Timeline for How do I get the size of a directory on the command line?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 13, 2024 at 10:08 | comment | added | Simon A. Eugster | @AdamTaylor Or show in MB: | awk '{total+=$1} END {printf("%.0f MB\n",total/1024/1024)}' | |
| Mar 28, 2023 at 12:19 | history | edited | AdminBee | CC BY-SA 4.0 | Minor grammar and formatting improvements |
| Mar 28, 2023 at 11:38 | history | edited | Leon Chang | CC BY-SA 4.0 | why I don't use du command |
| Sep 10, 2022 at 16:21 | comment | added | Leon Chang | try printf("%.0f\n",total) | |
| Sep 9, 2022 at 14:13 | comment | added | Adam Taylor | This command could output things like 2.6957e+09 | |
| May 22, 2019 at 21:31 | comment | added | bballdave025 | This is great, because you don't get the overhead required to store the files, but only the size of the files themselves. | |
| Sep 24, 2018 at 11:08 | comment | added | anton_rh | I would use -not -type d to sum not only sizes of ordinary files (-type f) but also sizes of symbolic links and so on. | |
| Dec 6, 2016 at 13:12 | review | Late answers | |||
| Dec 6, 2016 at 13:42 | |||||
| Dec 6, 2016 at 12:57 | review | First posts | |||
| Dec 6, 2016 at 14:07 | |||||
| Dec 6, 2016 at 12:53 | history | answered | Leon Chang | CC BY-SA 3.0 |