I am a new student to bash scripting, and I am stumped on an assignment question. I was wondering if there is an easy way to determine whether a users' input is an integer or not. More specifically, if a user is prompted to input an integer, is there a quick check to validate?
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6 Answers
One way is to check whether it contains non-number characters. You replace all digit characters with nothing and check for length -- if there's length there's non-digit characters.
if [[ -n ${input//[0-9]/} ]]; then
echo "Contains letters!"
fi
Another approach is to check whether the variable, evaluated in arithmetic context, is equal to itself. This is bash-specific
if [[ $((foo)) != $foo ]]; then
echo "Not just a number!"
fi
1 Comment
Future
If input is 5b for example, it will fail check.
This is kind of a kludge, it's using -eq for something other then what it was intended, but it checks for an integer, if it doesn't find an int it returns both an error which you can toss to /dev/null and a value of false.
read input
if [[ $input ]] && [ $input -eq $input 2>/dev/null ]
then
echo "$input is an integer"
else
echo "$input is not an integer or not defined"
fi
2 Comments
Xofo
This assumes /dev/null exists
Cosmin Lehene
It also assumes the argument is not empty
You can test by using Regular expression
if ! [[ "$yournumber" =~ ^[0-9]+$ ]] ;
then exec >&2; echo "error: Not a number"; exit 1
fi
1 Comment
Craig Hicks
This is the most literal, and therefore easiest to remember.
I found this post http://www.unix.com/shell-programming-scripting/21668-how-check-whether-string-number-not.html that talks about this.
If your input does not need to check if there is a +/- on the number, then you can do:
expr $num + 1 2> /dev/null
if [ $? = 0 ]
then
echo "Val was numeric"
else
echo "Val was non-numeric"
fi
2 Comments
Logu
Simplest for integers. But doesn't support decimal fractions like 1.1, 1.2 etc
fbicknel
expr "$num" + 1 > /dev/null 2>&1 will avoid the value of $num + 1 being echoed to the stdout. Quotes around $num are there for safety. (in case $num has something with spaces in it like Foo Bar.)Here is another way of doing it. It's probably a bit more elaborate than needed in most cases, but would handle decimals also. I had written the below code to get rounded number. It also checks for numeric input in the process.
#--- getRound -- Gives number rounded to nearest integer -----------------------
# usage: getRound <inputNumber>
#
# echos the rounded number
# Best to use it like:
# roundedNumber=`getRound $Number`
# check the return value ($?) and then process further
#
# Return Value:
# 2 - if <inputNumber> is not passed, or if more arguments are passed
# 3 - if <inputNumber> is not a positive number
# 0 - if <inputNumber> is successfully rounded
#
# Limitation: Cannot be used for negative numbers
#-------------------------------------------------------------------------------
getRound (){
if [ $# -ne 1 ]
then
exit 2
fi
#--- Check if input is a number
Input=$1
AB=`echo A${Input}B | tr -d [:digit:] | tr -d '.'`
if [ "${AB}" != "AB" ] #--- Allow only '.' and digit
then
exit 3
fi
DOTorNone=`echo ${Input} | tr -d [:digit:]` #--- Allow only one '.'
if [ "${DOTorNone}" != "" ] && [ "${DOTorNone}" != "." ]
then
exit 3
fi
echo $Input | awk '{print int($1+0.5)}' #--- Round to nearest integer
}
MyNumber=`getRound $1`
if [ $? -ne 0 ]
then
echo "Empty or invalid input passed"
else
echo "Rounded input: $MyNumber"
fi
Comments
This one works for me, handling empty input case.
if [ $input -eq $input 2>/dev/null -o $input -eq 0 2>/dev/null ]
then
echo Integer
else
echo Not an integer
fi
1 Comment
0andriy
${input:-0} -eq ${input:1} should work as well, I suppose. (It’s about empty argument only)