Hi there
I have a file with a variable amount of rows but the 45th, 46th and 47th charachter of each line is the status field which is a three digit code ie 001, 002, 003 etc. My question is this..I need to strip all the records/lines with 002's out of the file completely and put them into another file leaving the original file with everything but the 002's. Sorry for the Newbieness of this question but im a bit stuck on this one
any help on this would be greatly appreciated
Cheers
Gary
Which OS and shell are you using, and can we see a sample set of the data?
Something like
grep -v '.\{44\}002.*' infile > outfile
Should have the desired effect.
Cheers
ZB
EDIT:
A safer bet would be including "^" in the expression, i.e.
grep -v '^.\{44\}002.*' infile > outfile
example file
xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxx,001,xxx
xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxx,002,xxx
xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxx,004,xxx
xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxx,007,xxx
xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxx,002,xxx
xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxxxxxxxx,xxx,002,xxx
os = solaris9
shell = ksh
I tried the command
grep -v '.\{44\}002.*' infile > outfile
but this just removes the 002's completely and populates the outfile with whatever is left, how can I can i create a second outfile (say outfile2) using the same method but this time putting just the 002's in there, as i say, i need tio keep the extracted 002's in a seperate file
cheers
To have just the 002's don't use the -v option to grep.
i.e.
grep '.\{44\}002.*' infile > outfile2
So we've got
grep '.\{44\}002.' infile > 002s_only
grep -v '.\{44\}002.' infile > everything_else
I'd suggest you have a read through the grep manual page, also "man 7 regex" on Linux, "man 5 regexp" on HP-UX (not too sure about Solaris) will give info on Regular Expressions. (else there's always google!).
Study regular expressions. If you intend to work at the Unix command line, they will prove invaluable.
Cheers
ZB
thanks, that works great
Hi again, ok I thought this was resolved but they now want to extract the 002's and the 003's into a file and the rest into another file, so basically I need to add some sort of AND operator into this command
grep '.\{44\}002.*' infile > outfile
so i sort of want it to do
grep '.\{44\}002 AND 003.*' infile > outfile
obviously this means that when I use the -v to extract all non 002/003 lines i will need to use this AND operator aswell
is this possible ?
Try:
grep '.\{44\}00[23].*' infile > outfile
Thanks perderabo, but what if the strings were completely different ie, didnt have a common 00 at the front. I have another status code of 55 (with a space after to make up the 3rd character ie
xxxxxx,002,xxx
xxxxxx,003,xxx
xxxxxx,55 ,xxx
I cant really use the '00' as a prefix and I presume I cant put [55002003] into the command ??
I have been advised by somebody else to use | inside some () to get multiple values into the condition ie
grep '.\{44\}\(002|55\).*' infile > outfile
although this doesnt seem to work, does anybody know if im on the right track ??
ive tried it with egrep instead of grep and also without escaping the ()'s all to no avail
any help would be greatly appreciated
I'm starting to think that awk would solve these problems easier.
You've got 7 comma seperated fields in the input file.
The "status" is in the 6th field.
Therefore
awk 'BEGIN { FS="," } { if ( $6 ~ /55|002|003/ ) { print $0 } }' infile > outfile
Would place any record with status 55, 002 or 003 into "outfile".
Modify the ( $6 ~ /55|002|003/ ) part and insert the statuses (or is that statii?!) of the records you want.
Cheers
ZB
works perfectly thankyou, out of interest how would i do a "not" version of this script so that everything other than 002,003,55 get put in a file
cheers
Change the if statement to
... if ( $6 !~ /55|002|003/ ) ...
Note: ~ means "matches"
!~ means "doesn't match"
Cheers
ZB
Hi,
I am using the following AWK command to strip some lines from a file based on a pattern. But this is creating an additional blank line at the position where the pattern matched. Please advise how to avoid the blank line. Thanks.
awk 'BEGIN { FS="," } { if ( $6 ~ /55|002|003/ ) { print $0 } }' infile > outfile
Regards,
Tipsy.
Sorry. It works fine.
Thanks,
Tipsy.