Shell Programming and Scripting

Hi all,

What am I doing wrong here?

Code:

$ cat test_sed.ksh
#!/usr/bin/ksh
var="sed -e \'6s/9/6/\' testfile.txt > testfile.txt.2"
$var
$ ./test_sed.ksh
sed: command garbled: \'6s/9/6/\'

Thank you!

Try:

Code:

eval $var

Though the output you might expect will not be in the variable, as the substitution is send to stdout and that is redirected to the file "testfile.txt.2". If you want both, the file written and the output of the sed on the screen, check out if you have the tee command available on your machine. In this case replace the line would look like:

Code:

var="sed -e '6s/9/6/' testfile.txt | tee testfile.txt.2"

Escaping the slashes is not necessary.

I am only interested in the file, I don't need the output on screen.

I guess I could do something like this:

Code:

function test123 {
#echo test
sed -e '6s/9/6/' testfile.txt > testfile.txt.2
}

test123

But I'd still like to know why using a variable with sed and escaping some characters is not working.

When you escaped the single quote characters you stopped Shell from interpreting them and forced them to be passed to sed - which did not know what to do with them in that context.

If you were using a sed command file, you would not have the single quotes because they are only for Shell in this context.

Thank you. I will use a sed command file.