删除 SED 中具有特定出现模式的行？
请花几秒钟。我想删除第六次出现
sed -i '/^
$/ d/6' file 。然而，结果是
sed: -e expression #1, char 9: extra characters after
错误的。
Please spend a few second. i would like to delete the line which has the sixth <ul> occurance.
i google the way of doing it and i tried
sed -i '/^<ul>$/ d/6' file .
However, it turn out a
sed: -e expression #1, char 9: extra characters after
error.
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AFAIK 你不能用
sed轻松做到这一点。您可以这样做，例如将所有行连接到一个字符串，然后删除第 6 次出现，然后拆分字符串，但使用awk更容易：测试结果在这里：https://ideone.com/3YpVu
注意：它会丢弃第 6 个整行
，并假设每行只能找到一个。华泰
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 withawk: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
perl -p -i -e "s/^$\n//g" 应该删除与您给出的 perl 正则表达式匹配的任何行在任何现代 UNIX 系统上。否则，如果您正在计算出现次数并终止第 6 次出现 的行，我知道没有任何命令可以在一行中执行此操作。perl -p -i -e "s/^<perl-regex-here>$\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.