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Python:从RestAPI下载zip文件

python-requests flask python

Murtuza Z · 技术社区 · 6 年前

第三方RestAPI提供服务器日志文件,目前我们正在使用curl命令下载日志文件,比如

curl -H "X-Auth-Token: XXXXXXXXXXXXXXXXXXXXXXXXXX" https://host_address/api/v3.0/admin/logs -o logs.zip

但我正在尝试使用Flask/Python创建简单的仪表板,

下面是我的Python/Flask代码:

@app.route('/download/server/logs')
def download_log():
    import requests
    from flask import send_file
    res = requests.get('http://<rest_api_host>/v1.2/admin/logs', stream=True)
    return send_file(
        res.content,
        attachment_filename='console_log.zip',
        mimetype='application/zip'
    )

但是当我从浏览器中点击那个网址时,我得到了下面的错误信息,

Traceback (most recent call last):
...
...
...
  File "/Users/admin/Documents/project/__init__.py", line 940, in download_console_logs
    res.content,
  File "/Users/admin/Documents/project/venv3/lib/python3.6/site-packages/requests/models.py", line 823, in content
    self._content = bytes().join(self.iter_content(CONTENT_CHUNK_SIZE)) or bytes()
  File "/Users/admin/Documents/project/venv3/lib/python3.6/site-packages/requests/models.py", line 745, in generate
    for chunk in self.raw.stream(chunk_size, decode_content=True):
  File "/Users/admin/Documents/project/venv3/lib/python3.6/site-packages/urllib3/response.py", line 436, in stream
    data = self.read(amt=amt, decode_content=decode_content)
  File "/Users/admin/Documents/project/venv3/lib/python3.6/site-packages/urllib3/response.py", line 384, in read
    data = self._fp.read(amt)
  File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/http/client.py", line 449, in read
    n = self.readinto(b)
  File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/http/client.py", line 497, in readinto
    self._close_conn()
  File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/http/client.py", line 403, in _close_conn
    fp.close()
AttributeError: 'NoneType' object has no attribute 'close'

我使用PyCharm设置了断点/调试器,可以看到这一点 res.content

这是一个简单的图表,它解释了我要做什么,

我采取了下面的方法,它很有效地解决了我的问题。

@app.route('/download/server/logs')
def download_log():
    import requests
    from flask import Reponse
    res = requests.get('http://<rest_api_host>/v1.2/admin/logs', stream=True)
    return Response(
        res.iter_content(chunk_size=1024),
        direct_passthrough=True
    )

2 回复 | 直到 6 年前

Roberto Caboni Prince Rabadiya 4 年前

try:
    srcFileName="filename"
    with open(srcFileName, 'wb') as f:
        f.write(srcFileName)
except Exception as e:
    print(e)

Aakash Handa 4 年前

从本地文件夹下载

def download_Object():
    try:
        srcFileName="bulk_file.zip"
        with open(srcFileName, 'wb') as f:
            f.write(srcFileName)
    except Exception as e:
        print(e)

Nimeshka Srimal 6 年前

我不确定你是否给出了正确的错误信息。但是我认为如果你把一个字符串传递给 send_file()

如果你包起来 res.content 里面 io.BytesIO() 发送文件() 方法,我相信它应该有效。

@app.route('/download/server/logs')
def download_log():
    import requests
    from flask import send_file
    import io
    res = requests.get('http://<rest_api_host>/v1.2/admin/logs', stream=True)
    return send_file(
        io.BytesIO(res.content),
        attachment_filename='console_log.zip',
        mimetype='application/zip'
    )

SUBHAM MAJAVADIYA 6 年前

如果要下载文件,请尝试烧瓶的响应方法:

import requests
from flask import Response
@app.route('/download/server/logs')
def download_log():
res = requests.get('http://<rest_api_host>/v1.2/admin/logs')
return Response(res.content,headers = dict(res.headers))