[SOLVED] 10 command not found

Hello,
I am working on learning if-else statement, no matter what I do, it is throwing error. I am working on Ubuntu.

#! /bin/bash
count=10
if [$count -eq 10]
then
echo "condition is true"
fi


Kindly help

ubuntu@ip-172-31-92-85:~/suman$ cat ifelse.sh
#! /bin/bash
count=10
if [$count -eq 10]
then
echo "condition is true"
fi
ubuntu@ip-172-31-92-85:~/suman$ ./ifelse.sh
./ifelse.sh: line 3: [10: command not found
ubuntu@ip-172-31-92-85:~/suman$

You need white space after the [, otherwise the shell thinks you want to run a program named [10. Also white space before the ].

Perfect, thank you, I have another question please, in the same lines as the operator. I see that the string equals the operator '='. What is assignment operator represented as?

you might want to use www.shellcheck.net to check your srcipt.
also please use [code]here comes your code[/code] tags to post scripts. That will make it more readable.

Originally Posted by spalisetty Perfect, thank you, I have another question please, in the same lines as the operator. I see that the string equals the operator '='. What is assignment operator represented as?

The assignment operator is =. Inside the [ ... ] construct, use -eq to check equality. For arithmetic comparisons, you can also use the double parentheses:

if ((count==11))
then echo it's equal
else echo it's not
fi

Note the absence of the $ sign. I think a single = works as well in this context. It could be that this is not POSIX-compliant.

You should probably bookmark this https://www.tldp.org/LDP/abs/html/comparison-ops.html

Thank You Chris