供指导？

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)

问题答案：

在requestsbug跟踪器上询问了您的问题;
他们的建议是使用流式上传。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,
    )

我不能保证它会按原样运行，因为我无法实际测试它，但是它应该很接近。我链接到的错误跟踪器注释还提到，发送多个标头可能会导致问题，因此，如果实际上指定的标头是必需的，则此方法可能无效。

关于块编码：这应该是您的第二选择。您的代码未指定'rb'为的模式open(...)，因此更改该模式可能应使上述代码正常工作。如果没有，您可以尝试一下。

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,
)

如何上传超过2147483647字节的字符串块？

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