如何仅显示aws s3 ls命令中的文件? [英] How to display only files from aws s3 ls command?

问题描述

我正在使用aws cli使用以下命令来列出s3存储桶中的文件(文档):

I am using the aws cli to list the files in an s3 bucket using the following command (documentation):

aws s3 ls s3://mybucket --recursive --human-readable --summarize

此命令为我提供以下输出:

This command gives me the following output:

2013-09-02 21:37:53   10 Bytes a.txt
2013-09-02 21:37:53  2.9 MiB foo.zip
2013-09-02 21:32:57   23 Bytes foo/bar/.baz/a
2013-09-02 21:32:58   41 Bytes foo/bar/.baz/b
2013-09-02 21:32:57  281 Bytes foo/bar/.baz/c
2013-09-02 21:32:57   73 Bytes foo/bar/.baz/d
2013-09-02 21:32:57  452 Bytes foo/bar/.baz/e
2013-09-02 21:32:57  896 Bytes foo/bar/.baz/hooks/bar
2013-09-02 21:32:57  189 Bytes foo/bar/.baz/hooks/foo
2013-09-02 21:32:57  398 Bytes z.txt

Total Objects: 10
   Total Size: 2.9 MiB

但是，这是我想要的输出:

However, this is my desired output:

a.txt
foo.zip
foo/bar/.baz/a
foo/bar/.baz/b
foo/bar/.baz/c
foo/bar/.baz/d
foo/bar/.baz/e
foo/bar/.baz/hooks/bar
foo/bar/.baz/hooks/foo
z.txt

如何仅显示文件列表而忽略日期，时间和文件大小?

How can I omit the date, time and file size in order to show only the file list?

推荐答案

您不能仅使用aws命令来执行此操作，但是您可以轻松地将其通过管道传递给另一个命令以去除不需要的部分.您还需要删除--human-readable标志以使输出更易于使用，并且需要--summarize标志以删除末尾的摘要数据.

You can't do this with just the aws command, but you can easily pipe it to another command to strip out the portion you don't want. You also need to remove the --human-readable flag to get output easier to work with, and the --summarize flag to remove the summary data at the end.

尝试一下:

aws s3 ls s3://mybucket --recursive | awk '{print $4}'

考虑文件名中的空格:

aws s3 ls s3://mybucket --recursive | awk '{$1=$2=$3=""; print $0}' | sed 's/^[ \t]*//'

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