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Cannot extract libraries using sed


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# 8  
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.
# 9  
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]}'

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