sed join lines together - regex

Question

what would be the sed (or other tool) command to join lines together in a file that do not end w/ the character '0'?
I'll have lines like this
412|n|Leader Building Material||||||||||d|d|20||0
which need to be left alone, and then I'll have lines like this for example (which is 3 lines, but only one ends w/ 0)
107|n|Knot Tying Tools|||||Knot Tying Tools
|||||d|d|0||0
which need to be joined/combined into one line
107|n|Knot Tying Tools|||||Knot Tying Tools|||||d|d|0||0

Answer 1

sed ':a;/0$/{N;s/\n//;ba}'
In a loop (branch ba to label :a), if the current line ends in 0 (/0$/) append next line (N) and remove inner newline (s/\n//).
awk:
awk '{while(/0$/) { getline a; $0=$0 a; sub(/\n/,_) }; print}'
Perl:
perl -pe '$_.=<>,s/\n// while /0$/'
bash:
while read line; do
if [ ${line: -1:1} != "0" ] ; then
echo $line
else echo -n $line
fi
done

Answer 2

awk could be short too:
awk '!/0$/{printf $0}/0$/'
test:
kent$ cat t
#aasdfasdf
#asbbb0
#asf
#asdf0
#xxxxxx
#bar
kent$ awk '!/0$/{printf $0}/0$/' t
#aasdfasdf#asbbb0
#asf#asdf0
#xxxxxx#bar

Answer 3

The rating of this answer is surprising ;s (this surprised wink emoticon pun on sed substitution is intentional) given the OP specifications: sed join lines together.
This submission's last comment
"if that's the case check what #ninjalj submitted"
also suggests checking the same answer.
ie. Check using sed ':a;/0$/{N;s/\n//;ba}' verbatim
sed ':a;/0$/{N;s/\n//;ba}'
does
no one
ie. 0
people,
try
nothing,
ie. 0
things,
any more,
ie. 0
tests?
(^D aka eot 004 ctrl-D ␄ ... bash generate via: echo ^V^D)
which will not give (do the test ;):
does no one ie. 0
people, try nothing, ie. 0
things, any more, ie. 0
tests? (^D aka eot 004 ctrl-D ␄ ... bash generate via: echo ^V^D)
To get this use:
sed 'H;${z;x;s/\n//g;p;};/0$/!d;z;x;s/\n//g;'
or:
sed ':a;/0$/!{N;s/\n//;ba}'
not:
sed ':a;/0$/{N;s/\n//;ba}'
Notes:
sed 'H;${x;s/\n//g;p;};/0$/!d;z;x;s/\n//g;'
does not use branching and
is identical to:
sed '${H;z;x;s/\n//g;p;};/0$/!{H;d;};/0$/{H;z;x;s/\n//g;}'
H commences all sequences
d short circuits further script command execution on the current line and starts the next cycle so address selectors following /0$/! can only be /0$/!! so the address selector of
/0$/{H;z;x;s/\n//g;} is redundant and not needed.
if a line does not end with 0 save it in hold space
/0$/!{H;d;}
if a line does end with 0 save it too and then print flush (double entendre ie. purged and lines aligned)
/0$/{H;z;x;s/\n//g;}
NB ${H;z;x;s/\n//g;p;} uses /0$/ ... commands with an extra p to coerce the final print and with a now unnecessary z (to empty and reset pattern space like s/.*//)

Answer 4

A typically cryptic Perl one-liner:
perl -pe 'BEGIN{$/="0\n"}s/\n//g;$_.=$/'
This uses the sequence "0\n" as the record separator (by your question, I'm assuming that every line should end with a zero). Any record then should not have internal newlines, so those are removed, then print the line, appending the 0 and newline that were removed.
Another take to your question would be to ensure each line has 17 pipe-separated fields. This does not assume that the 17th field value must be zero.
awk -F \| '
NF == 17 {print; next}
prev {print prev $0; prev = ""}
{prev = $0}
'

Answer 5

if ends with 0 store, remove newline..
sed '/0$/!N;s/\n//'