Shell Programming and Scripting

Hi everyone,

I'm new here (not really, I've been visiting this forum, but in a passive way)
But I'm new in sed...

I have a little problem with the following command:

/identification/{
N
s/identification *\n*and *\n*tracking *\n*system/Identification System/
}

Anybody that can explain me, what those *\n* mean. I know this is a substitution command, and I know that the \n is a newline symbol. And I also know that the * means, the precedent may appear one or more time. But does it have another meaning within this context?? This command works properly, but if I try this command without the *, then it doesn't work.

Code:

/identification/{
N
s/identification *\n*and *\n*tracking *\n*system/Identification System/
}

* means 0 or mor occurance of the preceding character.
\n means new line as you know
N - reads the next line and appends it to the existing cotents in the pattern space
- without the *, it must have the space and the \n for sure.

ok..so it says, whereever it finds the word identification, it reads the next line into the pattern space. so it appends the next line to the pattern space ( so your pattern space will have "identification....\n next line "

and then you are trying to substitute identification *\n*and *\n*tracking *\n*system
with Identification System in the pattern space.


cheers,
Devaraj Takhellambam

Hey, thanks for the reply

Indeed, the * means 0 or more occurance of the preceding character, and not 1 or more time... I was a little bit frustrated...

I found the answer, (while I was taking a big? break) and wanted to tell that this topic may be closed...

What I wanted to hear actually, is that the words identification, and, tracking & system were all spread over the text. Those words weren't always together, and that's why the *\n* justified the whole case...

But really thanks for your answer, I appreciate it very well

See ya

uhm, sorry you did tell it

No problem. The mind plays trick at times so it is always good to read and understand.