Adding Two Variable Thru Script

I have some variables which have stored numerical values like
Quote:
A1=10 A2=20 A3=25
B1=10 B2=15 B3=12 and so on
As well as another set of variables which stored these variables like
Quote:
var1=A1 var2=A2 var3=A3
var4=B1 var5=B2 var6=B3 and so on
I have to calculated things using second set of variable.
In the script, there is line
Code:

read number

user will enter number manually, suppose 500
I have to calculate using
Code:

expr 500 \* ${$var1} / 100

and output of this multiplied by
Code:

expr $above_output \* ${$var2} /100

and output of this multiplied by
Code:

expr $above_output \* ${$var3} /100

But not able to deal with two levels of variable. The calculation will be hard when it comes for three level and four level.

Please help.

For some reason this statement reports "FALSE" when in fact it shouldn't. Any idea why?

Code:

#!/bin/bash

var1=$(nc -dvzw5 thebes.openshells.net 22)
var2='Connection to thebes.openshells.net 22 port [tcp/ssh] succeeded!'
if [ "$var1" = "$var2" ]; then
echo TRUE
else
echo FALSE
fi

So I created this big script and want to ask for username and set a variable, however if the username exist in the username variable then continue with script otherwise ask the user to enter his name.

Flow:
-- start script "Enter User Name"
-- ask user to input username
-- if username is blank or different value then put that username in $USERNAME variable
-- then continue with the rest of the script

-- next time this user runs the script it will have his name loaded so he just presses enter to continue with the script.

-- If it is NOT the same user then, new user can erase existing input and enter a new name.

Hope it makes sense...

Thanks in advance for the help.

Hi,

I've used unix before, but have just started to play with CYGWIN (1.1.3) on windows XP (32 bit). I'm having trouble getting 'echo' to behave as I would expect.

I've created a script called runstuff which contains these lines....
Code:

#!/bin/bash
# This line works as expected....
echo "the user is <$USER> and the hostname is <$HOSTNAME>"
MYVAR1=bill
MYVAR2=fred
# This lines doesn't.....
echo "my first variable is<$MYVAR1> and my second is <$MYVAR2>"

The output I get is this...
Quote:
$ runstuff
the user is <Mike> and the hostname is <MikesLaptop>
> and my second is <fredl

In the first echo statement it is just echoing a couple of external variables - this works as expected. In the 2nd echo example the two variables are internal, but the displayed output isn't what I'd expect, my guess is that the first part of the output "my first variable is <bill" has somehow been overwritten by the remainder of the string. Or am I missing something?

I have tried exporting the variables, and I've experimented with wrapping them in quotes, but the results are always the same.

Any help gratefully received.

Hello all,
I am trying to execute Code:

ssh $a@$b mkdir abc

in a shell script where, a and b are variables
a=username and b=server.ip
It is giving me an error Quote:
ssh connect to host port 22 connection refused
why am I getting the error?? ssh does not accept variables? Can anyone help.
Thanks in Advance

Hello,

I apologize in advance for my limited UNIX scripting knowledge. I am new to it and really want to learn.

I am trying to write a bash script that updates a config file based on user input.

What is the best method for accomplishing this? I need it to prompt the user for two variables, find the location in the config file, and insert new text with the two variables.

I was thinking I could use sed to find the text in the config file where the new text must be inserted before and replace it with the new text, two variables, and same ending text as before. Example:

echo "1st variable?"
read variable1

echo "2nd variable?"
read variable2

sed -e "s|<the spot in the config file that needs new config>|sometext...$variable2_somemoretext...$variable1\n<the spot in the config file that needs new config>|g" config > config2


This works except variable2 is not inserted into the replace string, only variable1 is. This is the result in config2:

sometext..._somemoretext...$variable1
<the spot in the config file that needs new config>


If anyone can tell me why variable2 isn't working in the sed replace string, or if there's a much better way for accomplishing what I'm after, I'd appreciate any help I can get.

Hello,
Host machine: Windows seven 64 bit
Guest machine on Virtual box: Oracle linux 6.5(64 bit)

If there are multiple files sourced, each containing many variables,and each in a location not known to the person using a user,
is there a way to know the file name where the variable is set?

Many thanks

Hy All,

I have two nodes running RHEL 6.4 which are in cluster. I have clustered httpd service. Now I have certain application suppose X which reside in SAN partition. Now my job is to only automate this script to run by analyzing httpd status. I have successfully created the script apptrigger.sh whose content is given below,

-bash-4.1$ cat apptrigger.sh
a=0
b=0
c=0
d=0
a="service httpd status"
b=`eval $a`
##### CHECKING WHETHER HTTPD SERVICE IS RUNNING OR NOT #####
if [ "$b" != "httpd is stopped" ]; then
c="bash /switchapp/switch/ezlinkenterprise/bin/ezlink-cluster.sh status"
d=`eval $c`
##### CHECKING WHETHER EZ IS RUNNING OR NOT #####
if [ "$d" != "EzTaskMgr is Running" ]; then
ezstart
fi
fi

I have created this script in say test user. Now when I put this script to automate or run every minute through test user and defined its value in crontab -e of test user whose content is as shown below
$crontab -l
* * * * * /bin/bash /home/test/apptrigger.sh

but this script is running as root user so I cannot get the o/p
I think some environment variable need to be set. Anyone can guide me with this step? Thanks for your views and suggestion.

books say that PATH variable contains the locations where the shell searches the commands to execute them.

I created a shell script and saved it in /home/user/Downloads directory and then i ran this script from the /home/user/Downloads directory. It did not execute.

however when i did set the path
export PATH=$PATH:/bin/sh within /home/user/Downloads directory, the shell script executed. Does it not mean that PATH variable is holding the locations of the shell program rather than the commands that are to be executed?

everytime i have to give the location of the shell i.e /bin/sh in the PATH variable no matter from where i am executing my script

I'm running a bash script on a Raspberry Pi to control LED pixels. I'm stuck on a basic copy command.

I have a variable called nexttemp which contains the string B_254_TieDyeKaleidoscope.fseq

That file, in the same directory, is a large binary file that I'm trying to copy to a file called MasterFseq.fseq

If I do a

cp nexttemp MasterFseq.fseq

then MasterFseq.fseq contains the B_254_TieDyeKaleidoscope.fseq file name.

I want instead to have copied the binary file to MasterFseq.fseq.

It's my understanding that a $nexttemp will point to the binary file and not the name of it but when I do a

cp $nexttemp MasterFseq.fseq

the $nexttemp is thrown away and I'm told there is no destination file name, eg.

root@FPP:/home/pi/media/sequences# echo cp $nexttemp MasterFseq.fseq
+ echo cp MasterFseq.fseq
cp MasterFseq.fseq

How do I get the binaray file, which is named in nexttemp, copied to MasterFseq.fseq?

John