Tim here.
If you're using ed or sed rather than ex/vi/vim, it's pretty easy
depending on what constitutes a "word" (do you include punctuation
as word-separation, or just white-space? What about contractions
like "can't" or abbreviations like "Dr."?). You can use
6s/[^[:space:]]\{1,\}/replacement/6
6s/[[:alpha:]]\{1,\}/replacement/6
in both cases. The first 6 is the line-number, the 2nd 6 is the 6th
word in the line.
In ex/vi, it's a bit more complex because they don't support the
numeric flag at the end to indicate the 6th one, so you'd have to do
something atrocious like
6s/\(\([^[:space:]]\{1,\}[[:space:]]\{1,\}\)\{5\}\)[^[:space:]]*/\1replacement/p
(that "5" is the 5 words before the one you want to replace). It
would generally be easier to just retype the line.
It might be easier to prompt for a replacement:
6s/[^[:space:]]\{1,\}/xxx/gc
and then answer "no" 5 times, then answer "yes" for the 6th one.
If you're running the ex-mode of vim instead of vi, vim provides more
powerful regular expressions so you can do
6s/\(\W*\zs\w\+\)\{6}/replacement
Vim's regex also makes it easy to define what constitutes a
word-vs-not-a-word (using "\w" and "\W") or
whitespace-vs-non-whitespace (using "\s" and "\S"). Depending on how
you define a "word" the ed/sed/ex/vi versions get a lot messier.
Hope this helps,
-Tim (the vi/vim/ed/ex geek behind the @ed1conf account on Twitter)
On December 21, 2019, Linux for blind general discussion wrote:
> On line 6 of a file what is the command to delete word 6 on that
> line? A long time ago when the bsd learn utility was working on a
> system I studied the advanced ex lessons and unfortunately forgot
> that syntax.
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