If condition not detecting operator

Question

Today I have started to learn to write bash shell scripts, and the prof has tasked us to create a simple script calculator using three inputs: two of which are numeric inputs and one is an operator input.

First, to my understanding, when a variable is assigned from reading input, it is treated as a string. so then by comparing the variable op to the string version of + it should evaluate n1+n2. however this is not the case, and I get the following errors:

./q2.txt: line 9: [+: command not found

./q2.txt: line 13: [+: command not found

#!/bin/bash
echo "Enter the operator"
read op
echo "Enter the first number"
read n1
echo "Enter second number"
read n2

if ["$op" == "+"]
then
    n3=$((n1+n2))

elif [$op = "-"]
then
    n3 = $((n1-n2))

fi
echo "Answer: $n3"

exit 0

The `echo` + `read` could be written in one statement as: `read -r -p 'Enter number> ' n1`. Also, you even not need the comparison at all, just use `n3=$(($n1 $op $n2))` :) — clt60, Mar 01 '16 at 17:04

score 1 · Accepted Answer · answered Mar 01 '16 at 16:54

1

the if syntax must left a space before and after condition [ condition ]

so your script will be

#!/bin/bash
echo "Enter the operator"
read op
echo "Enter the first number"
read n1
echo "Enter second number"
read n2

if [ "$op" == "+" ]
then
    n3=$((n1+n2))

elif [ $op = "-" ]
then
    n3 = $((n1-n2))

fi
echo "Answer: $n3"

exit 0

Then it will works

answered Mar 01 '16 at 16:54

ClaudioM

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oh my gosh i cant believe it was that trivial. thanks – user3170251 Mar 01 '16 at 16:55
you're welcome @user3170251 – ClaudioM Mar 01 '16 at 16:56

If condition not detecting operator

1 Answers1