在UNIX中将行转置为列 [英] Transpose rows into column in unix

问题描述

我有下面给出的输入文件

I have input file which is given below

输入文件

10,9:11/61432568509
118,1:/20130810014023
46,440:4/GTEL
10,9:11/61432568509
118,1:/20130810014023
46,440:4/GTEL

我正在寻找的输出.

10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL
10,9:11/61432568509,118,1:/20130810014023,46,440:4/GTEL

我已经尝试使用awk命令，但是没有得到想要的输出.有人可以帮我吗?

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

推荐答案

使用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

如 Ed Morton ，以下变体更便携:

$ 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

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