BASH script help

[hellomoto]

I am trying to grasp some basic BASH scripting, im getting a bit confused regarding variables.

VAR="uptime"
$VAR

This would assign the variable as uptime the print the uptime.
$3 hours, 15 minutes

VAR="uptime"
echo "Your computer has been on for $VAR"

I would expect this to output as follows:
$Your computer has been on for 3 hours, 15 minutes

However what I get is:
$Your computer has been on for uptime

Would I be right in saying $VAR is being read as a "string" and not as a "variable". If this is the case how do I change it to be read as a variable?

[CharlesA]

Put it in the varable like so:

VAR=$(uptime)

The one you have set in the OP will just echo uptime to the terminal, not run the command.

The only problem is that uptime prints out like this:

charles@atlantis:~$ uptime
 12:42:21 up 1 day,  5:40,  1 user,  load average: 0.00, 0.00, 0.00

[hellomoto]

this fixed the problem, thankyou

[SeijiSensei]

VAR=`uptime`

Using the "backtick" operator (to the left of 1 in the upper row of the keyboard) evaluates the expression inside the backticks and passes the result to VAR. An alternative to using the backticks is an expression like this:

VAR=$(uptime)

[CharlesA]

VAR=$(uptime)

That is the preferred way, since it's easier to nest than backticks.

[honeybear]

VAR=`uptime`

Using the "backtick" operator (to the left of 1 in the upper row of the keyboard) evaluates the expression inside the backticks and passes the result to VAR. An alternative to using the backticks is an expression like this:

VAR=$(uptime)

the

 VAR=`uptime`

is usually adviced, because it is more powerful"" and less messy

[honeybear]

Question:

VAR=`cat FILE`
if [ "$VAR" != "" ] ; then

or

if [ "`cat FILE`" != "" ] ; then

which one is better/working ?

[Vaphell]

imo it's the same in this case
if i am not mistaken `` is less desirable when you want to process the output of a command in multiple steps - `` is like macro so in reality it's the command itself substituted for the variable, thus making it that each time variable is used, the command is run.
$(...) on the other hand stores the output in a variable, so the command is only run once. I may be mistaken though so take that with a grain of salt.

[CharlesA]

See the thread here (and links therein) for more info as to why $(...) is prefered to backticks.