更改用户UNIX给定的特定行和列的值

时间:2016-03-31 10:55:23

标签: shell unix awk replace

我有这个文件:

 16492674422392|Alberto|Parra|female|1985-09-22|2012-09-01T01:30:59.228+0000|190.96.12.239|Chrome
> 16492674424948|Peng|Chen|female|1984-07-26|2012-09-23T00:51:52.900+0000|1.4.10.198|Internet
> Explorer
> 16492674425075|Changpeng|Xu|female|1984-03-27|2012-10-02T03:55:00.946+0000|1.50.15.119|Firefox
> 16492674425398|Prince|Kobayashi|male|1989-08-07|2012-09-30T03:30:41.772+0000|14.101.89.18|Chrome
> 16492674426410|Yang|Wei|male|1980-07-01|2012-10-01T13:11:48.528+0000|27.144.204.193|Firefox

我希望用户:

  1. 选择一个id(id是第一列)
  2. 选择一列
  3. 将值更改为用户选择的一个值。
  4. 我用:

    ./tool.sh 16492674426410 3 replacement
    

    作为输入,我运行的代码是:

    awk -v antik1=$1 -v antik2=$2 '
    {
        sub(antik1, antik2);
        print;
    }' persons.dat.txt
    

    此脚本不允许用户选择列和ID。我该如何修改它以便按我的意愿工作?

1 个答案:

答案 0 :(得分:0)

试试这个测试版本:

#!/bin/bash --

awk -v anid="${1}" -v antik1="${2}" -v antik2="${3}" '    
BEGIN {
  FS="|";
  OFS="|";
}
{
  if ($1 == anid) {
    $antik1=antik2;
  }
  print;
}' persons.dat.txt

测试:

$ ./tool.sh 16492674426410 3 replacement
16492674422392|Alberto|Parra|female|1985-09-22|2012-09-01T01:30:59.228+0000|190.96.12.239|Chrome
16492674424948|Peng|Chen|female|1984-07-26|2012-09-23T00:51:52.900+0000|1.4.10.198|Internet Explorer
16492674425075|Changpeng|Xu|female|1984-03-27|2012-10-02T03:55:00.946+0000|1.50.15.119|Firefox
16492674425398|Prince|Kobayashi|male|1989-08-07|2012-09-30T03:30:41.772+0000|14.101.89.18|Chrome
16492674426410|Yang|replacement|male|1980-07-01|2012-10-01T13:11:48.528+0000|27.144.204.193|Firefox