[SOLVED] Simple shell script conundrum

[Exocet]

Hey, folks. I've been using Linux for a while now, and I'm just dipping my toe into bash shell scripts, after flirting with them on and off for a while. Apart from this, my only experience with programming anything is some batch files in the old DOS days.

So I'm having a problem with the logic of shell scripting right now, and this is the problem. This script executes perfectly fine, and is a very simple matter indeed:

#!/bin/bash

#evaluates whether the first argument on commandline is a directory
#if false, set bgfolder to default directory
if [ -d "$1" ]; then
    ${bgfolder=$1}
else
    ${bgfolder=$HOME/Pictures/custom_backgrounds/1680}
fi
#print result to stdout
echo "$bgfolder"

exit 0

The last echo command is just a test for the snippet, of course. It works this way in a larger script that changes my wallpaper automatically according to a random sort through a directory. It works well. But of course I need to be able to exploit functions to make use of a script of considerable size, so I tried this:

#!/bin/bash

function pix_dir
{
#evaluates whether the first argument on commandline is a directory
#if false, echo default directory
if [ -d "$1" ]; then
    ${bgfolder=$1}
else
    ${bgfolder=$HOME/Pictures/custom_backgrounds/1680}
fi
#print result to stdout
echo "$bgfolder"
}

pix_dir

exit 0

Now, theoretically, this should do the same thing as the first example, but it passes over the first portion of the if statement and sets the variable to the default picture folder.

Can anyone tell me why this might bet the case? I've tried unquoting the $1 parameter in the conditional statement, and attempted less elegant means of directory testing (e.g. [ $1 != */* ]). These do not work inside a function either.

Any ideas?

[HROAdmin26]

It looks like you cannot get to the command line args directly from inside the function. A simple test would be to echo the $1 variable in the function - it is likely unset.

Try declaring the variable first - which is a good scripting practice.

#!/bin/bash

FIRST_ARG=$1

function (
#Test if FIRST_ARG is a dir
)

exit

Bash Scripting Guide

[Exocet]

Thanks for your quick reply.

Actually, I think I tried that earlier, but I gave it another go:

#!/bin/bash

bgfolder=$1

function pix_dir
{
#evaluates whether the first argument on commandline is a directory
#if false, echo default directory
if [ -d "$1" ]; then
    ${bgfolder=$1}
else
    ${bgfolder=$HOME/Pictures/custom_backgrounds/1680}
fi
#print result to stdout
echo "$bgfolder"
}

pix_dir

exit 0

Well, when I tested it with the root directory, it accepted the argument, echoing "/", but it wouldn't return the default directory when I typed something apart from a directory. It just gave me the $1 argument back, even if it was a random string. Where I'm trying to get to is for it to go to default for anything other than a directory for argument 1, even if there is no argument 1, which in this case the code returns a blank line instead.

My 'set -x' output looks like this:

+ pix_dir
+ '[' -d '' ']'
+ hjkhkj
/home/user/bin/walltest_func.sh: line 16: hjkhkj: command not found
+ echo hjkhkj
hjkhkj

And when I actually provide a directory, it shows this:

+ pix_dir
+ '[' -d '' ']'
+ /
/home/user/bin/walltest_func.sh: line 16: /: is a directory
+ echo /
/

exit 0
+ exit 0

Thanks for looking...

[HROAdmin26]

You will need to review your script more closely...

/home/user/bin/walltest_func.sh: line 16: hjkhkj: command not found

Assuming you put in that garbage as the "argument", you have a variable stated in the script such that bash is trying to execute it as a command.

[Exocet]

Yes, I was trying to get the script to ignore anything other than a directory, even a blank line or random string, which is what I gave it at the commandline. It was intentionally not a real command. Perhaps the directory evaluation isn't designed to handle this, although it seemed reasonable at the time. Do you have a suggestion for getting this done?

My purpose, ultimately, is to be able to have a default directory for the sort, but be able to add a directory to the script's link for evaluation when I execute it from a gnome-panel. This works without the function exactly as I wished, but not with the function.

[HROAdmin26]

#!/bin/bash

TEST_DIR="$1"
cd /	# Change to / to give context

if [ -d "$1" ]; then
	echo "Yeehaw"
else
	echo "Better luck next time"
fi

exit 0

Works fine for me. You need to either always use a full path, or account for a relative path.

/tmp/junk.sh /
Yeehaw

/tmp/junk.sh askdasd
Better luck next time

/tmp/junk.sh media
Yeehaw

/tmp/junk.sh media/cdrom
Yeehaw

/tmp/junk.sh media/askjdasj
Better luck next time

** Sorry, I'm done. Good luck with your script.

[Exocet]

Thanks for your help! I appreciate it. Later...

[HROAdmin26]

Added one thing:

#!/bin/bash

TEST_DIR="$1"
cd /	# Change to / to give context

if [ -d "$TEST_DIR" ]; then
	BGFOLDER="$TEST_DIR"
	echo "Yeehaw, BGFOLDER is $BGFOLDER"
else
	BGFOLDER="/some/sorry/folder"
	echo "Better luck next time - BGFOLDER is $BGFOLDER"
fi

exit 0

/tmp/junk.sh media/cdrom
Yeehaw, BGFOLDER is media/cdrom

/tmp/junk.sh media/askjdasj
Better luck next time - BGFOLDER is /some/sorry/folder

[Exocet]

Excellent! Knowledge and experience to the rescue!

That last construction did it, thanks. Been chewing this over for a day or two now, and I never thought of assigning the variables in that way.

From either no argument or random string, now:

+ pix_dir
+ '[' -d '' ']'
+ bgdir=/home/user/Pictures/custom_backgrounds/1680
+ echo /home/user/Pictures/custom_backgrounds/1680
/home/user/Pictures/custom_backgrounds/1680

exit 0
+ exit 0

Thank you.

And I appreciate the heads-up on the tutorial site as well.