Delete the line with a sepific occurance pattern in SED?

Please spend a few second. i would like to delete the line which has the sixth <ul> occurance. i google the way of doing it and i tried

sed -i '/^<ul>$/ d/6' file .

However, it turn out a

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sed: -e expression #1, char 9: extra characters after 

error.

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AFAIK you can't do that easily with sed. You can do that like joining all the lines to one string, then delete that 6th occurence, then split the string, but it's easier with awk:

awk '/<ul>/ {ul++} ul == 6 { getline } 1' INPUTFILE > TMPFILE && mv TMPFILE INPUTFILE

Test results are here: https://ideone.com/3YpVu

Note: it discards the whole line with the 6th <ul>, and assumes that there are only one <ul> can be found per line.

HTH

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awk '/<ul>/ && ++count == 6 {next} {print}' filename

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perl -p -i -e "s/^<perl-regex-here>$\n//g" <filename> should get rid of any line matching the perl regular expression you give on any modern unix system. Otherwise, if you are counting occurences and kill the line with the 6th occurrence of , I know of no command that will do that in one line.