How to display only files from aws s3 ls command?

Question

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

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

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

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

Answer 1

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.

Try this:

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

Edit: to take spaces in filenames into account:

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

Answer 2

Use the s3api with jq (AWS docu aws s3api list-objects):

This mode is always recursive.

$ aws s3api list-objects --bucket "bucket" | jq -r '.Contents[].Key'
a.txt
foo.zip
foo/bar/.baz/a
[...]

You can filter sub directories by adding a prefix (here foo directory). The prefix must not start with an /.

$ aws s3api list-objects --bucket "bucket" --prefix "foo/" | jq -r '.Contents[].Key'
foo/bar/.baz/a
foo/bar/.baz/b
foo/bar/.baz/c
[...]

jq Options:

• -r = Raw Mode, no quotes in output
• .Contents[] = Get Contents Object Array Content
• .Key = Get every Key Field (does not produce a valid JSON Array, but we are in raw mode, so we don't care)

Addendum:

You can use pure AWS CLI, but the values will be seperated by \x09 = Horizontal Tab (AWS: Controlling Command Output from the AWS CLI - Text Output Format)

$ aws s3api list-objects --bucket "bucket" --prefix "foo/" --query "Contents[].Key" --output text
foo/bar/.baz/a   foo/bar/.baz/b   foo/bar/.baz/c   [...]

AWS CLI Options:

• --query "Contents[].Key" = Query Contents Object Array and get every Key inside
• --output text = Output as Tab delimited Text with now Quotes

Addendum based on Guangyang Li Comment:

Pure AWS CLI with New Line:

$ aws s3api list-objects --bucket "bucket" --prefix "foo/" --query "Contents[].{Key: Key}" --output text
foo/bar/.baz/a
foo/bar/.baz/b
foo/bar/.baz/c
[...]

Answer 3

A simple filter would be:

aws s3 ls s3://mybucket --recursive | perl -pe 's/^(?:\S+\s+){3}//'

This will remove the date, time and size. Left only the full path of the file. It also works without the recursive and it should also works with filename containing spaces.