passing a veriable in a command substitution

dwj79 · 04-01-2009, 10:03 AM

Hi,

im trying to do the following:

Code:

MYVAR="filename.php"
CMSUSER= $(stat --format=%U $MYVAR)

Anyway the command substitution statent doesn't seem to recognize the variable.
how can i make it work?

thanx in advance!

acid_kewpie · 04-01-2009, 10:12 AM

the substituion works fine, but you have a space after CMSUSER= which means you're making that value null, and then looking to execute the output of the stat command and another command in itself, also not making sense. just remove the space.

druuna · 04-01-2009, 10:14 AM

Hi,

There's a space between = and $(...... Remove the space:

CMSUSER= $(stat --format=%U $MYVAR)

vs

CMSUSER=$(stat --format=%U $MYVAR)

Hope this helps

dwj79 · 04-02-2009, 02:39 AM

Thanx,

im kinda newbie to bash :P and sometimes it's difficult to deal with such a strict syntax

thanx again!

acid_kewpie · 04-02-2009, 02:46 AM

It might seem strict, but it's not really, it's just illogical. As bash is a shell and not a programming language, each line is a command in it's own right, not a line of code. So whitespace has to matter in certain contexts in a way that a single line of java needn't, as you're actually splitting one operation into two different things with the whitespace.