vi, awk, sed - Substitute / Replace Few First , Last Characters in all lines of Linux file

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I have a text file called dat.txt and it having some text data which is highly confidential. I wanted to replace some few characters with x to make it invalid. See different options with vi, awk, sed.

My input file is-

$ cat dat.txt

1-CCJGL1-CCJGL1-CCJGL

1-BSDF0Q1-BW;LKJ1-BWP30Q

1-LKJ<MN1-1LKJMG1-1W13LG

1-2<MBMV1-NVNBVKJH21HMRE

1-2EW*&Y1-(878761-2AJKGY

Using vi editor:

Open the file in vi and run following commands.

1. To replace last 6 characters with x

:%s/......$/xxxxxx/g

2. To replace first 6 characters with x

:%s/^....../xxxxxx/g

Using awk:

awk '{

    start_str=substr($0,1,6)

    gsub(/./,"x",start_str)

    finish_str=substr($0,6)

    print start_str finish_str

}' dat.txt

Output

xxxxxxGL1-CCJGL1-CCJGL

xxxxxxF0Q1-BW;LKJ1-BWP30Q

xxxxxx<MN1-1LKJMG1-1W13LG

xxxxxxBMV1-NVNBVKJH21HMRE

xxxxxx*&Y1-(878761-2AJKGY

awk '{

    len=length($0)

    start_str=substr($0,len-5,len)

    finish_str=substr($0,1,len-6)

    gsub(/./,"x",start_str)

    print finish_str start_str

}' dat.txt

Output

1-CCJGL1-CCJGL1xxxxxx

1-BSDF0Q1-BW;LKJ1-xxxxxx

1-LKJ<MN1-1LKJMG1-xxxxxx

1-2<MBMV1-NVNBVKJHxxxxxx

1-2EW*&Y1-(878761-xxxxxx

Using sed:

sed -r "s/^(.{0})(.{6})/\1xxxxxx/" dat.txt

Output

xxxxxxL1-CCJGL1-CCJGL

xxxxxx0Q1-BW;LKJ1-BWP30Q

xxxxxxMN1-1LKJMG1-1W13LG

xxxxxxMV1-NVNBVKJH21HMRE

xxxxxx&Y1-(878761-2AJKGY

sed -r "s/(.{0})(.{6})$/\1xxxxxx/" dat.txt

Output

1-CCJGL1-CCJGL1xxxxxx

1-BSDF0Q1-BW;LKJ1-xxxxxx

1-LKJ<MN1-1LKJMG1-xxxxxx

1-2<MBMV1-NVNBVKJHxxxxxx

      1-2EW*&Y1-(878761-xxxxxx

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