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

我注意到，为了使用botocore.exceptions. clienterror捕获异常，我们需要安装botocore。botocore占用36M的磁盘空间。如果我们使用aws lambda函数，这尤其会产生影响。如果我们只是使用异常，那么我们可以跳过使用额外的库!

我正在验证文件扩展名为'.csv' 如果桶不存在，这将不会抛出异常! 如果桶存在但对象不存在，则不会抛出异常! 如果桶为空，则抛出异常! 如果桶没有权限，就会抛出异常!

代码看起来像这样。请分享你的想法:

import boto3
import traceback

def download4mS3(s3bucket, s3Path, localPath):
    s3 = boto3.resource('s3')

    print('Looking for the csv data file ending with .csv in bucket: ' + s3bucket + ' path: ' + s3Path)
    if s3Path.endswith('.csv') and s3Path != '':
        try:
            s3.Bucket(s3bucket).download_file(s3Path, localPath)
        except Exception as e:
            print(e)
            print(traceback.format_exc())
            if e.response['Error']['Code'] == "404":
                print("Downloading the file from: [", s3Path, "] failed")
                exit(12)
            else:
                raise
        print("Downloading the file from: [", s3Path, "] succeeded")
    else:
        print("csv file not found in in : [", s3Path, "]")
        exit(12)

假设您只是想检查一个键是否存在(而不是悄悄地覆盖它)，首先进行这个检查。也会检查错误:

import boto3

def key_exists(mykey, mybucket):
    s3_client = boto3.client('s3')
    try:
        response = s3_client.list_objects_v2(Bucket=mybucket, Prefix=mykey)
        for obj in response['Contents']:
            if mykey == obj['Key']:
                return 'exists'
        return False  # no keys match
    except KeyError:
        return False  # no keys found
    except Exception as e:
        # Handle or log other exceptions such as bucket doesn't exist
        return e

key_check = key_exists('someprefix/myfile-abc123', 'my-bucket-name')
if key_check:
    if key_check == 'exists':
        print("key exists!")
    else:
        print(f"S3 ERROR: {e}")
else:
    print("safe to put new bucket object")
    # try:
    #     resp = s3_client.put_object(Body="Your string or file-like object",
    #                                 Bucket=mybucket,Key=mykey)
    # ...check resp success and ClientError exception for errors...

在Boto3中，如果您正在检查文件夹(前缀)或使用list_objects的文件。您可以使用响应字典中的“Contents”是否存在来检查对象是否存在。这是另一种避免try/except捕获的方法，就像@EvilPuppetMaster建议的那样

import boto3
client = boto3.client('s3')
results = client.list_objects(Bucket='my-bucket', Prefix='dootdoot.jpg')
return 'Contents' in results

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。

有一种简单的方法可以检查文件是否存在于S3桶中。我们不需要为此使用异常

sesssion = boto3.Session(aws_access_key_id, aws_secret_access_key)
s3 = session.client('s3')

object_name = 'filename'
bucket = 'bucketname'
obj_status = s3.list_objects(Bucket = bucket, Prefix = object_name)
if obj_status.get('Contents'):
    print("File exists")
else:
    print("File does not exists")