Print all columns except first using AWK - linux

Question

I have a file which contains file list. The file looks like this
$ cat filelist
D src/layouts/PersonAccount-Person Account Layout.layout
D src/objects/Case Account-Record List.object
I want to cut first two Columns and print only file names with along directory path. This list is dynamic. File name has spaces in between. So I can't use space as delimiter. How to get this using AWK command?
The output should be like this
src/layouts/PersonAccount-Person Account Layout.layout
src/objects/Case Account-Record List.object

Answer 1

Can you try this once:
bash-4.4$ cat filelist |awk '{$1="";print $0}'
src/layouts/PersonAccount-Person Account Layout.layout
src/objects/Case Account-Record List.object
else if you want to remove 2 columns it would be:
awk '{$1=$2="";print $0}'
This will produce the below output:
bash-4.4$ cat filelist |awk '{$1=$2="";print $0}'
Account Layout.layout
Account-Record List.object

Answer 2

Try this out:
awk -F" " '{$1=""; print $0}' filelist | sed 's/^ //c'
Here sed is used to remove the first space of the output line.

Answer 3

print only file names with along directory path
awk approach:
awk '{ sub(/^[[:space:]]*[^[:space:]][[:space:]]+/,"",$0) }1' filelist
The output:
src/layouts/PersonAccount-Person Account Layout.layout
src/objects/Case Account-Record List.object
----------
To extract only basename of the file:
awk -F'/' '{print $NF}' filelist
The output:
PersonAccount-Person Account Layout.layout
Case Account-Record List.object

Answer 4

This will do exactly what you want for your example :
sed -E 's/(.*)([ ][a-zA-Z0-9]+\/[a-zA-Z0-9]+\/[a-zA-Z0-9. -]+)/\2/g' filelist
Explanation :
Its matching your path (including spaces if there were any ) and then replacing the whole line with that one match. Easy peasy lemon squeezy :)
Regards!

Answer 5

A simple grep
grep -o '[^[:blank:]]*/.*' filelist
That's zero or more non-blank characters followed by a slash followed by the rest of the string.
This will not match any lines that don't have a slash

Answer 6

Here is a portable POSIX shell solution:
#!/bin/sh
cat "$#" |while read line; do
echo "${line#* * }"
done
This loops over each line of the given input file(s) (or else standard input) and prints the line without the first two spaces or the text that exists before them. It is not greedy.
Unlike some of the other answers here, this will preserve spacing (if any) in the rest of the line.
If you want that as a one-liner:
while read L < filelist; do echo "${L#* * }"; done
This will fail if the uppermost directory's name starts with a space. To work around that, you need to peel away the leading ten characters (which I assume are static):
#!/bin/sh
cat "$#" |while read line; do
echo "${line#??????????}"
done
As a one-liner, in bash, this can be simplified by using substrings:
while read L < filelist; do echo "${L:10}"; done