I have input file which is given below
Input file
10,9:11/61432568509
118,1:/20130810014023
46,440:4/GTEL
10,9:11/61432568509
118,1:/20130810014023
46,440:4/GTEL
Output which i am looking for.
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
I have tried with awk command, but i am not getting desired output. can anyone help me in this?
awk -F"" '{a[$1]=a[$1]FS$2}END{for(i in a) print i,a[i]}' inputfile
devnull
Using awk:
$ awk 'ORS=(NR%3==0)?"\n":","' inputfile
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
EDIT: As commented by sudo_O and Ed Morton, the following variant is more portable:
$ awk 'ORS=(NR%3?",":RS)' inputfile
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
With pr:
$ pr -ats, file --columns 3
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
With args and tr:
$ xargs -n3 < file | tr ' ' ,
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
Here is how to do it with paste:
paste -d, - - - < file
Output:
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
if each of your "data block" has 3 lines, you could do:
sed -n 'N;N;s/\n/,/g;p' file
if you love awk:
awk 'NR%3{printf "%s,",$0;next}7' file
> sed 'N;N;s/\n/,/g' your_file
A short awk version
awk 'ORS=NR%3?",":RS' file
Shortened, thanks to iiSaymour
One way with awk:
$ awk -v RS= -F'\n' 'BEGIN{OFS=","}{for (i=1;i<=NF; i=i+3) {print $i,$(i+1),$(i+2)}}' file
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
It defines each field to be a line. Hence, it prints them on blocks of three.
来源：https://stackoverflow.com/questions/19001405/transpose-rows-into-column-in-unix