grep -v: How to exclude only the first (or last) N lines that match?

Another possible solution is to use bashs own utilities:

count=1
found=0
cat execute-commons-fileupload.sh | while read line
do 
   if [[ $line == *"myPattern"* ]]
   then 
      if [ $found -eq $count ]
      then 
         echo "$line"
      else 
         found=$(($found+1))
      fi
   else 
     echo "$line"
   fi
done

By setting count, you can change the count of occurences of your pattern which you want to remove.

For me personally, this seems to be easier extendable, since you can easily add other conditions to the if statement (but this might be caused by my marginal knowledge of sed).

To do this you might have to use awk.

The simple way I know is this:

cat file | awk '{ $1=""; print}'

You can skip multiple columns too:

cat file | awk '{ $1=$2=$3=""; print}'

If you want to skip the last column and you're not sure how much columns you will have:

cat file | awk '{ $NF=""; print}'

Tested on Ubuntu 16.04 (GNU bash, version 4.3.48)

Best.

You could use awk to ignore the first n lines that match (e.g. assuming you wanted to remove only the 1st and 2nd match from the file):

n=2
awk -v c=$n '/PATTERN/ && i++ < c {next};1' infile

---

To ignore the last n lines that match:

awk -v c=${lasttoprint} '!(/PATTERN/ && NR > c)' infile

where ${lasttoprint} is the line number of the nth+1 to last match in your file. There are various ways to get that line no. (e.g. print only the line number for each match via tools like sed/awk, then tail | head to extract it)... here's one way with gnu awk:

n=2
lasttoprint=$(gawk -v c=$((n+1)) '/PATTERN/{x[NR]};
END{asorti(x,z,"@ind_num_desc");{print z[c]}}' infile)

Perhaps reduce the chances of filtering out your data by using a more accurate grep command. For example:

grep -v -F -x 'str1'

For lines that are exatctly str1. Or maybe:

grep -v '^str1.*str2$'

For lines that start with 'str1' and end with 'str2'.

sed provides a simpler way:

... |  sed '/some stuff/ {N; s/^.*n//; :p; N; $q; bp}' | ...

This way you delete first occurrence.

If you want more:

sed '1 {h; s/.*/iiii/; x}; /some stuff/ {x; s/^i//; x; td; b; :d; d}'

, where count of i is count of occurrences (one or more, not zero).

Multi-line Explanation

sed '1 {
    # Save first line in hold buffer, put `i`s to main buffer, swap buffers
    h
    s/^.*$/iiii/
    x
}

# For regexp what we finding
/some stuff/ {
    # Remove one `i` from hold buffer
    x
    s/i//
    x
    # If successful, there was `i`. Jump to `:d`, delete line
    td
    # If not, process next line (print others).
    b
    :d
    d
}'

In addition

Probably, this variant will work faster, 'cos it reads all rest lines and print them in one time

sed '1 {h; s/.*/ii/; x}; /a/ {x; s/i//; x; td; :print_all; N; $q; bprint_all; :d; d}'

As result

You can put this code into your .bashrc (or config of your shell, if it is other):

dtrash() {
    if [ $# -eq 0 ]
    then
        cat
    elif [ $# -eq 1 ]
    then
        sed "/$1/ {N; s/^.*n//; :p; N; $q; bp}"
    else
        count=""
        for i in $(seq $1)
        do
            count="${count}i"
        done
        sed "1 {h; s/.*/$count/; x}; /$2/ {x; s/i//; x; td; :print_all; N; $q; bprint_all; :d; d}"

    fi
}

And use it this way:

# Remove first occurrence
cat file | dtrash 'stuff' 
# Remove four occurrences
cat file | dtrash 4 'stuff'
# Don't modify
cat file | dtrash