extracting records from a data file

[Joe Robinson_2]

I'm trying to pull a number of records that contain 'X' as the seventh character of the string. I've tried using wild cards (??????X) in a grep statement, as well as [A-Z]....X, but am looking for a simpler, more direct way that I can use in the future as well.

Along with that, I'm curious if there would be a way to direct records that don't have 'X' in them to another file...if the 2 files record count adds up to the original, I'm in like Flynn.

Thanks,
Joe Robinson

[Jeff Machols]

You already have the best way to extract them grep ??????X

to get the opposite, all you have to do is grep -v ??????X

[Joe Robinson_2]

Not sure why that syntax won't work here...that was the first two things I thought of.

[Curtis Larson_1]

could always try

sed '/[.]\{6\}X/p'

and those without a X at the 7th character:

sed '/[.]\{6\}X/d'

[Jeff Machols]

you may need to anchor the grep by doing this

grep "^??????X" file
grep -v "^??????X" file

[Jeff Machols]

This will work also

cat file | cut -c 7 | grep X
cat file | cut -c 7 | grep -v X

[Curtis Larson_1 · Accepted solution]

and with grep

grep '.\{6\}X'

[Jeff Machols]

One thing you will want to add to Curtis' is

grep '^.\{6\}X' file

otherwise you will get it if X is the 8th, 9th, etc

the ^ will anchor it to the begining of the line

[Curtis Larson_1]

and with awk

awk '/^......X/ {print $0 >> "linesWithX";next} {print $0 >> "linesWithoutX";}'