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Zip files on S3 with lambda

Lowell Llames answered on November 12, 2022 Popularity 6/10 Helpfulness 2/10

answer Zip files on S3 with lambda

Zip files on S3 with lambda

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def createZipFileStream(bucketName, bucketFilePath, jobKey, fileExt, createUrl=False):
    response = {} 
    bucket = s3.Bucket(bucketName)
    filesCollection = bucket.objects.filter(Prefix=bucketFilePath).all() 
    archive = BytesIO()

    with zipfile.ZipFile(archive, 'w', zipfile.ZIP_DEFLATED) as zip_archive:
        for file in filesCollection:
            if file.key.endswith('.' + fileExt):   
                with zip_archive.open(file.key, 'w') as file1:
                    file1.write(file.get()['Body'].read())  

    archive.seek(0)
    s3.Object(bucketName, bucketFilePath + '/' + jobKey + '.zip').upload_fileobj(archive)
    archive.close()

    response['fileUrl'] = None

    if createUrl is True:
        s3Client = boto3.client('s3')
        response['fileUrl'] = s3Client.generate_presigned_url('get_object', Params={'Bucket': bucketName,
                                                                                    'Key': '' + bucketFilePath + '/' + jobKey + '.zip'},
                                                              ExpiresIn=3600)

    return response

Popularity 6/10 Helpfulness 2/10 Language python

Source: stackoverflow.com

Tags: lambda python zip

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