python之TypeError : must be convertible to a buffer, 不是 HTTPResponse

我有一个从 api 连接获取 pdf 文件的函数。

我的代码:

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

获取标签:

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)` 

运行上面的代码会抛出错误:

<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 字节文件。 什么是正确的方法。

请您参考如下方法:

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())