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I saw this question but I couldn't implement what I'm looking for.

Here is what I want:

  • I have a file with a few lines ending with 4.1.01.97 (same number for all lines - but number can change from time to time)
  • I want to open vim and change the number into another

I tried vim [file_name] +:%s/97/98/g +:wq which works fine but it wouldn't help if the file will change.

I next tried getting desired number with grep: cat [file_name] | tail -1 | grep -E '.[0-9][0-9]$'

I want to use the result of this in my vim command... Something like using vim [file_name] +:%s/97/98/g +:wq with the grep command inside ``, instead the 97.

How can I do that?

If this it possible without vim, that would also be helpful :)

Also, can my regular expression for the grep be improved?

This is how the file looks like: 

A_,4.1.01.97
B_,4.1.01.97
C_,4.1.01.97
D_,4.1.01.97
E_,4.1.01.97
F_,4.1.01.97
G_,4.1.01.97
H_,4.1.01.97

This is how I want the file to look like: 

A_,4.1.01.98
B_,4.1.01.98
C_,4.1.01.98
D_,4.1.01.98
E_,4.1.01.98
F_,4.1.01.98
G_,4.1.01.98
H_,4.1.01.98
CIsForCookies
  • 199

1 Answers1

1

You could use sed :

sed 's/97/98/g' /path/to/file.txt

Use sed -i to modiffy the file instead of printing to stdout:

sed -i 's/97/98/g' /path/to/file.txt

Edit:

Since you also wanted to know if your regex can be improved, here a suggestion to get the last number of the end of your lines:

tail -1  /path/to/file.txt | grep -o '[0-9]\{1,\}$'

If you want to obtain the highest integer found at the end of all lines in the file, you could: 

grep -o '[0-9]\{1,\}$' /path/to/file.txt | sort -n | tail -1

You could replace it in the file as such:

sed -i 's/[0-9]\{1,\}$/<your replacement here>/g' /path/to/file.txt

Later Edit:

You may use this one liner:

pattern=`tail -1 /path/to/file.txt | grep -o '[0-9]\{1,\}$'` && sed -i "s/[0-9]\{1,\}$/${pattern}/" /path/to/file.txt
AnythingIsFine
  • 230