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I have files with hundreds of lines of varying length. I want to find each line with the string "New" and print the first 7 characters and the 10th from the last character.

For example, cat file1.txt

1234567 New line with irrelevant info x end line

2345678 irrelevant line

3456789 New line with different irrelevant info y end line

4567890 irrelevant line

5678901 New line with yet more irrelevant info z end line

And my output would be:

1234567 x 

3456789 y

5678901 z

Choose one you like:

awk solution:

awk '/New/{ print substr($0, 1, 7), substr($0, length-9, 1) }' file1.txt

sed solution:

sed -rn '/New/ s/^(.{7}).*(.).{9}$/1 2/p' file1.txt

Sample output (for both approaches):

1234567 x

3456789 y

5678901 z

POSIXly:

Assuming the lines contain at least 10 characters (if not, the behaviour is unspecified for the second substr(), you can add a && length >= 10 or && length >= 17 after /New/ to skip the lines that have fewer than 10 or 17 characters):

awk '/New/ {print substr($0, 1, 7), substr($0, length - 9, 1)}'

or assuming the lines contain at least 17 characters (the lines that don't will be skipped):

sed -n '/New/ s/^(.{7}).*(.).{9}$/1 2/p'

This is one "brutal" gawk solution that does the job, using null as field separator FS and output field separator OFS , meaning each char of inputfile is considered to be a field for awk.

awk '/New/{print $1,$2,$3,$4,$5,$6,$7," ",$(NF-9)}' FS="" OFS="" file1

1234567 x

3456789 y

5678901 z

More solutions with grep / sed will follow.