CODEWITHSUNDEEP
bash
Luke
Luke

Reputation: 639

List of files after a date

I want to get all files after a specific date. I tried with: 

ls -ltr | awk {'print $6'} | sed s/-//g | awk {'if ($1-20110415 > 0 )  {print $1}'}

which works 50% fine. The last command prints only date of file. How to print date of file and filename? In the awk $8 is the filename, but I don't know how to transfer till last print in the command line.

Thank you, Luke

Upvotes: 14

Views: 18722

Answers (6)

Jeff Silverman
Jeff Silverman

Reputation: 724

I ran an experiment on my web server looking for files newer than a certain time.

$ find . -name "*.html" -newermt 2023-09-13T01:47 -ls | wc -l
1214
$ find . -name "*.html" -newermt 2023-09-13T01:47:30 -ls | wc -l
1177
$ find . -name "*.html" -newermt 2023-09-13T01:47:40 -ls | wc -l
1087
$ find . -name "*.html" -newermt 2023-09-13T01:47:50 -ls | wc -l
1016
$ find . -name "*.html" -newermt 2023-09-13T01:48 -ls | wc -l
929
$ find . -name "*.html" -newermt 2023-09-13T01:49 -ls | wc -l
665
$ find . -name "*.html" -newermt 2023-09-13T01:49:10 -ls | wc -l
665
$ find . -name "*.html" -newermt 2023-09-13T01:49:12 -ls | wc -l
665
$ find . -name "*.html" -newermt 2023-09-13T01:49:20 -ls | wc -l
634
$ find . -name "*.html" -newermt 2023-09-13T01:49:21 -ls | wc -l
624
$ find . -name "*.html" -newermt 2023-09-13T01:49:21.5 -ls | wc -l
616
$ find . -name "*.html" -newermt 2023-09-13T01:49:21.6 -ls | wc -l
616
$ find . -name "*.html" -newermt 2023-09-13T01:49:21.8 -ls | wc -l
612
$ find . -name "*.html" -newermt 2023-09-13T01:49:21.9 -ls | wc -l
611
$ find . -name "*.html" -newermt 2023-09-13T01:49:22 -ls | wc -l
610

$ find . -name "*.html" -newermt "13-Sep-2023 1:49" | wc -l
665
$ find . -name "*.html" -newermt "13-sep-2023 1:49:21" | wc -l
624
$ find . -name "*.html" -newermt "13-sep-2023 1:49:21.6" | wc -l
616
$ find . -name "*.html" -newermt "13-sep-2023 1:49:21.7" | wc -l
614
$ find . -name "*.html" -newermt "13-sep-2023 1:49:21.75" | wc -l
613
$ find . -name "*.html" -newermt "13-sep-23 1:49:21.75" | wc -l
613
$ find . -name "*.html" -newermt "09/13/23 1:49:21.75" | wc -l
613
$ find . -name "*.html" -newermt "09/13/2023 1:49:21.75" | wc -l
613
$ find . -name "*.html" -newermt "13/09/2023 1:49:21.75" | wc -l
find: I cannot figure out how to interpret ‘13/09/2023 1:49:21.75’ as a date or time

There might be other formats for time.

Upvotes: 0

Dirjit
Dirjit

Reputation: 87

ls -ltr --time-style=long-iso <path> | awk '$6 >= "2018-10-10" {print $6,$8}'

This gives me all files on the specified path that have been modified on 2018-10-10 or later.

--time-style=long-iso is a good way of making sure the time includes the year, month and date. This helped me exclude old files if you keep multi year files.

Upvotes: 2

Konrad Kiss
Konrad Kiss

Reputation: 6877

If you are looking for a way to find updated files and process them, you could touch a file after your process ends and use find's -newer : 

-newer file
    File was modified more recently than file.

So when you need an updated files list, you'd want to do this:

find /path/to/dir -newer touched.file

This would recursively list all files in /path/to/dir that have a modification date that is later than that of touched.file. Don't forget to touch that touched.file again when you finish processing the updated files or make note of which file to test on your next iteration.

Upvotes: 4

anubhava
anubhava

Reputation: 784918

Try find command like this:

find /my/path -mtime -1  # to get files modified in last 1 day

find /my/path -mtime -1.5  # to get files modified in last 1.5 day

Upvotes: 25

ZombieSheep
ZombieSheep

Reputation: 29953

How about using find with one of its various time-based switches, then using the printf to specify what fields you wish to display? You can find all the options in the man pages.

EDIT - I'm not on a suitable environment to give you an example right now - I'll try to update the answer tomorrow

Upvotes: 0

Oct
Oct

Reputation: 1525

What about this slightly modified version ?

ls -ltr | awk {'print $6 " " $8 '} | sed s/-//g | awk {'if ($1-20110426 > 0 ) {print $1 " " $2}'}

Seems to do the trick for me... ?

Upvotes: 0

Related Questions