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Word replacement using regexp in vi or vim

Hi.

Suppose I have a file containing the lines:

ID_SalesID mSalesId;
ID_RefID mRefId;
ID_CouponId mCouponId;

What pattern/regexp substitute command (in vi/vim) can I use such that I'm left with

mSalesId(NULL),
mRefId(NULL),
mCouponId(NULL),

Thanks.;

Hugh McCurdy

Two commands which I'll attach as "code" so fixed spacing will be used.

:1,$s/^.* m/m/
:1,$s/;/(NULL),/

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HonorGod

The

:1,$

could be replaced by

%

Superdave

%s/^$[ ^I]*$[^ ]* $[^;]*$;\1\2(NULL),/

Where the ^I 9 characters in is actually a tab.
This will do it in one expression and allow and preserve indenting.

ASKER CERTIFIED SOLUTION

Harisha M G

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Harisha M G

The above works this way:
%s - Search all lines
/ - delimiter
.* - find anything
\s\+ - followed by at least one space
$\S*$ - and then a word not containing space, which is group 1 (\1)
; - followed by a semi-colon
/ - delimiter
\1 - group 1
(NULL), - text, verbatim
/ - delimiter

And if you want to preserve the indentation:

%s - Search all lines
/ - delimiter
^ - from beginning of line
$\s*$ - find the spaces and call it group 1 (\1)
.* - then anything
\s\+ - followed by at least one space
$\S*$ - then a word and call it group 2 (\2)
; - followed by a semi-colon
/ - delimiter
\1\2 - group 1, group2
(NULL), - text, verbatim

%s/^\(\s*\).*\s\+\(\S*\);/\1\2(NULL),/

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oliversk

ASKER

Just what I was looking for.