What is this sed command doing?

Question

Can someone explain me what this sed command is trying to achieve?

sed 's/^[[:space:]]*//;s/[[:space:]]*$//

I understand that it is searching for whitespace characters from the beginning of the line, but I am not able to understand the part starting from //;s/[[:space:]]*$//

Answer 1

There are two s commands there: sed can take a list of commands seperated by semicolons. You could put the same thing in a script file like so:

s/^[[:space:]]*//
s/[[:space:]]*$//

Also, you're parsing the s command wrongly: s isn't search, it is substitute: it takes two arguments, the string to match and the string to replace it with (s/find/replace/).

So, s/^[[:space:]]*// means

s                (substitute)
/^[[:space:]]*/  (leading whitespace)
//               (with an empty string)

and the second command s/[[:space:]]*$// means substitute trailing whitespace with an empty string.

Answer 2

As far as I know it's just a substitute chain. First it removes all leading whitespace via s/^[[:space:]]// and it moves on to apply s/[[:space:]]$// to remove all trailing whitespace.