如何使用 vi 删除文本文件中每行的前 5 个字符？

Question

How do I remove the first 5 characters in each line in a text file?
I have a file like this:

   4 Alabama
   4 Alaska
   4 Arizona
   4 Arkansas
   4 California
  54 Can
   8 Carolina
   4 Colorado
   4 Connecticut
   8 Dakota
   4 Delaware
  97 Do
   4 Florida
   4 Hampshire
  47 Have
   4 Hawaii

I'd like to remove the number and the space at the beginning of each line in my txt file.

Answer 1

:%s/^.\\{0,5\\}// should do the trick. It also handles cases where there are less than 5 characters.

Answer 2

Use the regular expression ^..... to match the first 5 characters of each line. use it in a global substitution:

:%s/^.....//

Answer 3

As all lines are lined up, you don't need a substitution to solve this problem. Just bring the cursor to the top left position (gg), then: CTRL+vGwlx

Answer 4

Since the text looks like it's columnar data, awk would usually be helpful. I'd use V to select the lines, then hit :! and use awk:

:'<,'>! awk '{ print $2 }'

to print out the second column of the data. Saves you from counting spaces altogether.

Answer 5

I think easiest way is to use cut.

just type cut -c n- <filename>

Answer 6

Try

:s/^.....//

You probably don't need the "^" (start of line), and there'd be shortcuts for the 5 characters - but simple is good :)

Answer 7

:%s/^.\\{0,5\\}//g用于全局，因为我们要为每一行删除每行的前 5 列。

Answer 8

In my case, to Delete first 2 characters Each Line I used this :%s/^.\\{0,2\\}// and it works with or without g the same.

I am on a VIM - Vi IMproved 8.2 , macOS version , Normal version without GUI.