Bash Scripting: Echo to Standard Error

colonel_t · 04-03-2007, 02:10 AM

Hi all.

I'm trying to print out all arguments passed to a script on a single line sent to STDERR

My script:

#!/bin/bash
echo $* >&2

Am I missing something? I know that >&2 should send STDERR to whatever is after it, but do I need to specify that the line is an error or something?

Thanks

Centinul · 04-03-2007, 06:02 AM

Try

Code:

#!/bin/bash
echo "$*" >&2

Hope this helps!

mikemcdonough · 01-20-2011, 01:22 PM

Code:

#!/bin/bash
echo $* 1>&2

lineus · 07-11-2011, 01:12 PM

$ cat myscript.sh
#!/bin/bash
# not sure how portable this is, but it works on rhel 5.4

echo "$*" >/dev/stderr

$ ./myscript.sh unix is fun
unix is fun
$ ./myscript.sh unix is fun 2>~/testing_dev_stderr
$
$ cat ~/testing_dev_stderr
unix is fun

OR

$ cat myscript.sh
#!/bin/bash
# perhaps a more portable way

echo "$*" >/dev/fd/2

$ ./myscript.sh unix is fun
unix is fun
$ ./myscript.sh unix is fun 2>~/testing_dev_stderr
$
$ cat ~/testing_dev_stderr
unix is fun

I suppose the bottom line is that STDERR should be
available to you in the form of a file ( like just about everything else in unix/linux )

chrism01 · 07-12-2011, 01:16 AM

Code:

#output to stdout aka chan 1
# 2 methods: NB default is stdout (chan 1) anyway ...
echo A 
echo A >1

# output to stderr aka chan 2
echo B >2

# output to both
# 2>&1 => output stderr (chan 2)to same chan (&) as stdout (chan 1)
echo C >1 2>&1
echo C 2>&1

# chan 0 is stdin