I wrote this code:
echo -n "Enter a number1 "
echo -n "Enter a number2 "
read R1
read R2
while [ "$R1" < "$R2"]
do
if [ $((R1 % 2)) -eq 0 ]; then
$R3=$R1
echo "Number is $R3"
else
echo "Nothing"
fi
done
I don't understand why it always give me this error bash: 8]: No such file or directory
You should use -lt instead of <.
while [ "$R1" -lt "$R2" ]
< is interpreted as input redirection in bash.
Or you can use double square brackets to interpret those inside as arithmetic operation:
while [[ "$R1" < "$R2" ]]
What happens since < "$R2" is intrepreted as read from "$R2". Since you don't have a file with such a name, it complains.
[ (test command) command doesn't have < operator. You have to use -lt instead:
while [ "$R1" -lt "$R2" ]
There's a POSIX extenstion which supports it with a slash:
while [ "$R1" \< "$R2" ]
If you are using bash you bash then you can also use built-in [[ ..]] which has support for <, > etc.
while [[ "$R1" < "$R2" ]]
See also:
What is the difference between test, [ and [[ ?
After re-writing your code to put the loop inside if:
#!/bin/bash
echo -n "Enter a number1 "
read R1
echo -n "Enter a number2 "
read R2
if [[ "$R1" < "$R2" ]]
then
for((i=R1;i<R2;i++));
do
if [[ $((i % 2)) -eq 0 ]]; then
echo "Number is $i"
fi
done
else
echo "Nothing"
fi
Related
I'm writing a function in shell script to check if the two numbers are Palindromes but I am getting an error, It is showing error in line 18 command not found. Please help me how can I remove this error.
#!/bin/bash
echo "Enter two number:"
read a
read b
for num in $a $b;
do
x=$x$sep$num
sep=" "
done
y=$x
num1=$a
num2=$b
rem=""
rev=0
for word in $y;
do
checkPalindrome $word
if [ $? -eq 0 ]
then
echo "$word is palindrome"
fi
done
checkPalindrome() {
local s=$1
for i in $s ;
do
while [ $i -gt 0]
do
rem=$(($i%10));
rev=$(($rev*10+$rem));
i=$(($i / 10));
done
done
if [[ $rev -eq $num1 && $rev -eq $num2 ]]
then
return 0;
else
return 1;
fi
}
You need to provide your checkPalindrome() definition before you use it, as below:
#!/bin/bash
checkPalindrome() {
local s=$1
for i in $s
do
while [ "$i" -gt 0 ]
do
rem=$((i%10))
rev=$((rev*10+rem))
i=$((i / 10))
done
done
if [[ $rev -eq $num1 && $rev -eq $num2 ]]
then
return 0
else
return 1
fi
}
echo "Enter two number:"
read -r a
read -r b
for num in $a $b
do
x="$x$sep$num"
sep=" "
done
y="$x"
num1="$a"
num2="$b"
rem=""
rev=0
for word in $y;
do
if checkPalindrome "$word"
then
echo "$word is palindrome"
fi
done
You could consider the input as string (you don't have to restrict it to be an integer)
#!/bin/bash
is_palindrome() {
local arg=$1 i j
for ((i = 0, j = ${#arg} - 1; i < j; ++i, --j)); do
[[ ${arg:i:1} = "${arg:j:1}" ]] || return
done
}
read -r -p 'Enter two words: ' a b
for word in $a $b; do
is_palindrome "$word" && echo "$word is palindrome"
done
I want to just insert number between two values, and otherwise the script repeated until correct number.
This is my script and it does not work correctly:
validation(){
read number
if [ $number -ge 2 && $number -ls 5 ]; then
echo "valid number"
break
else
echo "not valid number, try again"
fi
}
echo "insert number"
validation
echo "your number is" $number
If you are using Bash, you are better off using the arithmetic expression, ((...)) for readability and flexibility:
if ((number >= 2 && number <= 5)); then
# your code
fi
To read in a loop until a valid number is entered:
#!/bin/bash
while :; do
read -p "Enter a number between 2 and 5: " number
[[ $number =~ ^[0-9]+$ ]] || { echo "Enter a valid number"; continue; }
if ((number >= 2 && number <= 5)); then
echo "valid number"
break
else
echo "number out of range, try again"
fi
done
((number >= 2 && number <= 5)) can also be written as ((2 <= number <= 5)).
See also:
Test whether string is a valid integer
How to use double or single brackets, parentheses, curly braces
Your if statement:
if [ $number -ge 2 && $number -ls 5 ]; then
should be:
if [ "$number" -ge 2 ] && [ "$number" -le 5 ]; then
Changes made:
Quoting variables is considered good practice.
ls is not a valid comparison operator, use le.
Separate single-bracket conditional expressions with &&.
Also you need a shebang in the first line of your script: #!/usr/bin/env bash
if [ $number -ge 2 && $number -ls 5 ]; then
should be
if [[ $number -ge 2 && $number -le 5 ]]; then
see help [[ for details
Try bellow code
echo "Enter number"
read input
if [[ $input ]] && [ $input -eq $input 2>/dev/null ]
then
if ((input >= 1 && input <= 4)); then
echo "Access Granted..."
break
else
echo "Wrong code"
fi
else
echo "$input is not an integer or not defined"
fi
2 changes needed.
Suggested by Sergio.
if [ "$number" -ge 2 ] && [ "$number" -le 5 ]; then
There is no need of break. only meaningful in a for, while, or until loop
while :; do
read option
if [[ $option -ge 1 && $option -lt 4 ]]; then
echo "correct"
c
break
else
echo "Incorrect option selected,choose an option between [1-4]"
fi
done
If the user input contains a single quote such as in
$ echo "Hello my name Foo I'm 18" | ./test.sh
It fails with error syntax error: operand expected (error token is "'")
Code:
#!/bin/sh
while read line
do
str=$line
for ((i = 0; i < ${#str}; i++))
do
if [[ "${str:$i:1}" -gt 4 ]] <-- fails
then
echo 'foo'
elif [[ "${str:$i:1}" -lt 4 ]] <--- fails
then
echo 'bar'
else
echo 'not a number'
fi
done
done
How can I make my code work when the user types in an apostrophe/single quote?
You can use:
#!/bin/bash
while read -r line; do
str=$line
for ((i = 0; i < ${#str}; i++)); do
ch="${str:$i:1}"
if [[ $ch == [0-9] && $ch -gt 4 ]]; then
echo 'foo'
elif [[ $ch == [0-9] && $ch -lt 4 ]]; then
echo 'bar'
else
echo 'not a number'
fi
done
done
Use /bin/bash instead of /bin/sh
Check if character is a digit using $ch == [0-9] before using -gt or -lt operators to avoid operand expected error.
Actually I am a new bash learner. I can use one condition in bash command. But how to use multiple condition in bash? I can use if statement like this:
read a
if [ $a = "y" ] ; then
echo "YES"
elif [ $a = "Y" ] ; then
echo "YES"
else
echo "NO"
fi
I am finding something like this:
read a b c
if [ $a -eq $b and $b -eq $c ] ; then
echo "EQUILATERAL"
elif [ $a -eq $b or $b -eq $c ] ; then
echo "ISOSCELES"
else
echo "SCALENE"
fi
I just want to know, what to use instead of and and or?
Use && for and (|| for or)
read a b c
if [ "$a" == "$b" ] && [ "$b" == "$c" ] ; then
echo "EQUILATERAL"
elif [ "$a" == "$b" ] || [ "$b" == "$c" ] ; then
echo "ISOSCELES"
else
echo "SCALENE"
fi
Use && and || to have multiple conditions. Additionally, change the square brackets to parentheses. Additionally change the -eq to == since you're comparing numbers and not strings. This works:
#!/bin/bash
read a b c
if (( $a == $b )) && (( $b == $c )); then
echo "EQUILATERAL"
elif (( $a == $b )) || (( $b == $c )) ; then
echo "ISOSCELES"
else
echo "SCALENE"
fi
In addition to the prior answers, the correct way to use compound expression in a single [ or test (they are the same) clause is to use -a (for and) and -o (for or).
(e.g. testing if both file1 and file2 are readable):
if [ -r "$file1" -a -r "$file2 ]
then
# do something with the files
fi
Using test itself:
if test -r "$file1" -a -r "$file2
then
# do something with the files
fi
The portable way of doing this inside test brackets is to use -a and -o. Beware however that -eq is a numeric comparison, so you need to make sure your variables are numeric before comparing them. Something like this:
#! /bin/sh
read a b c
expr "$a" : '[0-9][0-9]*$' \& "$b" : '[0-9][0-9]*$' \& "$c" : '[0-9][0-9]*$' >/dev/null || exit
if [ $a -eq $b -a $b -eq $c ] ; then
echo "EQUILATERAL"
elif [ $a -eq $b -o $b -eq $c ] ; then
echo "ISOSCELES"
else
echo "SCALENE"
fi
in bash I need to compare two float numbers, one which I define in the script and the other read as paramter, for that I do:
if [[ $aff -gt 0 ]]
then
a=b
echo "xxx "$aff
#echo $CX $CY $CZ $aff
fi
but I get the error:
[[: -309.585300: syntax error: invalid arithmetic operator (error token is ".585300")
What is wrong?
Thanks
Using bc instead of awk:
float1='0.43255'
float2='0.801222'
if [[ $(echo "if (${float1} > ${float2}) 1 else 0" | bc) -eq 1 ]]; then
echo "${float1} > ${float2}"
else
echo "${float1} <= ${float2}"
fi
use awk
#!/bin/bash
num1=0.3
num2=0.2
if [ -n "$num1" -a -n "$num2" ];then
result=$(awk -vn1="$num1" -vn2="$num2" 'BEGIN{print (n1>n2)?1:0 }')
echo $result
if [ "$result" -eq 1 ];then
echo "$num1 greater than $num2"
fi
fi
Both test (which is usually linked to as [)and the bash-builtin equivalent only support integer numbers.
Use bc to check the math
a="1.21231"
b="2.22454"
c=$(echo "$a < $b" | bc)
if [ $c = '1' ]; then
echo 'a is smaller than b'
else
echo 'a is larger than b'
fi
I would use awk for that:
e=2.718281828459045
pi=3.141592653589793
if [ "yes" = "$(echo | awk "($e <= $pi) { print \"yes\"; }")" ]; then
echo "lessthanorequal"
else
echo "larger"
fi
The simplest solution is this:
f1=0.45
f2=0.33
if [[ $f1 > $f2 ]] ; then echo "f1 is greater then f2"; fi
which (on OSX) outputs:
f1 is greater then f2
Here's another example combining floating point and integer arithmetic (you need the great little perl script calc.pl that you can download from here):
dateDiff=1.9864
nObs=3
i=1
while [[ $dateDiff > 0 ]] && [ $i -le $nObs ]
do
echo "$dateDiff > 0"
dateDiff=`calc.pl $dateDiff-0.224`
i=$((i+1))
done
Which outputs
1.9864 > 0
1.7624 > 0
1.5384 > 0