Multiple commands in a variable

[ayjaycee]

OK, this one is doing my head in X-)

I want to pass (possible) multiple commands to a script, for it to execute. I'm using an option/value syntax :- "-c "command ; command ; command" to set a variable which I then try to execute later on.

Unfortunately, only the 1st command in the string is run & the rest are ignored. I've tried the following test to see what's happening :-

# cat test
#!/bin/sh
cmd="pwd ; uname -a"
pwd ; uname -a
$cmd
`$cmd`

which results in :-

# sh -vx ./test
#!/bin/sh
cmd="pwd ; uname -a"
cmd=pwd ; uname -a

pwd ; uname -a
+ pwd
/usr/local/bin/sysadm
+ uname -a
SunOS ncwcs36 5.5.1 Generic_103640-31 sun4d sparc SUNW,SPARCserver-1000

$cmd
+ pwd ; uname -a
/usr/local/bin/sysadm

`$cmd`
+ pwd ; uname -a
+ /usr/local/bin/sysadm
./test: /usr/local/bin/sysadm: cannot execute

so, issuing command ; command works OK,
issuing $cmd just runs the 1st command &
issuing `$cmd` runs the 1st command & generates an error msg :-0

Any ideas/hint/tips on this? TandA

http://www.tanda.f2s.com

Day by day, the penguins steal my sanity.

 

[ayjaycee]

It's amazing - I post the question & then solve it before getting an answer - must be picking up some psychic vibes

Doing eval $cmd seems to do the trick. TandA

http://www.tanda.f2s.com

Day by day, the penguins steal my sanity.

 

[crowe]

Also, you could put your commands into a little function and then just call the function.

COMMAND() {
pwd
uname -a
}

Then just call COMMAND from the command line.

crowe