删除文件中几个栏位的问题，awk和sed怎么来实现

psmzs

有一个格式
"222","a1a","db222","555","ss2","22"
"ac","cc","de2e","5e55","dbb","c "
"11122","c12","dhh","bn","667","23 "
"b1c2","7789","db222","av5","22","  22"
................................

类似于这种格式的，我想删除第一列和第二列，就像下面这样
"db222","555","ss2","22"
"de2e","5e55","dbb","c "

用sed 和awk怎么来实现？

zsszss0000

awk -vFS="," '{$1="";$2=""}1' file

1.   "db222" "555" "ss2" "22"
   
2.   "de2e" "5e55" "dbb" "c "
   
3.   "dhh" "bn" "667" "23 "
   
4.   "db222" "av5" "22" "  22"

复制代码

Herowinter

回复 1# psmzs


    sed  -r 's/^([^,]*,){2}//' urfile

psmzs

回复 2# zsszss0000


   我还想其他的保留原样，你这输出后分隔是空格了，如果OFS设置为逗号的话，前面还带着两个逗号啊，

hjfeng1988

回复 4# psmzs

1. awk -F ',' '{for(i=3;i<NF;i++){printf $i FS}print $NF}' file

复制代码

yestreenstars

1. with open('input.txt') as f:
   
2.     for each_line in f:
   
3.         fields = each_line.split(',')
   
4.         print ','.join(fields[2:]),

复制代码

psmzs

回复 6# yestreenstars


    大神你这是什么语法。。。。

yestreenstars

回复 7# psmzs

Python……