Variable variable assignment error -"command not found"

Question

When I run the script I get the following errors. What am I doing wrong here please? Any help appreciated- Bash Newbie

Error:

line 12: 0=1: command not found
line 13: 0=1: command not found

My Script:

count_raw=0
avg_raw=0

$count_raw=1
$avg_raw=1

echo "count_raw=$count_raw"
echo "avg_raw=$avg_raw"

remove the dollar before count_raw and avg_raw

Avinash Raj
– Avinash Raj

2015-07-23 10:25:59 +00:00
Commented Jul 23, 2015 at 10:25 — Avinash Raj
– Avinash Raj, Commented Jul 23, 2015 at 10:25
That fixed it thanks-

Sascha
– Sascha

2015-07-23 10:59:49 +00:00
Commented Jul 23, 2015 at 10:59 — Sascha
– Sascha, Commented Jul 23, 2015 at 10:59

Donbhupi · Accepted Answer · 2015-07-24 05:14:32Z

1

= is an assignment operator when found free and $ holds value (not only in USA but in bash too) of a variable.

So when you say: $var=1, you're essentially trying to type a random string (0=1 in your case) in bash and bash doesn't like that. Look at the one-liner below that shows one example where you'd type in $var=1 and bash would be able to process it:

var=1; if [[ $var=1 ]]; then printf "Congrats! You have learned the difference between variable assignment and variable comparison in the ${var}st attempt.\n"; fi;

answered Jul 24, 2015 at 5:14

Donbhupi

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1 Comment

Sascha Over a year ago

Very good, amusing and also answered the question to a T. Thanks Donbhupi

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