Perl one liner用于在变量文件中查找和替换

Question

I'm trying to find <li ><a href='xxxxxxxx'>some_link</a></li> and replace it with nothing. To do this, I'm running the command below but it's recognizing $ as part of a regex.

perl -p -i -e 's/<li ><a href=.*$SOMEVAR.*li>\\n//g' file.html

I've tried the following things,
${SOMEVAR}
\\$SOMEVAR
FIND="<li ><a href=.*$SOMEVAR.*li>"; perl -p -i -e 's/$FIND//g' file.html

Any ideas? Thanks.

Answer 1

Bash only does variable substitution with double quotes.

This should work:

perl -p -i -e "s/<li ><a href=.*?$SOMEVAR.*?li>\n//g" file.html

EDIT Actually, that might act weird with the \\n in there. Another approach is to take advantage of Bash's string concatenation. This should work:

perl -p -i -e 's/<li ><a href=.*?'$SOMEVAR'.*?li>\n//g' file.html

EDIT 2: I just took a closer look at what you're trying to do, and it's kind of dangerous. You're using the greedy form of .* , which could match a lot more text than you want. Use .*? instead. I updated the above regexes.

Answer 2

如果“SOMEVAR”确实是一个外部变量，您可以将其导出到环境并引用它：

SOMEVAR=whatever perl -p -i -e 's/<li ><a href=.*$ENV{SOMEVAR}.*li>\n//g' file.html