Bash -Extract characters from string

Question

Suppose I have the following variable

$string = 'cat somecommandhere'

Now the output of the above cat command is

test="Thanks serverfault"

I want to extract the value between the quotes and store it in the same variable $string, including the quotes. I want string to contain value below:

$string = "Thanks serverfault"  

How can this be done with sed or grep but not awk. I haven't learned awk yet.

Answer 1

I'll use cut, since it's fairly easy to deducate that -f2 means "the second part" and -d'"' means cut on the double quotation mark.

$ output='test="Thanks serverfault"'
$ string=$( echo $output | cut -f2 -d'"' )
$ echo $string
Thanks serverfault

If you really want the quotes, you can just add them back yourself, it just makes the answer look more confusing than it needs to be.

If you really want to use sed (lucky for you, this keeps the quotes too):

$ output='test="Thanks serverfault"'
$ string=$( echo $output | sed 's/^test=//' )
$ echo $string
"Thanks serverfault"

Explanation of sed:

                           s/^test=//
  string find/replace mode ^ |     ^ replace with nothing
                             starts with test=

Answer 2

No need for sed; bash and sh have variable modifiers built in:

$ output='test="Thanks serverfault"'
$ string=\"${output#*\"}
$ string=${string%\"*}\"
$ echo $string
"Thanks serverfault"

# removes prefixes, % is suffixes.

This should be faster than calling external programs.