public/subjects/devops/file-details

file-details

Instructions

Create a script file-details.sh, which will allow you to see the files inside the folder hard-perms, this way:

Expected output:

$ ./files-details.sh
dr-------x 2023-02-07 15:34 0
-r------w- 2023-01-18 18:24 1
-rw----r-- 2023-01-18 18:24 2
drwxrwxrwx 2023-02-07 15:34 3
-r-x--x--- 2023-01-16 16:34 4
-r--rw---- 2023-01-18 18:24 5
-r--rw---- 2023-01-18 18:24 6
-r-x--x--- 2023-01-16 16:34 7
-rw----r-- 2023-01-18 18:24 8
-r------w- 2023-01-18 18:24 9
dr-------x 2023-02-07 15:34 A

Hints

• ls -l --time-style=TIME_STYLE

  • Time conversion specifiers you need to know:

    • %R, 24-hour hour and minute. Same as ‘%H:%M’.

  • Date conversion specifiers you need to know:

    • %F, full date in ISO 8601 format; like ‘%+4Y-%m-%d’ except that any flags or field width override the ‘+’ and (after subtracting 6) the ‘4’. This is a good choice for a date format, as it is standard and is easy to sort in the usual case where years are in the range 0000…9999.

• You can use sed to remove the first line of the output.

$ ./files-details.sh        # without using sed
total
-rw-rw-r-- 2022-12-20 03:10:18 README.md
$ ./files-details.sh        # using sed
-rw-rw-r-- 2022-12-20 03:10:18 README.md
$

• awk.

You can specify specific column names to display or include in the awk output using the field numbers. For example:

$ ls -l | awk '{print $1, $2, $3, $4, $5, $6, $7, $8, $9}' # print all the given fields
total 4
-rw-rw-r-- 1 user user 1989 dez 20 15:19 README.md
$ ls -l | awk '{print $1, $2, $4, $5, $7, $8}'             # print all the given fields
total 4
-rw-rw-r-- 1 user 2350 20 15:25
$

awk ‘{print $1}’ emp_records.txt awk {print $1, $6, $7, $8, $9}'

You have to use Man or Google to know more about commands flags, in order to solve this exercise! Google and Man will be your friends!

References

• Time conversion specifiers.
• Date conversion specifiers.
• awk.