UNIX for Dummies Questions & Answers

Using awk, is there a way to remove newlines in strings that don't contain 1_ at the beginning, but without setting RS="1_" in the BEGIN section and without using a END section as well?

input:
1_this is line 1
aaaaaaaa
bbbb
1_this is line 2
cccc
1_this is line 3
dddddddddddddd
eeeeeeeeeee
ffffffff

to get this output:
1_this is line 1|aaaaaaaabbbb
1_this is line 2|cccc
1_this is line 3|ddddddddddddddeeeeeeeeeeeffffffff


I think we cannot use RS in a sub function.

Code:

gawk '{

     if ($0 ~ /^"1_"/){
          sub (/$/,"|")
     }
     else{
          sub (/RS/,"")
     }
}1'

---------- Post updated at 08:34 PM ---------- Previous update was at 07:07 PM ----------

It works with that except that a "\n" is missing in the last line:

Code:

gawk 'BEGIN{FS="|"}

{
    if ($0 ~ /^1_/ && NR>1){
        print ""
    }
    printf ("%s", /^1_/ ? $0 FS : $0)
}1'

Try this:

Code:

gawk '/^1_this/ {
        if(NR != 1) printf("\n")
        printf("%s|", $0)
        next
}
        {printf("%s", $0)}
END {if(NR > 0) printf("\n")}' input

And without using an END section, is it doable...

Try this:

Code:

awk '/^1_/{if (x)print x;x="";}{x=(!x)?$0"|":x$0;}END{print x;}'

Thanks Don Cragun and spacebar !

But is it possible without using an END section?

No. With my script you need the END clause to print the trailing <newline>; with spacebar's script you need the END clause to print the entire last line.
Until you hit EOF, you don't know that there won't be another input line that needs to be joined to what you have already accumulated.

What is the reason for this constraint?

Regards,
Alister