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Using pamu's file and based on his approach:$ awk '!/^[,0-9]*$/ {print NR,$0}' file
7 28,f
8 ff,25
9 ff,ss

$ cat file
14,30
28,30
16,30
22,30
21,30
3,30
28,f
ff,25
ff,ss


$ awk '!/^[0-9]+,[0-9]+/{print NR,$0}' file
7 28,f
8 ff,25
9 ff,ss

$ cat file
3,30
16,y
x,30


$ awk -F "," '$1 !~ /[[:digit:]]/ || $2 !~ /[[:digit:]]/ { print "Line " NR ": " $0 }' file
Line 2: 16,y
Line 3: x,30

You were close. It's good you tried first....

This is the output of the second proposal - what is wrong, what are you missing?LAC|DN|EQN|CAT|OPTRCL|TRARSTR|LNATT|COS
040|24001001|920- 2- 0- 1|MS|3|ACTTRACL TRACLACT TRACLMOD|PB|CLIP

Added...

PLEASE use code tags for code and data as required by forum rules!

This has become uglier than what I went for, but at least for your special problem, it might work:awk '$1=="=" {header =...

Since you didn't use code tags, I'm not sure that I have correctly interpreted what your input looks like, but this might work:
awk '
/=/ { if(hc) f[hc] = s
h[++hc] = last
last...

Are you checking the java file in the below location ?

[XXXXX@YYYYYY bin]$ java
bash: /home/XXXXX/java1.5/bin/java: cannot execute binary file


---------- Post updated at 03:19 PM ----------...

If you have the perl with Date::Calc module, then you can use Mktime to calculate the epoch. Then subtract the both epoch and divide it by 60 (seconds) to find the minutes.


$time =...

You may have a go with this (although I feel someone will come up with a better solution soon):
perl -MTime::Local -lne 'if($. > 1) {
/(....)(..)(..)(..)(..)(..)/;...

But, on which computer?
I'm assuming that your remsh command works and that you can login when you provide the correct username and password.

The .rhosts file is used to give permissions to other...