Shell Programming and Scripting

Hey, I was trying to figure out how to launch a program from the command line, and it works if you pass it a config file. I was thinking about writing a script to dynamically create the config file and pass it to the command using something like command substitution (so I don't actually have to create a file).

I'm using Bash 3.2.39

Well I went and tried the following just to see if it would work:

Code:

command --config <(cat ./configfile)

But it just said that it can't find the file /dev/fd/63.

Well

Code:

cat <(cat ./configfile)

does work (dumps it to the screen), so I thought maybe the file descriptor was being closed before the commad could use it.

So I tried making a more persistent one:

Code:

exec 10< <(cat ./configfile)
command --config /dev/fd/10

But it just says that it cannot find /dev/fd/10, even though it is there, and i can

Code:

cat /dev/fd/10

just fine (albeit just once).

Shouldn't this be possible? Or does the program have to do something special to be compatable with named pipes?

I'm not super-knowledgable about bash, but I've read as much as I can find on the subject. Any insight is appreciated!

Thanks.

Um, I won't get into the technicalities of why you tried didn't work -- it's a bit complicated. Normally you want to send input from a file like this:

Code:

command < configfile

You can do this with named pipes:

Code:

mkfifo /var/tmp/mypipe.$$
cat configfile > /var/tmp/mypipe.$$ &
command --config /var/tmp/mypipe.$$

You have to put the cat into the background, because the pipe doesn't buffer any input.

While not as simple as I had hoped, it certainly does work.

I also found that using the Zsh shell, you can force process substitution to automatically use a temporary file instead of a pipe with =().

So:

Code:

command --config =( cat configfile )

works, but not in Bash.

Hey, that rocks!