How to display only files from aws s3 ls command?

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]*//'

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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
[...]

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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.

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Simple Way

aws s3 ls s3://mybucket --recursive --human-readable --summarize|cut -c 29-

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My Solution

List only files recursively using aws cli.

aws s3 ls s3://myBucket --recursive | awk 'NF>1{print $4}' | grep .

grep . - Clear empty lines.

Example: aws s3 ls s3://myBucket

                           PRE f5c10c1678e8484482964b8fdcfe43ad/
                           PRE f65b94ad31734135a61a7fb932f7054d/
                           PRE f79b12a226b542dbb373c502bf125ffb/
                           PRE logos/
                           PRE test/
                           PRE userpics/
2019-05-14 10:56:28       7754 stage.js

Solution: aws s3 ls s3://myBucket --recursive | awk 'NF>1{print $4}' | grep .

stage.js

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EDIT: After considering MultiDev's comment, that the previous solution won't work with objects that have spaces in them. I used s3api instead of s3

aws s3api list-objects --bucket mybucket --prefix myprefix --query 'Contents[].Key' | jq -rc '.[]'

prefix is optional

Using jq to get the raw elements (keys) from the returned array

Use something like --query 'Contents[].{Key: Key, Size: Size}' to get more info, then format the output further with jq

OLD Solution: aws s3 ls s3://mybucket --recursive | rev | cut -d" " -f1 | rev

I would suggest not depending on the spacing and fetching the 4th field.

You technically want the last field regardless of which position it was in.

So it's safer to use rev to your advantage...
rev reverses the string input char by char
so when you pipe the aws s3 ls out put to rev you have everything reversed, including the positions of the fields, so the last field always becomes the first field.
Instead of figuring out where the last field would be, you just rev, get first, then rev again because the characters in the field would be in reverse as well.

Example:

2013-09-02 21:32:57 23 Bytes foo/bar/.baz/a becomes a/zab./rab/oof setyB 32 75:23:12 20-90-3102

then cut -d" " -f1 would retrieve the first field a/zab./rab/oof

then rev again to get foo/bar/.baz/a

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Simple command would be

aws s3 ls s3://mybucket --recursive --human-readable --summarize |cut -d ' ' -f 8

If you need the timestamp, just update command field values.

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An S3 bucket may not only have files but also files with prefixes. In case you use --recursive it will not only list the files but also just the prefixes. In case you do not care about the prefixes and just the files within the bucket or just the prefixes within the bucket, this should work.

aws s3 ls s3://$S3_BUCKET/$S3_OPTIONAL_PREFIX/ --recursive | awk '{ if($3 >0) print $4}'

awk's $3 is the size of the file in case of prefix it would be 0. It could also be that the file is empty so it would skip empty files as well.

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How to display only files from aws s3 ls command?

1. Basic command

$ aws s3 ls s3://bucket --recursive

output :

2021-02-10 15:29:02          0 documents/
2021-02-10 15:29:02         18 documents/data/data.txt
2021-03-15 23:35:12          0 documents/data/my code.txt


2. To get only keys from s3 bucket containing spaces also.

$ aws s3 ls s3://bucket --recursive | awk '{ $1=$2=$3=""; print $0}' | cut -c4-

output : 

documents/
documents/data/data.txt
documents/data/my code.txt

3. Removing "documents/" from result

$ aws s3 ls s3://bucket --recursive | awk '$0 !~ /\/$/ { $1=$2=$3=""; print $0}' | cut -c4-

output :

documents/data/data.txt
documents/data/my code.txt

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For only the file names, I find the easiest to be:

aws s3 ls s3://path/to/bucket/ | cut -d " " -f 4

This will cut the returned output at the spaces (cut -d " ") and return the fourth column (-f 4), which is the list of file names.

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If your files don't have spaces, then this is the easiest way to do it:

aws s3 ls s3://mybucket  | cut -c32-

Output is:

1.txt.gz
2.txt.gz
3.txt.gz

Instead of:

2021-12-15 23:05:44         36 1.txt.gz
2021-12-15 23:05:45         37 2.txt.gz
2021-12-15 23:05:46         39 3.txt.gz

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Its just grep to filter by starting symbol. "^-" means line starts with '-' symbol. Directories, on other hand, start with the letter 'd'

ls -Al | grep "^-"

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