Question

我具有如下lambda函数

from __future__ import print_function

import urllib
import zipfile
import boto3
import io
import mimetypes

import logging

logger = logging.getLogger()
logger.setLevel(logging.INFO)


s3 = boto3.client('s3')
bucket = 'staging-bucket'

def lambda_handler(event, context):
    try:
        key = urllib.unquote_plus(event['Records'][0]['s3']['object']['key'].encode('utf8'))
        obj = s3.get_object(Bucket=bucket, Key=key)
        with io.BytesIO(obj["Body"].read()) as tf:
            # rewind the file
            tf.seek(0)
            # Read the file as a zipfile and process the members
            with zipfile.ZipFile(tf, mode = 'r') as zipf:
                for file in zipf.infolist():
                    fileName = file.filename
                    contentType, encoding = mimetypes.guess_type(fileName)
                    contentType = contentType or 'application/octet-stream'
                    filePath = "playable/staging/" + key.replace("package.zip", "") + fileName
                    putFile = s3.put_object(ACL = 'public-read', Bucket = "unzipped-bucket", Key = filePath, Body = zipf.read(file), ContentType = contentType)


    except Exception as e:
        logger.error('Error getting object {} from bucket {}. Make sure they exist and your bucket is in the same region as this function.'.format(key, bucket))
        raise e

    return

它从s3存储桶中提取 zip 文件，并将其提取到另一个s3存储桶

函数成功运行，但提取的文件名以zip文件名作为前缀，请参见下面的图片以供参考

源zip文件： package-1542108930.zip

源zip内容： source zip files

提取的文件夹内容： extracted files

我找不到python脚本中的错误，任何帮助将不胜感激。提前致谢。

Answer 1

我怀疑您的问题是此行：

filePath = "playable/staging/" + key.replace("package.zip", "") + fileName

请注意，您要删除字符串package.zip，但是（如从“前缀”中可以看到的）字符串实际上是package-1542108930.zip。

尝试：

filePath = "playable/staging/" + fileName

如果您根本不想使用任何名称。

如果要保留时间戳，则：

filePath = "playable/staging/" + key.replace("package-", "").replace(".zip", "") + fileName

AWS Lambda s3文件解压缩python中的文件名问题

1 个答案: