首页 > 解决方案 > Amazon Transcribe on S3 Upload:“[错误] BadRequestException:提供的 URI 未指向 S3 对象”

问题描述

我正在对一组媒体文件尝试 Amazon Transcribe,调整示例文档代码并使用本系列作为参考,以适应任何上传到我指定的媒体 S3 文件夹的内容,但我的测试文件存在问题。

上传存储桶/文件夹路径:

'MediaFileUri': https://us-west-2.console.aws.amazon.com/s3/buckets/upload-asr/mediaupload/file.mp4

我已验证该文件存在并且存储桶权限授予对 Amazon Transcribe 服务的访问权限。我可以使用相同的 URL 启动手动转录作业,但不能使用 SDK:我还使用上面的路径将它直接链接到函数中,但没有结果。我很欣赏这可能是一个 URL 路径问题,但没有看到太多关于该主题的内容,因此请检查是否存在明显错误。

import json
import time
import boto3
from urllib.request import urlopen


def lambda_handler(event, context):
    transcribe = boto3.client("transcribe")
    s3 = boto3.client("s3")

    if event:
        file_obj = event["Records"][0]
        bucket_name = str(file_obj['s3']['bucket']['name'])
        file_name = str(file_obj['s3']['object']['key'])
        file_type = file_name.split(".")[1]
        s3_uri = create_uri(bucket_name, file_name)
        job_name = context.aws_request_id


        transcribe.start_transcription_job(TranscriptionJobName = job_name,
                                            Media = {'MediaFileUri': s3_uri},
                                            OutputBucketName = "bucket-name",
                                            MediaFormat = file_type,
                                            LanguageCode = "en-US")

def create_uri(bucket_name, file_name):

CloudWatch 日志失败报告:

[ERROR] BadRequestException: An error occurred (BadRequestException) when calling the StartTranscriptionJob operation: 
The URI that you provided doesn't point to an S3 object. Make sure that the object exists and try your request again.

Traceback (most recent call last):
  File "/var/task/lambda_function.py", line 25, in lambda_handler
    LanguageCode = "en-US")
  File "/var/runtime/botocore/client.py", line 320, in _api_call
    return self._make_api_call(operation_name, kwargs)
  File "/var/runtime/botocore/client.py", line 623, in _make_api_call
    raise error_class(parsed_response, operation_name) 

类似: https ://forums.aws.amazon.com/thread.jspa?messageID=876906慪

标签: pythonamazon-web-servicesamazon-s3aws-lambdaaws-transcribe

解决方案


它适用于我使用这种格式:

Media={
    'MediaFileUri': f'https://s3-us-west-2.amazonaws.com/{BUCKET}/{KEY}'
},

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