TypeError: must be convertible to a buffer, not HTTPResponse

Question

我有一个从 api 连接获取 pdf 文件的功能。

我的代码：

label = fk.fetch_labels(oiids)
with open('a.pdf', 'wb') as handle:
    cont = label.raw
    print cont
    handle.write(cont)`

fetch_labels :

def fetch_labels(self, orderItemIds):
    headers = {'Accept': 'application/octet-stream'}
    url = "https://api.flipkart.net/sellers/orders/shipments"
    payload = {'orderItemIds':','.join(orderItemIds)}
    return self.session.get(url, headers=headers, params=payload, stream=True)`

在运行上面的代码中抛出错误：

<urllib3.response.HTTPResponse object at 0x7f1d8fa24d50>
Traceback (most recent call last):
  File "test.py", line 23, in <module>
    handle.write(cont)

TypeError：必须可转换为缓冲区，而不是 HTTPResponse `

当我使用 'wb' 将其写入 pdf 文件时，它只会创建一个 0 字节的文件。正确的方法是什么。

Answer 1

来自the documentation on streaming requests：

With requests.Response.iter_lines() you can easily iterate over streaming APIs such as the Twitter Streaming API. Simply set stream to True and iterate over the response with iter_lines()

然后：

iter_lines(chunk_size=512, decode_unicode=None, delimiter=None)
Iterates over the response data, one line at a time. When stream=True is set on the request, this avoids reading the content at once into memory for large responses.

Note This method is not reentrant safe.

因此您的代码可能会像这样使用它：

with open('a.pdf', 'wb') as handle:
    handle.writelines(label.iter_lines())

TypeError: must be convertible to a buffer, not HTTPResponse

TypeError: must be convertible to a buffer, not HTTPResponse

python

python-requests