使用 boto 将公共 URL 上可用的图像上传到 S3

Question

我在 Python Web 环境中工作，我可以使用 boto 的 key.set_contents_from_filename(path/to/file) 简单地将文件从文件系统上传到 S3。不过，我想上传一张已经在网络上的图片（比如https://pbs.twimg.com/media/A9h_htACIAAaCf6.jpg:large）。

我应该以某种方式将图像下载到文件系统，然后像往常一样使用 boto 将其上传到 S3，然后删除图像吗？

理想的情况是，如果有一种方法可以获取 boto 的 key.set_contents_from_file 或其他可以接受 URL 并将图像很好地流式传输到 S3 的命令，而无需将文件副本显式下载到我的服务器。

def upload(url):
    try:
        conn = boto.connect_s3(settings.AWS_ACCESS_KEY_ID, settings.AWS_SECRET_ACCESS_KEY)
        bucket_name = settings.AWS_STORAGE_BUCKET_NAME
        bucket = conn.get_bucket(bucket_name)
        k = Key(bucket)
        k.key = "test"
        k.set_contents_from_file(url)
        k.make_public()
                return "Success?"
    except Exception, e:
            return e

使用 set_contents_from_file，如上所述，我收到“字符串对象没有属性 'tell'”错误。将 set_contents_from_filename 与 url 一起使用，我得到一个 No such file or directory 错误。 boto storage documentation 没有提到上传本地文件，也没有提到上传远程存储的文件。

Answer 1

这是我使用requests 的方法，关键是在最初发出请求时设置stream=True，然后使用upload.fileobj() 方法上传到s3：

import requests
import boto3

url = "https://upload.wikimedia.org/wikipedia/en/a/a9/Example.jpg"
r = requests.get(url, stream=True)

session = boto3.Session()
s3 = session.resource('s3')

bucket_name = 'your-bucket-name'
key = 'your-key-name' # key is the name of file on your bucket

bucket = s3.Bucket(bucket_name)
bucket.upload_fileobj(r.raw, key)

Answer 2

好的，来自@garnaat，听起来 S3 目前不允许通过 url 上传。我设法通过仅将远程图像读入内存来将它们上传到 S3。这行得通。

def upload(url):
    try:
        conn = boto.connect_s3(settings.AWS_ACCESS_KEY_ID, settings.AWS_SECRET_ACCESS_KEY)
        bucket_name = settings.AWS_STORAGE_BUCKET_NAME
        bucket = conn.get_bucket(bucket_name)
        k = Key(bucket)
        k.key = url.split('/')[::-1][0]    # In my situation, ids at the end are unique
        file_object = urllib2.urlopen(url)           # 'Like' a file object
        fp = StringIO.StringIO(file_object.read())   # Wrap object    
        k.set_contents_from_file(fp)
        return "Success"
    except Exception, e:
        return e

也感谢How can I create a GzipFile instance from the “file-like object” that urllib.urlopen() returns?

Answer 3

对于使用官方“boto3”包（而不是原始答案中的旧“boto”包）的这个问题的 2017 年相关答案：

Python 3.5

如果您使用的是干净的 Python 安装，请先 pip 安装两个软件包：

pip install boto3

pip install requests

import boto3
import requests

# Uses the creds in ~/.aws/credentials
s3 = boto3.resource('s3')
bucket_name_to_upload_image_to = 'photos'
s3_image_filename = 'test_s3_image.png'
internet_image_url = 'https://docs.python.org/3.7/_static/py.png'


# Do this as a quick and easy check to make sure your S3 access is OK
for bucket in s3.buckets.all():
    if bucket.name == bucket_name_to_upload_image_to:
        print('Good to go. Found the bucket to upload the image into.')
        good_to_go = True

if not good_to_go:
    print('Not seeing your s3 bucket, might want to double check permissions in IAM')

# Given an Internet-accessible URL, download the image and upload it to S3,
# without needing to persist the image to disk locally
req_for_image = requests.get(internet_image_url, stream=True)
file_object_from_req = req_for_image.raw
req_data = file_object_from_req.read()

# Do the actual upload to s3
s3.Bucket(bucket_name_to_upload_image_to).put_object(Key=s3_image_filename, Body=req_data)

Answer 4

不幸的是，真的没有办法做到这一点。至少目前不是。我们可以向 boto 添加一个方法，例如 set_contents_from_url，但该方法仍然需要将文件下载到本地计算机然后上传。它可能仍然是一种方便的方法，但它不会为您节省任何东西。

为了做您真正想做的事情，我们需要在 S3 服务本身上有一些功能，允许我们将 URL 传递给它，并将 URL 存储到我们的存储桶中。这听起来像是一个非常有用的功能。您可能希望将其发布到 S3 论坛。

Answer 5

一个简单的 3 行实现，适用于开箱即用的 lambda：

import boto3
import requests

s3_object = boto3.resource('s3').Object(bucket_name, object_key)

with requests.get(url, stream=True) as r:
    s3_object.put(Body=r.content)

.get 部分的来源直接来自requests documentation

Answer 6

使用 boto3 upload_fileobj 方法，您可以将文件流式传输到 S3 存储桶，而无需保存到磁盘。这是我的功能：

import boto3
import StringIO
import contextlib
import requests

def upload(url):
    # Get the service client
    s3 = boto3.client('s3')

    # Rember to se stream = True.
    with contextlib.closing(requests.get(url, stream=True, verify=False)) as response:
        # Set up file stream from response content.
        fp = StringIO.StringIO(response.content)
        # Upload data to S3
        s3.upload_fileobj(fp, 'my-bucket', 'my-dir/' + url.split('/')[-1])

Answer 7

我已尝试使用 boto3 进行以下操作，它对我有用：

import boto3;
import contextlib;
import requests;
from io import BytesIO;

s3 = boto3.resource('s3');
s3Client = boto3.client('s3')
for bucket in s3.buckets.all():
  print(bucket.name)


url = "@resource url";
with contextlib.closing(requests.get(url, stream=True, verify=False)) as response:
        # Set up file stream from response content.
        fp = BytesIO(response.content)
        # Upload data to S3
        s3Client.upload_fileobj(fp, 'aws-books', 'reviews_Electronics_5.json.gz')

Answer 8

目前看来，S3 不支持远程上传。您可以使用以下类将图像上传到 S3。这里的上传方法首先尝试下载图像并将其保存在内存中一段时间​​，直到它被上传。为了能够连接到 S3，您必须使用命令 pip install awscli 安装 AWS CLI，然后使用命令 aws configure 输入一些凭据：

import urllib3
import uuid
from pathlib import Path
from io import BytesIO
from errors import custom_exceptions as cex

BUCKET_NAME = "xxx.yyy.zzz"
POSTERS_BASE_PATH = "assets/wallcontent"
CLOUDFRONT_BASE_URL = "https://xxx.cloudfront.net/"


class S3(object):
    def __init__(self):
        self.client = boto3.client('s3')
        self.bucket_name = BUCKET_NAME
        self.posters_base_path = POSTERS_BASE_PATH

    def __download_image(self, url):
        manager = urllib3.PoolManager()
        try:
            res = manager.request('GET', url)
        except Exception:
            print("Could not download the image from URL: ", url)
            raise cex.ImageDownloadFailed
        return BytesIO(res.data)  # any file-like object that implements read()

    def upload_image(self, url):
        try:
            image_file = self.__download_image(url)
        except cex.ImageDownloadFailed:
            raise cex.ImageUploadFailed

        extension = Path(url).suffix
        id = uuid.uuid1().hex + extension
        final_path = self.posters_base_path + "/" + id
        try:
            self.client.upload_fileobj(image_file,
                                       self.bucket_name,
                                       final_path
                                       )
        except Exception:
            print("Image Upload Error for URL: ", url)
            raise cex.ImageUploadFailed

        return CLOUDFRONT_BASE_URL + id

Answer 9

from io import BytesIO
def send_image_to_s3(url, name):
    print("sending image")
    bucket_name = 'XXX'
    AWS_SECRET_ACCESS_KEY = "XXX"
    AWS_ACCESS_KEY_ID = "XXX"

    s3 = boto3.client('s3', aws_access_key_id=AWS_ACCESS_KEY_ID,
                      aws_secret_access_key=AWS_SECRET_ACCESS_KEY)

    response = requests.get(url)
    img = BytesIO(response.content)

    file_name = f'path/{name}'
    print('sending {}'.format(file_name))
    r = s3.upload_fileobj(img, bucket_name, file_name)

    s3_path = 'path/' + name
    return s3_path

Answer 10

import boto
from boto.s3.key import Key
from boto.s3.connection import OrdinaryCallingFormat
from urllib import urlopen


def upload_images_s3(img_url):
    try:
        connection = boto.connect_s3('access_key', 'secret_key', calling_format=OrdinaryCallingFormat())       
        bucket = connection.get_bucket('boto-demo-1519388451')
        file_obj = Key(bucket)
        file_obj.key = img_url.split('/')[::-1][0]
        fp = urlopen(img_url)
        result = file_obj.set_contents_from_string(fp.read())
    except Exception, e:
        return e