Change file output format

Shell Programming and Scripting
1

I have a file which has following contents

usmtnz-dinfsi19
62
61
18400
18800
99.7
usmtnz-dinfsi19
62
61
18400
18800
99.7

i want the o/p to be like

date 
usmtnz-dinfsi19   62   61    18400    18800    99.7 
date 
usmtnz-dinfsi19   64   65    20000    23000    88.9 
date 
.........n

how can i do that ?

2

code:-

gawk '(NR%6){s=s" "$0 ;next} 
((NR+1)%6){"date"|getline d ; print d"\n"s" " $0 ; s="" }
'  infile.txt > outfile.txt

:D:D:D:D

---------- Post updated at 18:07 ---------- Previous update was at 18:04 ----------

or if you want the date format to be YYYY-MM-DD

nawk '(NR%6){s=s" "$0 ;next}
((NR+1)%6){"date  '+%Y-%m-%d'"|getline d ; print d"\n"s" " $0 ; s="" }
' file.txt > out.txt

:b::b::b::b:

3

Cool ,thank you very much now is it possible to use arithmetic operations on the output like

64   65    20000/1000    23000/1000    100 - 88.9

And one more thing can you please refer to a awk,nawk link where i can learn it ?

4

just cuz it's a handy trick:

pr -l1 -t -6 a

( But seriously, the "date" command is going to be the same practically for everyline unless you're running a file with a zillion lines. What value does that provide?! )

5

I didn't get what you wrote :frowning:

6 
cat a.txt | paste - - - - - - | sed 'i\
        date
        '
7

Thanx for the help

8

or better

cat a.txt | xargs -n 6 | sed 'i\
        date
        '

:D:D:D:D