vi/vim : complex substitution ?

I use vim. I have a lot of SQL queries to write, and am hoping there is some wild command I can use in vim to make this simpler.

From a file that is a list of fields, like the excerpt, for example:



I want to go to this:



I'm not sure if such a thing can be done with one line of command. Fortunately, all lines will be in the same format.. tablename.fieldname. so I know that I need to do something like:



If this is possible what would the proper syntax be?

%s/orderdetail\.\(.*\)/COUNT(DISTINCT(orderdetail.\1)) as \1,/g

'sed' is for this better, exactly the same sintax.

works beautifully.. for future reference..

can you (or anyone, of course)
explain this bit :

\. the \ masks the dot, dot in vi, sed, awk, grep, perl....
is not a dot, but a metacharacter replacing every oder
char, so a masked dot is a dot clear?
\( means: attention here is starting something
\) closes \)
.* any char, see (dot) below, the * is for NULL or INFINITY
iteration of precedenting char
finally
\1 show all chars found between \( and \)

ge:
echo aaa bbb 222 333 | sed -e 's/\(.*\) \(.*\) \(.*\) /\4 \1 \3 \2/'
prints
333 aaa 222 bbb

learn regular expression, it's a MUST in *nix
nota: they are used in the same manner in vi, sed, awk, grep, perl

excuse typos:

\) closes \(
...
s/\(.*\) \(.*\) \(.*\) /
should be
s/\(.*\) \(.*\) \(.*\) \(.*\)/

okay.. great thanks. I figured out most of it, but I was not sure why you were escaping the first dot.. now I know.

Some of the regular expression stuff I know, but just basic stuff. I love the power of vi and your posts have helped me with a lot of things I am going to do.. thank you!

I am planning on studying more on regular expressions, just have not gotten to that item on my list yet

Oh, and I forgot to add that I had also changed the command to this :