python - 如何在不创建临时本地文件的情况下将文件上传到 S3

Question

是否有任何可行的方法可以直接将动态生成的文件上传到亚马逊 s3，而无需先创建本地文件然后上传到 s3 服务器？我用蟒蛇。谢谢

score 26 · Accepted Answer

这是一个下载图像（使用请求库）并将其上传到 s3 的示例，而不写入本地文件：

import boto
from boto.s3.key import Key
import requests

#setup the bucket
c = boto.connect_s3(your_s3_key, your_s3_key_secret)
b = c.get_bucket(bucket, validate=False)

#download the file
url = "http://en.wikipedia.org/static/images/project-logos/enwiki.png"
r = requests.get(url)
if r.status_code == 200:
    #upload the file
    k = Key(b)
    k.key = "image1.png"
    k.content_type = r.headers['content-type']
    k.set_contents_from_string(r.content)

score 17 · Accepted Answer

您可以使用Python 标准库中的BytesIO。

from io import BytesIO
bytesIO = BytesIO()
bytesIO.write('whee')
bytesIO.seek(0)
s3_file.set_contents_from_file(bytesIO)

score 13 · Accepted Answer

boto库的Key对象有几个您可能感兴趣的方法：

有关使用 set_contents_from_string 的示例，请参阅boto 文档的存储数据部分，为了完整起见，粘贴在此处：

>>> from boto.s3.key import Key
>>> k = Key(bucket)
>>> k.key = 'foobar'
>>> k.set_contents_from_string('This is a test of S3')

score 2 · Accepted Answer

我假设你正在使用boto. boto'sBucket.set_contents_from_file()将接受一个StringIO对象，并且您为将数据写入文件而编写的任何代码都应该很容易适应写入StringIO对象。或者，如果您生成一个字符串，您可以使用set_contents_from_string().

score 2 · Accepted Answer

def upload_to_s3(url, **kwargs):
    '''
    :param url: url of image which have to upload or resize to upload
    :return: url of image stored on aws s3 bucket
    '''

    r = requests.get(url)
    if r.status_code == 200:
        # credentials stored in settings AWS_ACCESS_KEY_ID and AWS_SECRET_ACCESS_KEY
        conn = boto.connect_s3(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY, host=AWS_HOST)

        # Connect to bucket and create key
        b = conn.get_bucket(AWS_Bucket_Name)
        k = b.new_key("{folder_name}/{filename}".format(**kwargs))

        k.set_contents_from_string(r.content, replace=True,
                                   headers={'Content-Type': 'application/%s' % (FILE_FORMAT)},
                                   policy='authenticated-read',
                                   reduced_redundancy=True)

        # TODO Change AWS_EXPIRY
        return k.generate_url(expires_in=AWS_EXPIRY, force_http=True)

score 1 · Accepted Answer

在boto3中，有一种上传文件内容的简单方法，无需使用以下代码创建本地文件。我已经修改了 boto3 的 JimJty 示例代码

import boto3
from botocore.retries import bucket
import requests
from io import BytesIO
# set the values
aws_access_key_id=""
aws_secret_access_key=""
region_name=""
bucket=""
key=""

session = boto3.session.Session(aws_access_key_id=aws_access_key_id,aws_secret_access_key=aws_secret_access_key, region_name=region_name)
s3_client = session.client('s3')
#download the file
url = "http://en.wikipedia.org/static/images/project-logos/enwiki.png"
r = requests.get(url)
if r.status_code == 200:    
    #convert content to bytes, since upload_fileobj requires file like obj
    bytesIO = BytesIO(bytes(r.content))    
    with bytesIO as data:
        s3_client.upload_fileobj(data, bucket, key)

score 1 · Accepted Answer

我有一个 dict 对象，我想将其存储为 S3 上的 json 文件，而不创建本地文件。下面的代码对我有用：

from smart_open import smart_open

with smart_open('s3://access-key:secret-key@bucket-name/file.json', 'wb') as fout:
    fout.write(json.dumps(dict_object).encode('utf8'))

score 0 · Accepted Answer

您可以尝试使用smart_open（https://pypi.org/project/smart_open/）。我正是为此使用它：直接在 S3 中写入文件。

score 0 · Accepted Answer

0

鉴于静态加密现在是一个非常需要的数据标准，smart_open 不支持这个 afaik

于 2019-05-14T08:55:59.260 回答

score 0 · Accepted Answer

此实现是将图像列表（NumPy 列表、OpenCV 图像对象）直接上传到 S3 的示例

注意：您需要在上传文件时将图像对象转换为字节或缓冲区转换为字节，这样您就可以上传文件而不会出现损坏错误

#Consider you have images in the form of a list i.e. img_array
import boto3

s3 = boto3.client('s3')
res_url = []

for i,img in enumerate(img_array):
        s3_key = "fileName_on_s3.png"
        response = s3.put_object(Body=img.tobytes(), Bucket='bucket_name',Key=s3_key,ACL='public-read',ContentType= 'image/png')
        s3_url = 'https://bucket_name.s3.ap-south-1.amazonaws.com/'+s3_key
        res_url.append(s3_url)
#res_url is the list of URLs returned from S3 Upload

score -1 · Accepted Answer

boto3的更新：

aws_session = boto3.Session('my_access_key_id', 'my_secret_access_key')
s3 = aws_session.resource('s3')
s3.Bucket('my_bucket').put_object(Key='file_name.txt', Body=my_file)

score -2 · Accepted Answer

我遇到了类似的问题，想知道是否有最终答案，因为使用下面的代码，“starwars.json”会继续在本地保存，但我只想将每个循环的 .json 文件推送到 S3 中并且没有文件存储在本地。

for key, value in star_wars_actors.items():

response = requests.get('http:starwarsapi/' + value)



data = response.json()


with open("starwars.json", "w+") as d:
    json.dump(data, d, ensure_ascii=False, indent=4)



s3.upload_file('starwars.json', 'test-bucket',
               '%s/%s' % ('test', str(key) + '.json'))

python - 如何在不创建临时本地文件的情况下将文件上传到 S3

12 回答 12

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