UNIX for Beginners Questions & Answers

Hi MadeinGermany,

Thanks for the responses, nice idea(s). The line without => is not captured via your regex, however DonCragun has provided some details around this.

Code:

 /lib64/ld-linux-x86-64.so.2 (0x00007fc21e165000)

Hi DonCragun,

Thanks for the response.

The initial solution with awk/tr/sort works. Looks like it is successful as the lines which do not have => as delimiter are left untouched by tr anyway.

Code:

./copy_chroot_lib.sh ls echo | tr '=>' '\n' | awk '/\/lib/{print $1}' | sort -u
/lib64/ld-linux-x86-64.so.2   <------------ line without => is included
/lib/x86_64-linux-gnu/libc.so.6
/lib/x86_64-linux-gnu/libdl.so.2
/lib/x86_64-linux-gnu/libpcre.so.3
/lib/x86_64-linux-gnu/libpthread.so.0
/lib/x86_64-linux-gnu/libselinux.so.1

The awk solution is nice and works, as I only need to remove duplicates. Sed is also good, but a little complex to understand.

Can you please explain the following awk code?

Code:

 ./copy_chroot_lib.sh ls echo | awk '$(NF-1) ~ "^/lib" {o[$(NF- 1)]} END {for(f in o) print f}'

From what I understand, if the second last field starts with /lib, the field is put into an array named o. Then the array is traversed and the value of all array elements printed out. How does this remove duplicates?

Thanks.

The duplicates are removed because of the associative (string-indexed) array o[ ].
If the string $(NF-1) occurs the second time it defines the same array element again.
You can make it visible by assigning a counter value to the array element, and printing it at the end:

Code:

awk '$(NF-1) ~ "^/lib" {o[$(NF- 1)]++} END {for(f in o) print f, "==>", o[f]}'