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Passing Command Line Args in a Single Variable?

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http://www.unix.com – Hello All, I have a Bash Script and an Expect script that together will SSH to another server and do some stuff there... From within the Bash Script I process the Command Line Arguments, which are Required Args and Optional Args. When I call the Expect script from the Bash Script, I pass it all the required Args as there own variables: i.e. $ip_address, $username, $password, $option... Then, as I process any Optional Arguments I want to concatenate them into one Variable. Doing this will remove a whole bunch of If statements from my script that would, for example, if the optional argument (HowTos)

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Submitted by suki on 08/13/2012 – Made popular on 08/13/2012

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Hello All,

I am writing an Expect Script to execute some commands over ssh then exit the script. The script works just fine if I automate everything and assuming the correct password was entered.

So this Expect Script gets executed from a Bash script...

Cut | command line args 1 year 9 weeks ago

Hi,

Can you please hint me how to achieve the below?

Input: $./script.sh start 1 2

Internally inside the script i want to set a single variable with $2 and $3 value?

Output:

CMD=$1 ARGS=$2 $3

--VInodh

Hi, There is probably a very simple fix, but its wrecking my buzz

I call the following on the command line

R CMD BATCH '--args Y_filename="one.txt" out="two.txt"' brew.R

I want to call this from a bash script but replace one.txt with $1, and two.txt with $2 (ie the first two arguments from the command line)

Hi there. There is a shell script:

Code: #!/bin/bash ARGS=`(grep -i $1 old.txt || grep -i $1 new.txt) |awk 'BEGIN {FS=","}{print "\x27" $5 "\x27"}' | tr "\n" " "` /usr/bin/expect -- << EXPECTEND spawn -noecho myssh myhost.localnet log_user 0 expect root log_user 1 send "/home/one.pl $ARGS\r" set val $expect_ou

HI guys i am new to programing world, i have found one more wish with command line argument.

I have a shell script say one.sh ,

file_name=$1 //first argument file name(two.sh) //connect to ftp and download $file_name bash $file_name $2 $3 $4 //2nd 3rd and 4th arguments

say $file_name "actual name is two.sh, in two.sh

..... ..... chmod 777 script.sh

At this moment I have:

#!/bin/bash screen -p 'ScreenName' -x eval 'stuff '"'"$@"'"'\015' echo eval 'stuff '"'"$@"'"'\015'

But when I call my script as:

# script.sh asd "asd" 'asd'

my arguments passed as: asd asd asd

and I get output:

eval stuff 'asd asd asd'\015

I except a: asd "asd" 'asd'

How I can change my script to pass whole arguments line with all quotes?

How can I pass an optional argument to a bash script that will replace an existing variable in the script? For example:

#!/bin/bash #hostfinder.sh #Find hosts in current /24 network

subnet=$(hostname -i | cut -d.

I am writing a bash script that runs each of its arguments as a command. This works for commands in my PATH, but not for aliases. I can directly call an alias in the script, but I can't call an alias that has been passed as an argument.

The problem (I assume) is that aliases are expanded before variables.

I need to pass an argument from the command line to the "at" command.

I want "at" to run a script which requires a few arguments from the command line when it runs.