Redirect output to file specified by variable

Question

To keep it simple, at the top of my bash script is the following code

#!/bin/bash
#if no arguments passed
if [ -z "$1" ]; then
    DEBUG='1>/dev/null'
else
    DEBUG='1>&1'
fi

Then throughout the rest of the script are calls like this

echo Doing something here! $DEBUG

The point being if someone calls the script without arguments (./myScript.sh) it wont echo anything.
But if they do pass in a parameter (./myScript.sh whathaveyou) the echo statements should work normally.

Unfortunately, however echo Doing something here! $DEBUG is being evaluated as

echo Doing something here! "1>&1"

instead of

echo Doing something here! 1>&1

How can I make this work?

that other guy · Accepted Answer · 2013-10-08 23:37:32Z

2

Instead of trying to make your command evaluate text data as command syntax, you can make a conditional function:

if [ -z "$1" ]; then
  report() {
    true
  }
else
  report() {
    echo "$@"
  }
fi
report "Doing something here!"

If you want to run generic commands instead of just printing things, you can similarly use

if [ -z "$1" ]; then
  report() {
    "$@" > /dev/null
  }
else
  report() {
    "$@"
  }
fi

report echo "Here are you files:"
report ls

To implement it exactly as you specified, you'd have to use eval which has a number of security issues and isn't as convenient.

answered Oct 8, 2013 at 23:37

that other guy

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6 Comments

Hashbrown Over a year ago

So it's not possible? (well, without using eval) Thanks for an equally simple alternative!

that other guy Over a year ago

Not as described, no. But there are plenty of alternatives, such as this one or writing all your debug info to fd 42 and conditionally redirect that.

Hashbrown Over a year ago

yeah I was just asking in case someone comes here because of the title, and this work around isn't useful for them. I, myself, will be using your first example, definitely

that other guy Over a year ago

You could always edit your question to say something like "Adding a verbose/debug mode to a script" :P

Eran Ben-Natan Over a year ago

that other guy - big like to your solution. Hashbrown - why not using eval? it's kind of panacea to many similar shell problems

|

William Pursell · Accepted Answer · 2013-10-08 23:39:06Z

0

Why not just do:

#!/bin/bash

test -z "$1" && exec > dev/null

This will redirect the output of the script to the bit bucket if the first argument is empty or not passed. All commands (unless redirected) will then discard their output.

answered Oct 8, 2013 at 23:39

William Pursell

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4 Comments

that other guy Over a year ago

Perhaps because this would also silence all other commands in the script

William Pursell Over a year ago

If the commands to be used as debug are selective: if test -z "$1; then exec 3> /dev/null; else exec 3>&1; fi; ... debug-cmd >&3

William Pursell Over a year ago

@Hashbrown, huh? This does not re-evaluate anything. This only requires you to simplify the top portion of your code, but you've just selected a solution that requires you to modify your entire script and adds quite a bit of syntactic noise. Perhaps you commented on the wrong solution?

Hashbrown Over a year ago

sorry, yes, I mis-interpreted. In response to your answer; the reason I had done it this way was to make sure other echo lines would write normally. Only those denoted with $DEBUG would be moot under argumentless invokation

David T. Pierson · Accepted Answer · 2013-10-08 23:56:35Z

0

I tend to do something like this:

#!/bin/bash

# if $1 is not empty, set $enable_trace to 'true'
enable_trace=${1:+true}

function trace()
{
  if [ "$enable_trace" = 'true' ]; then
    echo "$@"
  fi
}

trace Doing something here

Introducing the $enable_trace parameter also allows you to turn the output on and off easily at different points within your script.

answered Oct 8, 2013 at 23:56

David T. Pierson

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