Replace every occurrence of the regular expression matching with a particular in VI editor?

Question

suppose I have a text file as follows.

create table "kevin".tb1 {
col1,
col2
}
create table "jhone".tb2 {
col1,
col2
}
create table "jake".tb3 {
col1,
col2

}

I need to obtain that text file as follows by replacing every table owner name occurrences replace witha same name called "informix".

out put should be like

create table "informix".tb1 {
col1,
col2
}
create table "informix".tb2 {
col1,
col2
}
create table "informix".tb3 {
col1,
col2
}

in vi editor ,

:%s/"kevin"/"informix"/g

I can replace them individually but I need to do all of them at once.

I modified. but I need to use regular expression to filter those words — Nwn sn, Dec 10 '18 at 11:18
`(?<=table ")(.*?)(?=")` try this regex and replace 'informix' — Code Maniac, Dec 10 '18 at 11:20
I need a solution which should work on VI editor. this is not working with VI editor — Nwn sn, Dec 10 '18 at 11:29
@CodeManiac The syntax `(?<=table ")` is from Perl/libpcre. In vim it's something like [`\@<=`](http://vimdoc.sourceforge.net/htmldoc/pattern.html#/\@=). — phd, Dec 10 '18 at 11:39

score 0 · Accepted Answer · answered Dec 10 '18 at 11:44

%s/\(create table\) "\i\+"/\1 "informix"/

Explanation:

% — run through every line in the file
s/ — search and replace
\(create table\) — match the text and store it in the backreference 1
"\i\+" — match any number (more than 1) of identifier characters inside double quotes
\1 "informix" — replace what is found with backreference 1 (text "create table"), a space and text "informix"

Replace every occurrence of the regular expression matching with a particular in VI editor?

1 Answers1