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S3 Bucket に含まれる key 一覧をファイルに出力する

More than 5 years have passed since last update.

仕事で S3 を利用するようになった。

S3 の Key は DB に保存しており通常は問題ないが、どこかでずれるとやっかい。

そこで boto を使って、 S3 の Key 一覧をとってみた。

HEAD しか投げてなさそうなので、効率は良いと思います。

#! /usr/bin/env python
# -*- coding: utf-8 -*-
対象のバケットに含まれるファイル一覧を TSV に出力する。
import sys
import os
import csv
from ConfigParser import SafeConfigParser
from getpass import getpass

from boto import connect_s3

AWS_CLI_CONFIG_PATH = os.path.expanduser('~/.aws/config')

def get_aws_config(config_path=AWS_CLI_CONFIG_PATH):
    aws cli の config から以下のキーを返す
    - aws_access_key_id
    - aws_secret_access_key'
    keys = ['aws_access_key_id', 'aws_secret_access_key']
    cfg = SafeConfigParser()
    with open(config_path, 'r') as fp:
    return tuple(cfg.get('default', x) for x in keys)

def get_bucket(aws_access_key_id, aws_secret_access_key, bucket_name):
    boto S3 bucket を返す
    if not aws_access_key_id and not aws_secret_access_key:
        aws_access_key_id, aws_secret_access_key = get_aws_config()
    return connect_s3(aws_access_key_id, aws_secret_access_key).get_bucket(bucket_name)

def write_tsv(aws_access_key_id, aws_secret_access_key, bucket_name, file_name):
    S3 bucket の key.name 一覧を file_name に TSV で書き出す。
    # 絶対ファイルパスの決定
    file_path = os.path.abspath(file_name)

    def _writerows(rows):
        with open(file_path, 'a') as fp:
            writer = csv.writer(fp, dialect='excel-tab')

    # header の書き出し
    _writerows([('key_name', )])

    # body の書き出し
    rows = []
    for key in get_bucket(aws_access_key_id, aws_secret_access_key, bucket_name).list():
        if len(rows) > 1000:
            rows = []

if __name__ == '__main__':
    if len(sys.argv) != 2:
        print('Please specify output filename.')

        print('Please input the aws_access_key_id/aws_secret_access_key and a target bucket name.')
        print('If you don\'t input the aws_access_key_id/aws_secret_access_key, then we use awscli config.')
        aws_access_key_id = getpass('aws_access_key_id: ')
        aws_secret_access_key = getpass('aws_secret_access_key: ')
        bucket_name = raw_input('target bucket name: ')

        if not aws_access_key_id and not aws_secret_access_key and not os.path.isfile(AWS_CLI_CONFIG_PATH):
            print('Please specify the aws_access_key_id/aws_secret_access_key or create awscli config.')

        print('Output: {}'.format(sys.argv[1]))

すてま 僕が所属している会社社員募集中みたいです。
なんか、 Python 書いてみてーなと思った人は応募してみてくださいな。

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