Yrurtj

If you have a reasonably short file grep alone might work:

grep -A5000 -m1 -e 'dog 123 4335' animals.txt

5000 is just my guess at "reasonably short", as grep finds the first match and outputs it together with the next 5000 lines (the file doesn't need to have that many). If you don't want the match itself you'll need to cut it off, e.g.

grep -A5000 -m1 -e 'dog 123 4335' animals.txt | tail -n+2

tac animals.txt | sed -e '/dog 123 4335/q' | tac

This line reads animals.txt in reverse order of lines and outputs up to and including the line with dog 123 4335 and then reverses again to restore proper order.

Again, if you don't need the match in the result, append tail. (You could also complicate the sed expression to discard its buffer before quitting.)

Assuming you want to match the whole line with your pattern, with GNU sed, this works:

sed -n '/^dog 123 4335$/ { :a; n; p; ba; }' infile

Standard equivalent:

sed -ne '/^dog 123 4335$/{:a' -e 'n;p;ba' -e '}' infile

With the following input (infile):

cat 13123 23424 

deer 2131 213132

bear 2313 21313

dog 123 4335

cat 13123 23424 

deer 2131 213132

bear 2313 21313

The output is:

cat 13123 23424 

deer 2131 213132

bear 2313 21313

Explanation:

• 
  /^dog 123 4335$/ searches for the desired pattern.


• 
  :a; n; p; ba; is a loop that fetches a new line from input (n), prints it (p), and branches back to label a :a; ...; ba;.

Update

Here's an answer that comes closer to your needs, i.e. pattern in file2, grepping from file1:

tail -n +$(( 1 + $(grep -m1 -n -f file2 file1 | cut -d: -f1) )) file1

The embedded grep and cut find the first line containing a pattern from file2, this line number plus one is passed on to tail, the plus one is there to skip the line with the pattern.

If you want to start from the last match instead of the first match it would be:

tail -n +$(( 1 + $(grep -n -f file2 file1 | tail -n1 | cut -d: -f1) )) file1

Note that not all versions of tail support the plus-notation.

In practice I'd probably use Aet3miirah's answer most of the time and alexey's answer is wonderful when wanting to navigate through the lines (also, it also works with less). OTOH, I really like another approach (which is kind of the reversed Gilles' answer:

sed -n '/dog 123 4335/,$p'

When called with the -n flag, sed does not print by default the lines it processes any more. Then we use a 2-address form that says to apply a command from the line matching /dog 123 4335/ until the end of the file (represented by $). The command in question is p, which prints the current line. So, this means "print all lines from the one matching /dog 123 4335/ until the end."

sed -e '1,/dog 123 4335/d' file1

If you need to read the pattern from a file, substitute it into the sed command. If the file contains a sed pattern:

sed -e "1,/$(cat file2)/d" file1

If the file contains a literal string to look for, quote all special characters. I assume the file contains a single line.

sed -e "1,/$(sed 's/[\/^$.*]/\&/g' file2)/d" file1

If you want the match to be the whole line, not just a substring, wrap the pattern in ^…$.

sed -e "1,/^$(sed 's/[\/^$.*]/\&/g' file2)$/d" file1

$ more +/"dog 123 4335" file1

With awk:

awk 'BEGIN {getline pattern < "other file"}

   NR == 1, $0 ~ pattern {next}; {print}' < "input file"

If the input is an lseekable regular file:

With GNU grep:

{ grep  -xFm1 'dog 123 4335' >&2

  cat; } <infile 2>/dev/null >outfile

With sed:

{ sed -n '/^dog 123 4335$/q'

  cat; } <infile >outfile

A GNU grep called w/ the -m option will quit input at the match - and it will leave its (lseekable) input fd immediately after the point it found its last match. So calling grep w/ -m1 finds the first occurrence of a pattern in a file, and leaves the input offset at precisely the right place for cat to write everything following the pattern's first match in a file to stdout.

Even without a GNU grep you can do the exact same thing w/ a POSIX compatible sed - when sed quits it is specified to leave its input offset right where it does. GNU sed is not standards-compliant in this way, though, and so the above will likely not work w/ a GNU sed unless you call it with its -u switch.

One way using awk:

awk 'NR==FNR{a[$0];next}f;($0 in a){f=1}'  file2 file1

where file2 contains your search patterns. First, all the contents of file2 are stored in the array "a". When the file1 is processed, every line is checked against the array, and printed only if is not present.

My answer for the question in the subject, without storing pattern in a second file.
Here is my test file:

$ cat animals.txt 

cat 13123 23424 

deer 2131 213132

bear 2313 21313

dog 123 4335

cat 13123 23424 

deer 2131 213132

bear 2313 21313

GNU sed:

 $ sed '0,/^dog 123 4335$/d' animals.txt 

 cat 13123 23424 

 deer 2131 213132

 bear 2313 21313

Perl:

$ perl -ne 'print unless 1.../^dog 123 4335$/' animals.txt

cat 13123 23424 

deer 2131 213132

bear 2313 21313

Perl variant with pattern in a file:

$ cat pattern.txt 

dog 123 4335

$ perl -ne 'BEGIN{chomp($p=(<STDIN>)[0])};print unless 1../$p/;' animals.txt < pattern.txt

cat 13123 23424 

deer 2131 213132

bear 2313 21313

How to print all lines after a match up to the end of the file?

Another way to put it is "how to delete all lines from the 1st one till the match (including)", and this can be sed-written as:

sed -e '1,/MATCH PATTERN/d'

Wth ed:

ed -s file1 <<< $'/dog 123 4335/+1,$p'

This sends one print command to ed in a here-string; the print command is limited in range to one after (+1) the dog 123 4335 match until the end of the file ($).

Since awk isn't expressly disallowed, here's my offering assuming 'cat' is the match.

awk '$0 ~ /cat/ { vart = NR }{ arr[NR]=$0 } END { for (i = vart; i<=NR ; i++) print arr[i]  }' animals.txt