AWS Lambda s3文件解压缩python中的文件名问题

Question

I have lambda function as below

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

It takes zip file from s3 bucket and extracts it to another s3 bucket

the function runs successfully, but the extracted filename has the zip file name as prefix, see below pictures for reference

Source zip file : package-1542108930.zip

Source zip contents: source zip files

Extracted folder contents: extracted files

I'm unable to find the bug in python script, any help will appreciated. Thanks in advance.

Answer 1

I suspect your problem is this line:

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

Note that you're removing the string package.zip but (as you can see from the "prefixes"), the string is actually package-1542108930.zip .

Try:

filePath = "playable/staging/" + fileName

If you simply do not want any name.

If you want to maintain the timestamp, then:

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