How to do this?

Jerrygreat

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Nov 21, 2008, 1:40:01 PM11/21/08

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Hello, friends,
I have a huge file like:

dbv file=/sja7/appl/oracle/ship/system01.dbf feedback=100 logfile=./.log
blocksize=8192
dbv file=/sja8/appl/oracle/ship/undotbs.dbf feedback=100 logfile=./.log
blocksize=8192
dbv file=/sja7/appl/oracle/ship/tools.dbf feedback=100 logfile=./.log
blocksize=8192
dbv file=/sja7/appl/oracle/ship/dtuser_01.dbf feedback=100 logfile=./.log
blocksize=8192
......

how can I change it to like bleow by using one command line

dbv file=/sja7/appl/oracle/ship/system01.dbf feedback=100
logfile=./system01.log blocksize=8192
dbv file=/sja8/appl/oracle/ship/undotbs.dbf feedback=100
logfile=./undotbs.log blocksize=8192
dbv file=/sja7/appl/oracle/ship/tools.dbf feedback=100 logfile=./tools.log
blocksize=8192
dbv file=/sja7/appl/oracle/ship/dtuser_01.dbf feedback=100
logfile=./dtuser_01.log blocksize=8192
.............

Thank you very much

Jerry
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xulxer

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Nov 21, 2008, 5:05:08 PM11/21/08

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Hi,

if you want to split the line in front of all logfile netries and join
this bit with all blocksize entries on a new line on a unix box, you can do:

%s~$logfile.*$\n$.*$~\r\1 \2~g

I don't know if you have to use \n instead \r on windows.

Kind regards

Chris

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William Anderson

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Nov 21, 2008, 5:09:45 PM11/21/08

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Jerry,

This should be pretty easy to accomplish with a simple regex. Something
like

%s#^$dbv file=/sja./appl/oracle/ship/$$.\+$$.dbf .*logfile=$[^
]*#\1\2\3./\2.log#

Bill

StarWing

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Nov 21, 2008, 5:45:00 PM11/21/08

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my idea, combine xulxer and william's

:%s#$/oracle/ship/$$.*$$.dbf .*$$logfile=$.\{-}\n$.*$#
\1\2\3\r\4./\2.log \5#g

On 11月22日, 上午6时09分, William Anderson <whand...@lasthonorableman.net>
wrote:

Cesar Romani

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Nov 21, 2008, 7:03:52 PM11/21/08

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:g/logfile/s//\r&/ | normal J

Regards,
Cesar

Tony Mechelynck

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Nov 21, 2008, 10:22:10 PM11/21/08

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On 21/11/08 23:05, xulxer wrote:
> Hi,
>
> if you want to split the line in front of all logfile netries and join
> this bit with all blocksize entries on a new line on a unix box, you can do:
>
> %s~$logfile.*$\n$.*$~\r\1 \2~g
>
> I don't know if you have to use \n instead \r on windows.
>
> Kind regards
>
> Chris

In all versions of Vim, you _match_ a line break with \n in the first
half of a :subst command, you _insert_ one with \r in the second half.

This means that one silly way to count lines (by replacing line breaks
with themselves) is

:%s/\n/\r/

Best regards,
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Agathoklis D. Hatzimanikas

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Nov 23, 2008, 4:33:16 AM11/23/08

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On Fri, Nov 21, at 02:45 StarWing wrote:
>
> my idea, combine xulxer and william's
>
> :%s#$/oracle/ship/$$.*$$.dbf .*$$logfile=$.\{-}\n$.*$#
> \1\2\3\r\4./\2.log \5#g

While regular expressions can do wonders, here is yet
another way to solve your problem using VimL:

let line = 1
while line < line('$')
let fline = split(getline(line))[:2]
let logfile = fnamemodify(fline[1], ":t:r")
call setline(line, join(fline))
call setline((line + 1), 'logfile=./'.logfile.'.log '.getline((line + 1)))
let line += 2
endwhile

Regards,
Ag.

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