I am trying to write a awk script to produce a custom output from ls -l something like this:
File xxx.txt has size of 100 blocks, was last modified on July 3 2013, is owned by Kohn. The user has read permission, has write permission and has execute permission.
Dir abc has size of 200 blocks, was last modified on July 1 2013, is owned by Kohn. The user has read permission, does not have write permission and has execute permission.
...
I find the most difficult task is to parse the first column $1 to get the permission and file/directory. Can you give me a hint how to solve this?
Sincerely, Kohn
Don't parse ls use stat
#!/bin/bash
myls() {
local filetype=$(stat -c "%F" "$1")
local format="${filetype^} %n has size of %b blocks, "
format+="was last modified on $(date -d "@$(stat -c "%Y" "$1")" "+%B %e, %Y"), "
format+="is owned by %U. "
format+="$(permissions "$1")"
stat -c "$format" "$1"
}
permissions() {
local user_perms=$(stat -c "%A" "$1")
local string="The user "
string+="$(has ${user_perms:1:1} r) read permission, "
string+="$(has ${user_perms:2:1} w) write permission, "
string+="$(has ${user_perms:3:1} x) execute permission."
echo "$string"
}
has() { [[ $1 == $2 ]] && echo "has" || echo "does not have"; }
for file; do
myls "$file"
done
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.