Transpose rows into column in unix

Question

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

Answer 1

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