I'm looking for a sed solution like the one here: Print everything after nth delimiter
sed -E 's/^([^\]*[\]){3}//' infile
but to extract the text before nth delimiter, instead of nth delimiter like the example. Something that works with all sed variants. And operate on all lines like the example.
The delimiter in this example is \ but could be for any other. Should work with any version of sed.
Why don't you use cut?
cut -d '\' -f 1-3 infile
With sed, instead of deleting the match, capture it and use a back-reference to replace the whole line with the captured group:
sed -E 's/(^([^\]*[\]){3}).*/\1/' infile
Though that would also print the trailing backslash... To avoid that you could run
sed -E 's/(^([^\]*[\]){2}[^\]*).*/\1/' infile
a shorter awk:
awk NF=3 FS='\\' OFS='\\'
NF is undefined behavior per POSIX so don't count on it doing any one thing across all awk variants. In most awks increasing NF will create empty fields at the the end (same as $NF=$7 to increase NF to 7 which IS portable to all awks so there's no point doing NF=7), in some awks decreasing it will remove fields from the end, in other awks decreasing it will do nothing. There's no portable way to decrease NF other than sub() or match() to remove some number of fields.
awk 'NF=3,$3=$3' ... seems to work. do you know any other implementations that don't truncate?
--posix or --re-intervals options. Oh, and I also exclude a script that does gsub(/^"|"$/,"") from "any awk" as that fails in mawk1 and tawk.
You could replace the nth delimiter with newline (which cannot otherwise occur in the pattern space) and then delete everything starting with that newline. Here for n == 3:
sed 's/delim/\
/3; P; d'
Or if the nth delimiter must be retained in the output:
sed 's/delim/&\
/3; P; d'
To skip the lines that don't have n delimiters:
sed -n 's/delim/\
/3; t1
d; :1
P'
P;d instead of s/\n.*//
.* choking on non-characters.
Using awk:
$ awk -v var=3 'BEGIN{FS=OFS="\\"}
(NF>=var){ split($0,arr,OFS);
$0="";
for (i=1; i<=var; ++i) $(NF+1)=arr[i];
print}'
To keep nth delimiter the following command may be used.
$ awk -v var=3 'BEGIN{FS=OFS="\\"}
(NF>=var){ for (i=1; i<=var; ++i) printf "%s%s", $i, OFS; print ""}'
$ nawk '(match($0, /^([^\\]*[\\]){3}/))
{ print substr($0,RSTART,RLENGTH)}'
With GNU awk:
The following command uses back-referencing a captured group. This is an awk command taken from this answer. Thanks to @don_crissti
$ awk -F "\\" -v col=3 '(NF>=col){print gensub(/(^([^\\]*[\\]){3}).*/, "\\1", "g")}'
sedimplementation you are using, ii) show us an example input file and iii) the output you expect. We cannot help you parse data you don't show. You mention a delimiter, what is that delimiter? Should we guess you have `` as the delmiter?sed -E 's/(^([^\]*[\]){3}[^\]*).*/\1/' infile. You can replace "\" with a letter like "e" or a number and keeps running:sed -E 's/(^([^e]*[e]){3}[^e]*).*/\1/' infile. So you are not willing to put the effort to elaborate your answer dude.