Skip file lines until a match is found, then output the rest

Similar to the answer of Avi, but without including the line with "LastHeaderLine".

sed -e '1,/LastHeaderLine/d'

Why not try awk for this? It would look like this:

awk 'NR == 1, /LastHeaderLine/ { next } { print }' myinputfile > myoutputfile

where NR == 1 is true for the first line, /LastHeaderLine/ matches your last header line. The comma operator lets the following function { next } fire for all sentences in the range of the two regular expression. In this case it will skip to the next line of input without further operation. For all other input lines it will print the lines to the standard output, which you can redirect using >.

Using sed:

sed -ne '/LastHeaderLine/,$p' <inputfile

will match everything from the regex match to the end of the file. 'p' prints the lines that match.

Edit:

On second thought, you don't want to print the line matching LastHeaderLine. This is difficult to do with sed. In perl, you could do the following:

perl -ne 'if ($flag) {print;} if (/LastHeaderFile/) {$flag=1;}' <inputfile

This would print only lines strictly following the regex match.