You got it - almost - right, but still wrong. ;-))

Here is how "variable expansion" (this is how the process is called) works:

if you prepend a string with the dollar sign ("$"), the string will be interpreted as name of a variable and the name will be replaced by the content of the variable. Example:

Code:

echo $myvar

This will replace the string "$myvar" with the content of the variable "myvar" and then execute the echo-command with it.

Note that you do NOT need this "expansion" when you assign a variable! Hence:

Code:

myvar="bla bla"
echo $myvar

Also notice, that this expansion notation is a shortcut: the complete construct is to enclose the name in curly braces. The following illustrates this:

Code:

myvar="this is a test"
echo $myvar
echo ${myvar}

Now, that leaves the question of how to assign one variable with the content of another. Can you guess it?

Code:

myvar="this is myvar"
newvar="$myvar"
echo $newvar
echo ${newvar}
newervar="${myvar}"
echo $newervar
echo ${newervar}

A word about quoting: you will notice that i enclosed the assignment into double quotes. This is because the shell automatically splits at word boundaries. This would be what happens without quoting:

Code:

myvar=multiple words

The shell would assign "multiple" to the variable "myvar" and be left over with the word "words", which would lead to an error message, because it simply makes no sense. The same with

Code:

newvar=$myvar

First the shell would replace "$myvar" with the content of the variable "myvar" ending with this line:

Code:

newvar=multiple words

and here what i said above applies. This is why you need quoting, which is a bigger topic in itself, so i am cutting it short here. Just ask if you would like to know more about this.

Also notice that expansion is not limited to simple expanding to the unaltered content. It is also possible to change on the expanded content of the variable:

Code:

newvar=${myvar}
echo ${myvar#????????}      # cut off the first 8 characters
echo ${myvar%??????}        # cut off the last 6 characters
echo ${myvar%??????}        # cut off the last 6 characters
echo ${myvar/words/WORDS}  # replace a certain part with something else

and so on. I suggest a good book about shell programming for this.

I hope this helps.

bakunin