How to check if a value exists from user input in bash?

Question

I want to input values like "2.1", "1.2", "143.22",... So something with this format: number1.number2

So far I have this code to separate number1 and number2.

echo -n "Enter value:  "
read answer

left=`echo $answer | awk -F "." '{print $1}'`      # number1
right=`echo $answer | awk -F "." '{print $2}'`     # number2

I don't want the user to input something like "1.2.3". How do I check if the user has entered an extra value?

I am thinking of something like this:

i=3

while true; do

    if [ -z "`echo $answer | awk -F "." '{print $i}'"]; then
        echo "Invalid input"

    i=$((i+1))
done

Will this work?

Hint: You could check for the number of fields, i.e. NF, using awk. — devnull
– devnull, Commented Jun 26, 2014 at 13:43
@devnull Will try. But will awk be able to distinguish the delimiter .? — Redson
– Redson, Commented Jun 26, 2014 at 13:46

marcello · Accepted Answer · 2014-06-26 14:04:46Z

2

You can simplify the test using an expression like this:

if [[ $answer =~ ^[0-9]+\.[0-9]+$ ]]; then ... fi

HTH, Marcello

edited Jun 26, 2014 at 14:04

answered Jun 26, 2014 at 13:56

marcello

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3 Comments

Redson Over a year ago

Can you give an explanation for the regular expression

user3442743 Over a year ago

^=start of string.....[0-9]+=any number from 1 to 9 and the + means one or more occurences..... \.=just a dot backslash escaped..... $= end of string

marcello Over a year ago

[0-9] is a digit (you could use [[:digit:]] instead); + means "one or more"; . is a wildcard, but we need it as a character, so we quote it using the `; ^` and $ means begin and end

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