Replacing Carriage returns without loosing EOL


 
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# 1  
Old 03-07-2008
Power Replacing Carriage returns without loosing EOL

Hello, I have read a few threads on this subject and tried a few things out, but still come up short.

There was one good example, then the last reply was something to the effect of 'Use Sed' & 'Read a book'...

Well I read a bunch of online tutorials on sed, awk, tr, but still can't get the right results.

input file>

1234|hello there buddy|8beers|2008-01-01
1234|this is a nice party, don't you think?|2champaigns|2008-01-01
1234|there are two things:
thing 1 & thing2|36shots|2008-01-01

desired output>
1234|hello there buddy|8beers|2008-01-01
1234|this is a nice party, don't you think?|2champaigns|2008-01-01
1234|there are two things: thing1 & thing2|36shots|2008-01-01

The closest I could get was with this command->

nawk '{ if ( $0 ~ /1234/ && NR > 1 ) { printf "\n"; } printf $0; } END { printf "\n"; }' party.txt > dump.txt

But after line 460664 I received an error:
awk: There are not enough parameters in printf statement ...

The only thing that looks different about that line is there is an ampersand '&' in one of the fields.

I tailed the rest of the file into the the nawk and it puked after another 1600 lines. I tried a couple more times and it still keeps running out of steam before the millionth row.

The other way I tried was ghetto style:
tr -d '\n' to remove all the carriage returns, then
sed s/2008-01-01/2008-01-01\n/g <party.txt> dump.txt

but sed pukes with a sed: Memory allocation failed after about a minute.

Ideally, I need a one liner...
Any suggestions?

Thanks.
# 2  
Old 03-07-2008
Try this:

Code:
nawk 'NR==1 || $0 !~ /^1234/{printf("%s",$0)} $0 ~ /^1234/{printf("\n%s",$0)} END{print ""}' party.txt > dump.txt

Regards
# 3  
Old 03-08-2008
Bug Beautiful

Thanks bud, this worked beautifully through 16 million lines...
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