Shell Programming and Scripting

Hi Experts..

I have requirement in which I have a text file say a.txt which looks like below:

ASD
1
2
3
4
5
ASD
9
8
7
6
5
ASD
1
2
3
4
5
.
.
.

Now what I want to do is I have read the file line by line and search for the occurance of ASD and then I need to go to the line after that ASD where the number is 5 and replace it with SAT.

Similarly I need to do this for all the ASD's in the text file.

Any help on how to do this will be helpful..

Thanks in advance.
Sathya V.

Code:

awk '/^ASD$/ {flag=1} /^5$/ && flag == 1 {sub(5, "SAT", $0); flag=0}'1 file

Hi Balaje..

Thanks a lot for ur reply..but the occurance of number 5 is not a static length,..it may be in the 5th line after the ASD or 10th line after the ASD.

Please let me know if anything can be done.

Thanks & Regards,
Sathya V.

Solution above will work like you want, it does not count line.
It test if it find ASD, then if it later find 5, it change the line, no mater of how many lines below.

/^5$/ this search for line with 5 and 5 only. If you have more than 5 on the line, then some like this may work /^5/

Hi,.

that thing has been taken care..

Now I have a text file which looks like
asdf 1234
cdef 2345
..
..
..
KLDM 0945
<EMPTY LINE>

I need to remove the last empty line from the file. It does not have any whitespaces. This text file is created using a perl script with /n in the print statement, so that last empty line is getting created.

I have tried options like:

awk 'NF > 0' INPUT > OUTPUT
sed '/^\s*$/d' INPUT > OUTPUT
grep -v '^\s*$' INPUT > OUTPUT

But nothing worked.

Please help.

Thanks & Regards,
Sathya V.

Try one of this:

Code:

awk 'NF'
awk '/./'
awk '!/^$/'
sed '/./!d'