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Problem with command argument

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http://www.unix.com – I'm having problem with taking in arguments and be able to use the content of the argument for something else. In this case I need to take in user names and check if the person is online using finger or who For example, my script name is getuser in shell I type getuser username1 username2 I want to be able to take username1 and check if he's online then take username2 and check if he's online. I'm currently just echo ing the argument to understand how to increment. Note that there may be more then 1 argument. I know that echo $* will print all the arguments echo $1 will print username1 e (HowTos)

robot1's picture
Submitted by robot1 on 10/01/2012

Stories similar to Problem with command argument

I have a following bash script I encountered on the web that prints the power set of a given lines of elements.

p() { [ $# -eq 0 ] && echo || (shift; p "$@") | while read r ; do echo -e "$1 $r\n$r"; done }

after the first && there is echo that does not have any argument.

Here is the code to test it:

p $(echo -e "1\n2\n3")

I have a shell script that accepts an argument and uses that in the script. So to run the script I type /path/to/script argument

The problem is when trying to setup that file to run using the at utility. I have tried

at -f /path/to/script argument 17:45

at -f '/path/to/script argument' 17:45

but neither of these options work.

My Bash-Script should accept arguments and options. Moreover arguments and options should be passed to another script.

The second part I've solved:

for argument in "$@"; do options $argument done

another_script $ox $arguments

function options { case "$1" in -x) selection=1 -y) selection=2 -h|--help) help_message;; -*) ox="$ox $1";; *) arguments="$arguments $1";; esa

bash scripting add arguments 13 weeks 23 hours ago

I have created a script which takes parameters and arguments.

This is my first question.

I'm trying to make a small script with options and I'm using getopts to do it.

I am trying to automate some of my stuff and facing a issue, the problem boils down to escaping the value that is being passed as a argument to the script.

myScript.sh

#!/bin/bash

loadPatch -name $1

where $1 is the first argument and can have value like 'p12.9.5-bug34'

Running it as

myScript p12.9.5-bug34

produces no result. I checked with echo $?

Unix Shell Scripting 39 weeks 5 days ago

I'm sorry if this doesn't go here, but I'm in depserate need of help with my last unix homework.

Anyways, I'm taking summer classes, and one of them is UNIX. I've understood everything thus far, but I'm having a killer time with how my instructor has worded the problems for shell scripting.

I'd like to use variable as condition in case statement. Something similar to:

#!/bin/sh

ALLOWED_SERVICES=tomcat6|james;

case $1 in $ALLOWED_SERVICES ) service $1 restart ;; * ) echo "Unsupported argument" ;; esac

This doesn't work.

I'm launching a java program from a shell script like this:

JAVA_OPTS= CLASSPATH= JMX_OPTS= java $JAVA_OPTS -cp $CLASSPATH $JMX_OPTS

And launched program expects user to type another argument. It can't be specified in the list of arguments the program was launched with.