Evaluate and assign variables inside ssh session in shell script

Question

I'm trying to increment indx and eindx values inside a ssh session.but none of commands are working..

#!/bin/bash
hosts=( #10.xx.xx.xx )
PdidPrefix1=$1
Runtime=$2
indx=$3
eindx=`expr $indx + $4`
logfilename=$5
for i in "${hosts[@]}"
do
echo $i
ssh centos@$i << EOF
for var in {1..4}
do
indx=`expr $eindx + 1` // not working
eindx=$((indx+eindx))  // not working
done
EOF
done
exit

tried using let also

let "indx=indx+1" //not working

Kindly suggest best way to handle variables.

Have you tried quoting the heredoc sigil, (e.g. 'EOF') to prevent variable expansion on the local host? — David C. Rankin
– David C. Rankin, Commented Oct 9, 2017 at 7:18
Quoting the sigil prevents both pathname and variable expansion. It is a standard way to prevent both. It appears the problem you are having is you want the variables expanded locally and then the results pass via ssh to the host at $i. While not the cause, using $((...)) arithmetic is correct, leave expr a + b in the past. Have you tried with $((...)) arithmetic in all instances where you are doing math? — David C. Rankin
– David C. Rankin, Commented Oct 9, 2017 at 8:05
@DavidC.Rankin used $((..)) expression ,however it did not work.If I am using EOF,it is throwing an error that it cannot resolve $var which i'm trying to use in another command — Svp57
– Svp57, Commented Oct 9, 2017 at 9:45

Ashim Paul · Accepted Answer · 2017-10-09 17:49:41Z

2

#!/bin/bash
hosts=( xx.10.20.30 )
PdidPrefix1=$1
Runtime=$2
indx=$3
eindx=`expr $indx + $4`
logfilename=$5
for i in "${hosts[@]}"
do
echo $i
ssh centos@$i "bash -s --  '$indx' '$eindx'" <<\EOF

echo "In remote shell with parameters $1 $2..."
indx=$1
eindx=$2

for (( var=1; var<=4; var++ ))
do

indx=`expr $eindx + 1`
eindx=$((indx+eindx))
echo "var=" $var " indx=" $indx " eindx=" $eindx
done
EOF
done
exit

answered Oct 9, 2017 at 17:49

Ashim Paul

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2 Comments

Svp57 Over a year ago

No Luck.If I use \EOF, inner for loop variable 'var' is not getting resolved.If I remove and try to run, I can see in debug,expr $eindx + 1 is getting executed but not getting assigned to indx variable .

Svp57 Over a year ago

my bad..bro..Its working...I copied some part of code snippet and trying..Thanks...

Ashim Paul · Accepted Answer · 2017-10-10 18:41:47Z

0

I have run the same script. And it shows the proper value.

answered Oct 10, 2017 at 18:41

Ashim Paul

1002 bronze badges

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