How to use a shell variable inside sed's s command?

Question

we know that we can get the value of a variable using $ sign:

x=3
echo $x
3

Is there other ways that we can use to get the value without using $ sign.

I'm asking this because $ sign is a special character to sed, and I get sed: -e expression #1, char 31: unterminated s' command error when trying to do this:

A=25 #lineNumber
destStr="192.168.1.3 192.168.1.4"
sed ""$A"s/Allow from .*/Allow from $destStr/" -i test2

and it prints the string destStr(not the value) when I use the escape character \:

sed ""$A"s/Allow from .*/Allow from \$destStr/" -i test2

using echo instead of sed:

25s/Allow from .*/Allow from  192.168.1.3
192.168.1.4
/ -i test2

Solution 1

As the final edits revealed, the problem is unrelated to the dollar sign, but is caused by the content of deststr, which is not 192.168.1.3 192.168.1.4 but rather two lines, one containing only 192.168.1.3 and the other containing only 192.168.1.4, both lines bveing terminated with a newline character. That is, the actual command after variable replacement is:

sed "25s/Allow from .*/Allow from  192.168.1.3
192.168.1.4
/" -i test2

Now sed interprets its command line by line, and thus the first command it tries to interpret is:

25s/Allow from .*/Allow from  192.168.1.3

which clearly is an unterminated s command, and thus reported by sed as such.

Now the solution you found, using echo, works because

echo $var

calls echo with two arguments (because the whitespace is not quoted, it is interpreted as argument delimiter), the first one being 192.168.1.3 and the second one being 192.168.1.4; both are forms that are not interpreted further by the shell.

Now echo just outputs its (non-option) arguments separated by a space, therefore you now get as command line:

sed "25s/Allow from .*/Allow from  192.168.1.3 192.168.1.4/" -i test2

as intended.

Note however that for command substitution, instead of backticks you should use $() whereever possible, since it's too easy to get backticks wrong. Therefore the follwing does what you want:

sed "$A s/Allow from .*/Allow from $(echo $destStr)/" -i test2

Note that I also took advantage of the fact that sed allows a space between address and command, to simplify the quoting. In situations where such an extra space is not possible, you can also use the following syntax:

sed "${A}s/Allow from .*/Allow from $(echo $destStr)/" -i test2

Also note that this relies on the fact that the non-space characters in destStr are interpreted neither by the shell, nor by sed if occurring in the replacement string.

Solution 2

I have managed to solve my problem using echo with ``:

sed ""$A"s/Allow from .*/Allow from `echo $destStr`/" -i test2

but I'm still want to know if there are other ways to get the value without using $ sign.

Solution 3

This is how you can do it within the script:

sed -e "${A}s/Allow from .*/Allow from ${destStr}/" -i test2

Using ${variable} let the bash interpret ${variable}s/... like you want it to.

Solution 4

Attempting to actually answer the stated question: "Is there other ways that we can use to get the value without using $ sign."

This isn't in the context of your use case, but you could do something like this:

$SOME_VAR="some value or something"
f() { echo ${!1}; }
some_command `f "SOME_VAR"`

You could use this to avoid using $ in the argument list for sed.

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Comments

[celtschk]

Which shell are you using? Your first version works just fine for me using bash.

[Nidal]

@celtschk, I'm using bash too.

[celtschk]

Do you get the error with exactly the values you've given? Especially: Does destStr have the literal value string?

[celtschk]

What is the output if you replace sed with echo (but leave the arguments unchanged)?

[Nidal]

@celtschk,see Updates,so I guess my problem was : the value contains "\n".

[celtschk]

So I get that you did not set destStr with the command above, but got it from somewhere else, and whereever you got it from inserted newline characters in between. That's your real problem (it is completely unrelated to the dollar sign), and that's also why your echo solution works: echo replaces that line feed with a simple space.

[Nidal]

it didn't work too, both ways didn't work