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

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")

其他回答

不仅是客户端,还有桶:

import boto3
import botocore
bucket = boto3.resource('s3', region_name='eu-west-1').Bucket('my-bucket')

try:
  bucket.Object('my-file').get()
except botocore.exceptions.ClientError as ex:
  if ex.response['Error']['Code'] == 'NoSuchKey':
    print('NoSuchKey')

我不太喜欢在控制流中使用异常。这是在boto3中工作的另一种方法:

import boto3

s3 = boto3.resource('s3')
bucket = s3.Bucket('my-bucket')
key = 'dootdoot.jpg'
objs = list(bucket.objects.filter(Prefix=key))
if any([w.key == path_s3 for w in objs]):
    print("Exists!")
else:
    print("Doesn't exist")

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)

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'))