bash integer number only - bash

I can't seem to allow single integer input only.
If someone puts abc, it will work.
But if someone puts in abc123 or 123abc it will still treat it as a valid integer
# input to be an integer.
validate_integer(){
if [ ! "$#" -eq "1" ]; then
error "Please enter one numberic value only"
return 1
elif [[ "$1" =~ ^[[:alpha:]]+$ ]]; then
error "Input must be a NUMBER"
return 1
else
return 0
fi
}

Change this line:
elif [[ "$1" =~ ^[[:alpha:]]+$ ]]; then
to this:
elif ! [[ "$1" =~ ^[[:digit:]]+$ ]]; then
There, ^[[:digit:]]+$ means the string must be composed of digits from beginning to end, and it must be 1 or more characters long. We negate this with !, to handle the case when the string does not satisfy this condition (not fully numeric).

Related

This script is supposed to run every 12 hours but outputs a "Command 200 not found" error on each test. Any ideas? [duplicate]

How do I compare a variable to a string (and do something if they match)?
Using variables in if statements
if [ "$x" = "valid" ]; then
echo "x has the value 'valid'"
fi
If you want to do something when they don't match, replace = with !=. You can read more about string operations and arithmetic operations in their respective documentation.
Why do we use quotes around $x?
You want the quotes around $x, because if it is empty, your Bash script encounters a syntax error as seen below:
if [ = "valid" ]; then
Non-standard use of == operator
Note that Bash allows == to be used for equality with [, but this is not standard.
Use either the first case wherein the quotes around $x are optional:
if [[ "$x" == "valid" ]]; then
or use the second case:
if [ "$x" = "valid" ]; then
Or, if you don't need an else clause:
[ "$x" == "valid" ] && echo "x has the value 'valid'"
a="abc"
b="def"
# Equality Comparison
if [ "$a" == "$b" ]; then
echo "Strings match"
else
echo "Strings don't match"
fi
# Lexicographic (greater than, less than) comparison.
if [ "$a" \< "$b" ]; then
echo "$a is lexicographically smaller then $b"
elif [ "$a" \> "$b" ]; then
echo "$b is lexicographically smaller than $a"
else
echo "Strings are equal"
fi
Notes:
Spaces between if and [ and ] are important
> and < are redirection operators so escape it with \> and \< respectively for strings.
To compare strings with wildcards, use:
if [[ "$stringA" == *"$stringB"* ]]; then
# Do something here
else
# Do something here
fi
I have to disagree one of the comments in one point:
[ "$x" == "valid" ] && echo "valid" || echo "invalid"
No, that is not a crazy oneliner
It's just it looks like one to, hmm, the uninitiated...
It uses common patterns as a language, in a way;
And after you learned the language.
Actually, it's nice to read
It is a simple logical expression, with one special part: lazy evaluation of the logic operators.
[ "$x" == "valid" ] && echo "valid" || echo "invalid"
Each part is a logical expression; the first may be true or false, the other two are always true.
(
[ "$x" == "valid" ]
&&
echo "valid"
)
||
echo "invalid"
Now, when it is evaluated, the first is checked. If it is false, than the second operand of the logic and && after it is not relevant. The first is not true, so it can not be the first and the second be true, anyway.
Now, in this case is the the first side of the logic or || false, but it could be true if the other side - the third part - is true.
So the third part will be evaluated - mainly writing the message as a side effect. (It has the result 0 for true, which we do not use here)
The other cases are similar, but simpler - and - I promise! are - can be - easy to read!
(I don't have one, but I think being a UNIX veteran with grey beard helps a lot with this.)
The following script reads from a file named "testonthis" line by line and then compares each line with a simple string, a string with special characters and a regular expression. If it doesn't match, then the script will print the line, otherwise not.
Space in Bash is so much important. So the following will work:
[ "$LINE" != "table_name" ]
But the following won't:
["$LINE" != "table_name"]
So please use as is:
cat testonthis | while read LINE
do
if [ "$LINE" != "table_name" ] && [ "$LINE" != "--------------------------------" ] && [[ "$LINE" =~ [^[:space:]] ]] && [[ "$LINE" != SQL* ]]; then
echo $LINE
fi
done
You can also use use case/esac:
case "$string" in
"$pattern" ) echo "found";;
esac
Bash 4+ examples. Note: not using quotes will cause issues when words contain spaces, etc. Always quote in Bash, IMO.
Here are some examples in Bash 4+:
Example 1, check for 'yes' in string (case insensitive):
if [[ "${str,,}" == *"yes"* ]] ;then
Example 2, check for 'yes' in string (case insensitive):
if [[ "$(echo "$str" | tr '[:upper:]' '[:lower:]')" == *"yes"* ]] ;then
Example 3, check for 'yes' in string (case sensitive):
if [[ "${str}" == *"yes"* ]] ;then
Example 4, check for 'yes' in string (case sensitive):
if [[ "${str}" =~ "yes" ]] ;then
Example 5, exact match (case sensitive):
if [[ "${str}" == "yes" ]] ;then
Example 6, exact match (case insensitive):
if [[ "${str,,}" == "yes" ]] ;then
Example 7, exact match:
if [ "$a" = "$b" ] ;then
Enjoy.
I would probably use regexp matches if the input has only a few valid entries. E.g. only the "start" and "stop" are valid actions.
if [[ "${ACTION,,}" =~ ^(start|stop)$ ]]; then
echo "valid action"
fi
Note that I lowercase the variable $ACTION by using the double comma's. Also note that this won't work on too aged bash versions out there.
I did it in this way that is compatible with Bash and Dash (sh):
testOutput="my test"
pattern="my"
case $testOutput in (*"$pattern"*)
echo "if there is a match"
exit 1
;;
(*)
! echo there is no coincidence!
;;esac
I was struggling with the same situation for a while, here is how I could resolve:
if [ "$var1" == "$var2" ]; then
#dowhateveryouwant
fi
Be careful with the spaces left before and after the comparison sign, otherwise it won't work or it'll give you an unexpected result.
I've spent so much time on using a single equal(=) sign but didn't work. I Hope it can help.
Are you having comparison problems? (like below?)
var="true"
if [[ $var == "true" ]]; then
# It should be working, but it is not...
else
# It is falling here...
fi
Try like the =~ operator (regular expression operator) and it might work:
var="true"
if [[ $var =~ "true" ]];then
# Now it works here!!
else
# No more inequality
fi
Bash regex operator =~ (official reference)
StackOverflow further examples (here)

Running bash script throws syntax error: operand expected [duplicate]

How do I compare a variable to a string (and do something if they match)?
Using variables in if statements
if [ "$x" = "valid" ]; then
echo "x has the value 'valid'"
fi
If you want to do something when they don't match, replace = with !=. You can read more about string operations and arithmetic operations in their respective documentation.
Why do we use quotes around $x?
You want the quotes around $x, because if it is empty, your Bash script encounters a syntax error as seen below:
if [ = "valid" ]; then
Non-standard use of == operator
Note that Bash allows == to be used for equality with [, but this is not standard.
Use either the first case wherein the quotes around $x are optional:
if [[ "$x" == "valid" ]]; then
or use the second case:
if [ "$x" = "valid" ]; then
Or, if you don't need an else clause:
[ "$x" == "valid" ] && echo "x has the value 'valid'"
a="abc"
b="def"
# Equality Comparison
if [ "$a" == "$b" ]; then
echo "Strings match"
else
echo "Strings don't match"
fi
# Lexicographic (greater than, less than) comparison.
if [ "$a" \< "$b" ]; then
echo "$a is lexicographically smaller then $b"
elif [ "$a" \> "$b" ]; then
echo "$b is lexicographically smaller than $a"
else
echo "Strings are equal"
fi
Notes:
Spaces between if and [ and ] are important
> and < are redirection operators so escape it with \> and \< respectively for strings.
To compare strings with wildcards, use:
if [[ "$stringA" == *"$stringB"* ]]; then
# Do something here
else
# Do something here
fi
I have to disagree one of the comments in one point:
[ "$x" == "valid" ] && echo "valid" || echo "invalid"
No, that is not a crazy oneliner
It's just it looks like one to, hmm, the uninitiated...
It uses common patterns as a language, in a way;
And after you learned the language.
Actually, it's nice to read
It is a simple logical expression, with one special part: lazy evaluation of the logic operators.
[ "$x" == "valid" ] && echo "valid" || echo "invalid"
Each part is a logical expression; the first may be true or false, the other two are always true.
(
[ "$x" == "valid" ]
&&
echo "valid"
)
||
echo "invalid"
Now, when it is evaluated, the first is checked. If it is false, than the second operand of the logic and && after it is not relevant. The first is not true, so it can not be the first and the second be true, anyway.
Now, in this case is the the first side of the logic or || false, but it could be true if the other side - the third part - is true.
So the third part will be evaluated - mainly writing the message as a side effect. (It has the result 0 for true, which we do not use here)
The other cases are similar, but simpler - and - I promise! are - can be - easy to read!
(I don't have one, but I think being a UNIX veteran with grey beard helps a lot with this.)
The following script reads from a file named "testonthis" line by line and then compares each line with a simple string, a string with special characters and a regular expression. If it doesn't match, then the script will print the line, otherwise not.
Space in Bash is so much important. So the following will work:
[ "$LINE" != "table_name" ]
But the following won't:
["$LINE" != "table_name"]
So please use as is:
cat testonthis | while read LINE
do
if [ "$LINE" != "table_name" ] && [ "$LINE" != "--------------------------------" ] && [[ "$LINE" =~ [^[:space:]] ]] && [[ "$LINE" != SQL* ]]; then
echo $LINE
fi
done
You can also use use case/esac:
case "$string" in
"$pattern" ) echo "found";;
esac
Bash 4+ examples. Note: not using quotes will cause issues when words contain spaces, etc. Always quote in Bash, IMO.
Here are some examples in Bash 4+:
Example 1, check for 'yes' in string (case insensitive):
if [[ "${str,,}" == *"yes"* ]] ;then
Example 2, check for 'yes' in string (case insensitive):
if [[ "$(echo "$str" | tr '[:upper:]' '[:lower:]')" == *"yes"* ]] ;then
Example 3, check for 'yes' in string (case sensitive):
if [[ "${str}" == *"yes"* ]] ;then
Example 4, check for 'yes' in string (case sensitive):
if [[ "${str}" =~ "yes" ]] ;then
Example 5, exact match (case sensitive):
if [[ "${str}" == "yes" ]] ;then
Example 6, exact match (case insensitive):
if [[ "${str,,}" == "yes" ]] ;then
Example 7, exact match:
if [ "$a" = "$b" ] ;then
Enjoy.
I would probably use regexp matches if the input has only a few valid entries. E.g. only the "start" and "stop" are valid actions.
if [[ "${ACTION,,}" =~ ^(start|stop)$ ]]; then
echo "valid action"
fi
Note that I lowercase the variable $ACTION by using the double comma's. Also note that this won't work on too aged bash versions out there.
I did it in this way that is compatible with Bash and Dash (sh):
testOutput="my test"
pattern="my"
case $testOutput in (*"$pattern"*)
echo "if there is a match"
exit 1
;;
(*)
! echo there is no coincidence!
;;esac
I was struggling with the same situation for a while, here is how I could resolve:
if [ "$var1" == "$var2" ]; then
#dowhateveryouwant
fi
Be careful with the spaces left before and after the comparison sign, otherwise it won't work or it'll give you an unexpected result.
I've spent so much time on using a single equal(=) sign but didn't work. I Hope it can help.
Are you having comparison problems? (like below?)
var="true"
if [[ $var == "true" ]]; then
# It should be working, but it is not...
else
# It is falling here...
fi
Try like the =~ operator (regular expression operator) and it might work:
var="true"
if [[ $var =~ "true" ]];then
# Now it works here!!
else
# No more inequality
fi
Bash regex operator =~ (official reference)
StackOverflow further examples (here)

Why is it saying integer expected? [duplicate]

This question already has answers here:
How do I test if a variable is a number in Bash?
(40 answers)
Closed 2 years ago.
#!/bin/bash
if [ "$1" == "" ] || [ "$2" == "" ] || [ "$3" == "" ]
then
echo This is empty or does not have all 3 parameters
exit
elif [ "$1" -lt 0 ]
then
echo This aint a number
exit
fi
Trying to run a script it is suppose first check if 3 positional parameters were entered,
secondly check if the input are numbers and then display the largest. I got the first if statement to work but when I input a string for the first parameter to test the elif statement an error that says integer expected.
you have to review the regex to check if string only contains digits, but you may try this:
#!/bin/bash
if [[ "$1" == "" ]] || [[ "$2" == "" ]] || [[ "$3" == "" ]]
then
echo "This is empty or does not have all 3 parameters"
exit
elif ! [[ "$1" =~ ^[0-9]+$ ]]
then
echo "This aint a number"
exit
fi
Note that [[ is actually a command/program that returns either 0 (true) or 1 (false). Any program that obeys the same logic (like all base utils, such as grep(1) or ping(1)) can be used as condition :
[[ -z STRING ]] #Empty string
[[ -n STRING ]] #Not empty string
[[ STRING == STRING ]] #Equal
[[ STRING != STRING ]] #Not Equal
[[ NUM -eq NUM ]] #Equal
[[ NUM -ne NUM ]] #Not equal
[[ NUM -lt NUM ]] #Less than
[[ NUM -le NUM ]] #Less than or equal
[[ NUM -gt NUM ]] #Greater than
[[ NUM -ge NUM ]] #Greater than or equal
[[ STRING =~ STRING ]] #Regexp
(( NUM < NUM )) #Numeric conditions
also to check if the string contains only digits/numerical cheracter
[[ "$1" =~ ^[0-9]+$ ]]
Please use a regex to check for number
#!/bin/bash
if [ $1 == "" ] || [ $2 == "" ] || [ $3 == "" ]
then
echo This is empty or does not have all 3 parameters
exit
elif ! [[ $1 =~ ^[0-9]+$ ]]
then
echo This aint a number
exit
fi

Check if string contains non digit characters

How can I check if a given string contains non numeric characters, examples :
x11z returns > 0
x$1 also returns > 0
1111~ also returns > 0
By character I mean everything not between 0-9. I saw similar threads but non of them talks about "non 0-9" except they show if its a-z or A-Z.
Just use a negated character class:
grep [^0-9]
This will match any non-numeric character, and not strings composed of only digits.
Just by using bash pattern matching:
[[ "$MY_VAR" =~ ^[^0-9]+$ ]] && echo "no digit in $MY_VAR"
Something like this:
if [[ "xf44wd" =~ [0-9]+ ]]; then
echo "contains $?"
else
echo "does no contains $?"
fi
or
if [[ ! "xf44wd" =~ [0-9]+ ]]; then
echo "does not contains $?"
else
echo "contains $?"
fi
Most of these suggestions return true if the first character is a digit, but don't find errors within the string. The function below returns true if the entire string is digits, false if any non-digits are detected in the string.
function isdigit () {
[ $# -eq 1 ] || return 1;
[[ $1 = *[^0-9]* ]] && return 1
return 0
}
Another option, a bash "containment" check
[[ "xf4fgh" = *[^0-9]* ]]
echo $?
0
[[ "1234" = *[^0-9]* ]]
echo $?
1

Is there an easy way to determine if user input is an integer in bash?

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?
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
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 "$input is not an integer or not defined"
fi
You can test by using Regular expression
if ! [[ "$yournumber" =~ ^[0-9]+$ ]] ;
then exec >&2; echo "error: Not a number"; exit 1
fi
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
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
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

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