Bash Script: form $(< file) ?

Jacky Quah · 10-28-2008, 08:13 AM

Bash Script: form $(< file)
I don't know the filename, the filename is stored in variable (say it $TEST), how can I do that ?

because $(< $TEST); show error...

burschik · 10-28-2008, 09:13 AM

Works for me. Does the file really exist? What error are you getting?

colucix · 10-28-2008, 09:25 AM

It works for me, too. Which is the error message? Also to debug your script you can try to run it as

Code:

bash -x script.sh

this will show you a detailed trace of all the executed commands and the related variable substitutions.

Jacky Quah · 10-28-2008, 10:38 AM

Ops sorry, I guess it's the problem is not that..

My bad habit...; with php...
I write something like this...
$T2=$(< $TEST);
lol. doesn't even realise what's wrong...

colucix · 10-28-2008, 12:33 PM

Quote:

Originally Posted by Jacky Quah

I write something like this...
$T2=$(< $TEST);
lol. doesn't even realise what's wrong...

Ok. I try to interpret in simple words what you're trying to do: you want to assign the content of the file, whose name is stored in the variable TEST, to the variable T2. Right?

The error in the statement above is that you cannot have $T2 to the left hand side of an assignement, but simply T2, that is the name of the variable.

When you write $T2 you mean the value of the variable T2 and the statement above looks like a comparison instead of an assignment. Hope it is clear now!