使用sed用bash变量替换整行

Question

I want to use sed to replace the entire line at a match with a bash variable. Replacing with plain text is fine.

$ echo -e "foo bar\nfooy bary\nfoot bart" | sed '/fooy/c\text'
foo bar
text
foot bart

If the text I want to replace with is stored in a bash variable, however, this does not work.

$ var="This is poetry."
$ echo -e "foo bar\nfooy bary\nfoot bart" | sed "/fooy/c\$var"
foo bar
$var
foot bart

How can I achieve this?

Answer 1

If you want to do this with sed (which you shouldn't, see below), use

sed "/fooy/c\\$var"

This is a shell expansion problem. \\$ in a doubly quoted string means a literal dollar, so $var is not expanded. GNU sed also accepts

sed "/fooy/c$var"

as long as you don't specify --posix , but BSD sed (as you may have on FreeBSD or MacOS X) will complain about it.

However, doing this with sed is not a good idea. Substituting shell variables into sed code is almost never a good idea, because it makes you vulnerable to code injection. Take, for example,

$ var=$(echo -e 'something\nd')
$ echo -e "foo bar\nfooy bary\nfoot bart" | sed --posix "/fooy/c\\$var"
something

That wasn't what we expected, was it? What happens here is that sed , since it is unable do distinguish what you want it to treat as data ( $var ) and code, treats $var as code, and therefore it treats the d after the \\n in it as a command (to delete the current line). A better way is to use awk :

$ var=$(echo -e 'something\nd')
$ echo -e "foo bar\nfooy bary\nfoot bart" | awk -v var="$var" '/fooy/ { print var; next } 1'
foo bar
something
d
foot bart

This sidesteps the code injection issue by not treating $var as code.

Answer 2

Don't escape the $ symbol. In bash, variables are referred by $variable-name . If you do escaping the $ symbol, then the bash won't consider it as a variable. It would treat $var as literal $var .

$ echo -e "foo bar\nfooy bary\nfoot bart" | sed "/fooy/c$var"
foo bar
This is poetry.
foot bart