Shell Programming and Scripting

This works perfect. Thanks a lot

---------- Post updated at 12:46 AM ---------- Previous update was at 12:02 AM ----------

I'm facing a error for few files.

Here is the error that i got -

awk: Input line xxxxx cannot be longer than 3,000 bytes.
The source line number is xxxxx.

Do u have any idea wat the problem here wud be?

Thanks!

Please don't bump this up with a new thread if we don't respond fast enough.

The error means exactly what it says: the input data is too long. Some implementations of awk have this annoying limit. If your system has nawk or gawk that should work better.

printf print without a newline
NF?$0:"\n" a conditional expression.
test_condition ? true_result : false_result
NF return 0 for blank lines. Zero is considered false.
$0 returns the whole line/record being considered
"\n" adds a newline to the output when NF is zero

---------- Post updated at 05:53 PM ---------- Previous update was at 04:46 PM ----------

I wrote my own long winded versions of the concise:

Code:

awk '{printf NF?$0:"\n"}'

This one is just a verbose version of the above version:

Code:

awk 'NF > 0{printf $0};NF == 0{printf "\n"}'

This one actually breaks the fields down properly in case you need to process them but outputs field and record separators manually:

Code:

awk 'BEGIN{RS="";FS=",[\n]*";OFS="";ORS=""}{for(i=1;i<NF;i++) print $i ",";print $NF "\n"}'

My system doesn't seem to support gawk. I tried the same command with nawk, and i got the same error. Is there any other way to handle this situation. Please help me on this. If im thru with this, my whole script would be ready to deploy.

Thanks!

---------- Post updated at 04:25 PM ---------- Previous update was at 03:21 PM ----------

I'm facing a error with few files.

awk: There are not enough parameters in printf statement


I tried to print the whole file(without doing any tailoring) with this statement - awk '{printf $0}' < xxx I got the same error, and i found the error reported line has a character '%' in it.

So, i guess printf considers the % symbol in file as a special character. But im not sure this wud be the problem.

Thanks

The correct way to print an arbitrary string with printf is printf "%s", $0