我想知道boto3中是否存在一个键。我可以循环桶内容并检查键是否匹配。

但这似乎太长了,也太过分了。Boto3官方文档明确说明了如何做到这一点。

也许我忽略了最明显的一点。有人能告诉我怎么做吗?


当前回答

如果你在一个目录或桶中有少于1000个,你可以获得它们的集合,然后检查这个集合中是否有这样的键:

files_in_dir = {d['Key'].split('/')[-1] for d in s3_client.list_objects_v2(
Bucket='mybucket',
Prefix='my/dir').get('Contents') or []}

即使my/dir不存在,这样的代码也可以工作。

http://boto3.readthedocs.io/en/latest/reference/services/s3.html#S3.Client.list_objects_v2

其他回答

如果您寻找一个与目录等效的键,那么您可能需要这种方法

session = boto3.session.Session()
resource = session.resource("s3")
bucket = resource.Bucket('mybucket')

key = 'dir-like-or-file-like-key'
objects = [o for o in bucket.objects.filter(Prefix=key).limit(1)]    
has_key = len(objects) > 0

这适用于父键或等同于文件的键或不存在的键。我尝试了上面喜欢的方法,但在父键上失败了。

S3_REGION="eu-central-1"
bucket="mybucket1"
name="objectname"

import boto3
from botocore.client import Config
client = boto3.client('s3',region_name=S3_REGION,config=Config(signature_version='s3v4'))
list = client.list_objects_v2(Bucket=bucket,Prefix=name)
for obj in list.get('Contents', []):
    if obj['Key'] == name: return True
return False

get()方法非常简单

import botocore
from boto3.session import Session
session = Session(aws_access_key_id='AWS_ACCESS_KEY',
                aws_secret_access_key='AWS_SECRET_ACCESS_KEY')
s3 = session.resource('s3')
bucket_s3 = s3.Bucket('bucket_name')

def not_exist(file_key):
    try:
        file_details = bucket_s3.Object(file_key).get()
        # print(file_details) # This line prints the file details
        return False
    except botocore.exceptions.ClientError as e:
        if e.response['Error']['Code'] == "NoSuchKey": # or you can check with e.reponse['HTTPStatusCode'] == '404'
            return True
        return False # For any other error it's hard to determine whether it exists or not. so based on the requirement feel free to change it to True/ False / raise Exception

print(not_exist('hello_world.txt')) 

Boto 2的Boto。s3.key。Key对象曾经有一个exists方法,通过执行HEAD请求并查看结果来检查Key是否存在于S3上,但它似乎已经不存在了。你必须自己动手:

import boto3
import botocore

s3 = boto3.resource('s3')

try:
    s3.Object('my-bucket', 'dootdoot.jpg').load()
except botocore.exceptions.ClientError as e:
    if e.response['Error']['Code'] == "404":
        # The object does not exist.
        ...
    else:
        # Something else has gone wrong.
        raise
else:
    # The object does exist.
    ...

load()对单个键执行HEAD请求,这是快速的,即使有问题的对象很大,或者bucket中有很多对象。

当然,您可能会检查对象是否存在,因为您计划使用它。如果是这种情况,您可以忘记load(),直接执行get()或download_file(),然后在那里处理错误情况。

FWIW,这里是我正在使用的非常简单的函数

import boto3

def get_resource(config: dict={}):
    """Loads the s3 resource.

    Expects AWS_ACCESS_KEY_ID and AWS_SECRET_ACCESS_KEY to be in the environment
    or in a config dictionary.
    Looks in the environment first."""

    s3 = boto3.resource('s3',
                        aws_access_key_id=os.environ.get(
                            "AWS_ACCESS_KEY_ID", config.get("AWS_ACCESS_KEY_ID")),
                        aws_secret_access_key=os.environ.get("AWS_SECRET_ACCESS_KEY", config.get("AWS_SECRET_ACCESS_KEY")))
    return s3


def get_bucket(s3, s3_uri: str):
    """Get the bucket from the resource.
    A thin wrapper, use with caution.

    Example usage:

    >> bucket = get_bucket(get_resource(), s3_uri_prod)"""
    return s3.Bucket(s3_uri)


def isfile_s3(bucket, key: str) -> bool:
    """Returns T/F whether the file exists."""
    objs = list(bucket.objects.filter(Prefix=key))
    return len(objs) == 1 and objs[0].key == key


def isdir_s3(bucket, key: str) -> bool:
    """Returns T/F whether the directory exists."""
    objs = list(bucket.objects.filter(Prefix=key))
    return len(objs) > 1