Simple Bash Questions

RAdams · 08-22-2006, 10:25 AM

These should be easy ones:

Let's say I have a bash script containing the command:
Code:
```
gedit /foo/bar
```
How can I make the script end after this command, but leave the gedit window open?
If I have a script with parameters, how can I detect if no parameter has been added? For example, if the correct usage of my script is:
[code]foobar parameter1 parameter2[code]
How can I detect that the parameters were given by the user?
On a releated note, how can I allow the script to assume a value for a parameter the user doesn't specify? For example, if I want paramter2 in the example above to be assumed as "baryum" if the user provides no input, how do I do that?
What is the meaning of life?

That's all for now. Thank you very much.

AAnarchYY · 08-22-2006, 10:28 AM

check out the article on shell scripting on ldp
http://www.tldp.org/LDP/abs/html/

macemoneta · 08-22-2006, 10:32 AM

1. Run the gedit as a background process, by ending the command with an '&'. This starts the gedit, but returns control to the script, which can then exit. For example:

Code:

gedit /foo/bar &

2. The way I prefer to do this is to check the variable against the null string:

Code:

if [ "$1" == "" ]
then
   echo "No parameters"
fi

3. You can provide defaults, by initializing a variable:

Code:

param2="baryum"
if [ "$2" != "" ]
then
   param2=$2
fi

4. There is no meaning to life. Enjoy it while you can - you're not getting out alive.

druuna · 08-22-2006, 10:39 AM

Hi,

AAnarchYY is correct, all the questions asked can be found there. Here's a bit more to get you started.

Quote:

Let's say I have a bash script containing the command:

Code:

gedit /foo/bar

How can I make the script end after this command, but leave the gedit window open?

Take a look at & ('start in background' operand)

Quote:

If I have a script with parameters, how can I detect if no parameter has been added? For example, if the correct usage of my script is:
[code]foobar parameter1 parameter2[code]
How can I detect that the parameters were given by the user?

getopts or $1, $2, $... are your friends, depending on complexity of parameters and/or options.

Quote:

On a releated note, how can I allow the script to assume a value for a parameter the user doesn't specify? For example, if I want paramter2 in the example above to be assumed as "baryum" if the user provides no input, how do I do that?

By using parameter expansion or by checking the parameter and setting it if a certain condition arises (if ... then ... elseif ... esle)

Quote:

What is the meaning of life?

Haven't you seen the movie yet, explains it all

AAnarchYY · 08-22-2006, 10:41 AM

42, everyone knows that

RAdams · 08-23-2006, 12:41 AM

Thanks everyone. Piddling around, I managed to make a script to accelerate the process of making more scripts (Read: I'm lazy).

Essentially, the script copies information from a template file into a script name the user specifies, using whatever editor the user chooses (this parameter has a default).

Usage Example:

Code:

sudo newscript foobar nano

Would create a new script containing the information in a template to /usr/local/bin/foobar, set the permissions, and edit it using nano.

The template is meant to be a world-readable-writable file, though its permissions could be manually changed as one pleases.

There's other features too, but you'll see them below:

Code:

#! /bin/sh

EDITOR="gedit"
TEMPLATE="/usr/local/bin/bashtemplate"

#Check if run as root
if [ "$UID" -ne 0 ] ; then
	echo "Sorry, root privileges are required to run this script."
	exit 0
fi

#Check to make sure the user provided a name for the script
if [ "$1" == "" ]
then
	echo "Error: no scriptname specified! (Usage: newscript [scriptname] [editor])"
	exit -1
fi

#Check to see if an editor has been specified. If not, use default.

if [ "$2" != "" ]
then
	EDITOR=$2
fi


#Setup the Script
echo "Creating script..."
cp -i $TEMPLATE /usr/local/bin/$1
chmod 755 /usr/local/bin/$1
$EDITOR /usr/local/bin/$1
exit 1

Yes, it's not much, but it let me get my feet wet in scripting with passed parameters.

A question: when running this command under su instead of sudo, and using gedit for the editor, the terminal returns:

Code:

(gedit:10423): GnomeUI-WARNING **: While connecting to session manager:
Authentication Rejected, reason : None of the authentication protocols specified are supported and host-based authentication failed.

Why is this? What can I do to stop this error? It doesn't keep the script from working; I'm just curious about it.

One more question: is there a standard location in a Linux filesystem to store code templates?

RAdams · 08-23-2006, 12:44 AM

Quote:

Originally Posted by AAnarchYY

42, everyone knows that

W,H,A,T,D,O,Y,O,U,G,E,T,W,H,E,N,Y,O,U,M,U,L,T,I,P,L,Y,S,I,X,B,Y,N,I,N,E

AAnarchYY · 08-23-2006, 06:44 AM

Quote:

Originally Posted by RAdams

What can I do to stop this error? It doesn't keep the script from working; I'm just curious about it.

Not really sure what all that is but to stop it just put a &> /dev/null at the end of the command.

Code:

$EDITOR /usr/local/bin/$1 &> /dev/null

Also, why did you have the exit return a 1 upon success and a 0 upon failure? A successful command returns a 0, while an unsuccessful one returns a non-zero value.

code templates? sounds like something under /usr/share to me.

RAdams · 08-23-2006, 09:48 AM

Quote:

Originally Posted by AAnarchYY

Not really sure what all that is but to stop it just put a &> /dev/null at the end of the command.

Code:

$EDITOR /usr/local/bin/$1 &> /dev/null

Also, why did you have the exit return a 1 upon success and a 0 upon failure? A successful command returns a 0, while an unsuccessful one returns a non-zero value.

code templates? sounds like something under /usr/share to me.

The exit codes are just a proprietary system I made up on the spot. I'll change it to standards, as I'm anal retentive about standards.

/usr/share? Makes sense to me.