Perl或Sed命令在匹配表达式之前删除内容

Question

I have a directory of html files in which I'd like to remove all content (and including) a matched expression. I have tried this (where XXXXXX us the string I am looking for:

perl -pi -e "$a++if s/.*XXXXXX.*//si;$a||s/.*//s" *.html

I am getting this error (where when I press enter, it is changing my command, see first line):

perl -pi -e "$a=1 if *.htmla && s/.*XXXXXX.*//is; s/.*//s if *.htmla" *.html
Bareword found where operator expected at -e line 1, near "*.htmla"
    (Missing operator before htmla?)
Bareword found where operator expected at -e line 1, near "*.htmla"
    (Missing operator before htmla?)
syntax error at -e line 1, near "="
syntax error at -e line 1, near "*.htmla
"
Execution of -e aborted due to compilation errors.

Any help on Perl command to use?

Answer 1

Try using

'

instead of

"

otherwise your shell will expand $a.

perl -pi -e "$a++if s/.*XXXXXX.*//si;$a||s/.*//s" *.html

should do the trick.

Answer 2

If you want to delete all lines up to a line, which matches some regular expression.

You can give a range to sed and delete from the first line until and including the match

sed -i -e '1,/match/d' file.html

If you want to remove the content from every line, which matches a regular expression

sed -i -e 's/.*match//' file.html

Answer 3

Quite with ' instead of " .

Variables are interpolated in " -quoted material; shells do this and so does Perl.

In this case, your shell which will substitute $a with its value before feeding the result to Perl. The result is illegal Perl syntax.