这里有一个对我有用的解决办法。需要注意的是，我事先知道密钥的确切格式，所以我只列出单个文件

import boto3

# The s3 base class to interact with S3
class S3(object):
  def __init__(self):
    self.s3_client = boto3.client('s3')

  def check_if_object_exists(self, s3_bucket, s3_key):
    response = self.s3_client.list_objects(
      Bucket = s3_bucket,
      Prefix = s3_key
      )
    if 'ETag' in str(response):
      return True
    else:
      return False

if __name__ == '__main__':
  s3  = S3()
  if s3.check_if_object_exists(bucket, key):
    print "Found S3 object."
  else:
    print "No object found."

这里有一个对我有用的解决办法。需要注意的是，我事先知道密钥的确切格式，所以我只列出单个文件

import boto3

# The s3 base class to interact with S3
class S3(object):
  def __init__(self):
    self.s3_client = boto3.client('s3')

  def check_if_object_exists(self, s3_bucket, s3_key):
    response = self.s3_client.list_objects(
      Bucket = s3_bucket,
      Prefix = s3_key
      )
    if 'ETag' in str(response):
      return True
    else:
      return False

if __name__ == '__main__':
  s3  = S3()
  if s3.check_if_object_exists(bucket, key):
    print "Found S3 object."
  else:
    print "No object found."

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

试试这个简单的方法

import boto3
s3 = boto3.resource('s3')
bucket = s3.Bucket('mybucket_name') # just Bucket name
file_name = 'A/B/filename.txt'      # full file path
obj = list(bucket.objects.filter(Prefix=file_name))
if len(obj) > 0:
    print("Exists")
else:
    print("Not Exists")

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

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'))