mjhjmtu

I need to replace all occurrences of "period asterisk" as it is shown here:

blah blah .*:.*:.* blah blah

with:

[0-9][0-9]:[0-9][0-9]:[0-9][0-9]

so that the end result looks like this:

blah blah [0-9][0-9]:[0-9][0-9]:[0-9][0-9] blah blah

I tried different variations of the following but it didn't work:

%s_ .*:.*:.* _ [0-9][0-9]:[0-9][0-9]:[0-9][0-9] _g

For Vim: :%s/.*:.*:.*/[0-9][0-9]:[0-9][0-9]:[0-9][0-9]/g

For sed: sed -e 's/.*:.*:.*/[0-9][0-9]:[0-9][0-9]:[0-9][0-9]/g'.

• 
  % means apply the substitution on all lines; is not needed in sed where by default the command is applied to all lines.


• 
  . is a period; a bare . means any character.


• 
  * is an asterisk; a bare * means 0 or more of the preceding atom.

You gave an example of a line with “period asterisk” repeated three times,
separated by colons. 
If your data will always look like that, you might as well stick
with AlexP’s answer. 
But your question says
that you need to replace all occurrences of “period asterisk”. 
If that statement of the problem is accurate, you should use

:%s/.*/[0-9][0-9]/g

which will find and replace “period asterisk”
even when it doesn’t appear in groups of three.

Using the nomagic modifier for the pattern in Vim:

:%s/M.*/[0-9][0-9]/g

The M will remove the specialness of both . and * in the pattern.

See :help magic in Vim.