Find a match in a field and print next n fields - bash

Question

BASH noob here.
I have a tab separated file structured like this:
ABC DEF x 123 456
GHI x 678 910
I need to match "x" and print x plus the following two fields:
x 123 456
x 678 910
I've tried a few things but the issue that throws me off is that "x" is not always in the same field. Can please somebody help?
Thanks in advance.

Answer 1

If you are working in bash, then bash provides parameter expansions with substring removal that are built-in. They (along with many more) are:
${var#pattern} Strip shortest match of pattern from front of $var
${var##pattern} Strip longest match of pattern from front of $var
${var%pattern} Strip shortest match of pattern from back of $var
${var%%pattern} Strip longest match of pattern from back of $var
So in your case you want to trim the longest path from the front up to x as the pattern, e.g.
while read line || [ -n "$line" ]; do
echo "x${line##*x}"
done
Where you read each line and then trim from the front until 'x' is found (you remove the 'x' as well), so you simply output "x....." where "....." is the rest of the line (restoring the 'x')
(for large data sets, you would want to use awk or sed for efficiency reasons)
Example Use/Output
Using your sample data in a heredoc, you could do:
$ while read line || [ -n "$line" ]; do
> echo "x${line##*x}"
> done << 'eof'
> ABC DEF x 123 456
> GHI x 678 910
> eof
x 123 456
x 678 910
You can just select-copy/middle-mouse-paste the following in your xterm to test:
while read line || [ -n "$line" ]; do
echo "x${line##*x}"
done << 'eof'
ABC DEF x 123 456
GHI x 678 910
eof
Using grep -o For Simplicity
You other option, is to use grep -o where the -o option returns the part of the line only-matching the expression you provide, so
grep -o 'x.*$' file
Is another simple option, e.g.
$ grep -o 'x.*$' << 'eof'
> ABC DEF x 123 456
> GHI x 678 910
> eof
x 123 456
x 678 910
Let me know if you have any further questions.

Answer 2

In case you need to match only tab separated field x:
pcregrep -o '(^|\t)\Kx(\t|$).*' file
awk 'n=match($0,/(^|\t)x(\t|$)/) {$0=substr($0,n); sub(/^\t/,""); print}' file
To print only the two following fields:
pcregrep -o '(^|\t)\Kx(\t[^\t]*){2}' file
awk 'n=match($0,/(^|\t)x\t[^\t]*\t[^\t]*/) {$0=substr($0,n,RLENGTH); sub(/^\t/,""); print}' file

Answer 3

Could you please try following, written and tested with shown samples in GNU awk.
awk '
match($0,/[[:space:]]+x[[:space:]]+[0-9]+[[:space:]]+[0-9]+$/){
val=substr($0,RSTART,RLENGTH)
sub(/^[[:space:]]+/,"",val)
print val
}
' Input_file
OR to match more than 1 set of digits after x with spaces try following.
awk '
match($0,/[[:space:]]+x[[:space:]]+([0-9]+[[:space:]]+){1,}[0-9]+/){
val=substr($0,RSTART,RLENGTH)
sub(/^[[:space:]]+/,"",val)
print val
}
' Input_file
Explanation: Adding detailed explanation for above.
awk ' ##Starting awk program from here.
match($0,/[[:space:]]+x[[:space:]]+[0-9]+[[:space:]]+[0-9]+$/){ ##Using match function to match regex here.
val=substr($0,RSTART,RLENGTH) ##Creating val which has sub string of matched regex(previous step) length.
sub(/^[[:space:]]+/,"",val) ##Substituting initial space with NULL in val here.
print val ##Printing val here.
}
' Input_file ##mentioning Input_file name here.

Answer 4

I need to match x and print x plus the following two fields:
Using awk without any regex:
awk 'BEGIN {FS=OFS="\t"} {for (i=1; i<=NF; ++i) if ($i == "x") break;
print $i, $(i+1), $(i+2)}' file
x 123 456
x 678 910
Or, using gnu sed:
sed -E 's/(^|.*\t)(x(\t[^\t]+){2}).*/\2/' file
x 123 456
x 678 910

Answer 5

If you want to remove everything before the "x", you can run a sed command like this:
sed 's/^.*x/x/g' file.txt
It finds all occurrences of the pattern ^.*x and replace it with x.
Breakdown of ^.*x:
^ means beginning of a line
.* a wild card pattern that can be more than one character
x the character "x"
Hence it replaces everything before and including "x" that are on the same line with the new pattern, just "x".
For more info on sed's find and replace command, see https://www.cyberciti.biz/faq/how-to-use-sed-to-find-and-replace-text-in-files-in-linux-unix-shell/.