I am writing a very simple comparison in shell bash script, but I never get it correct:
count=0
if [ expr $count / 4 = 0 ];
then
echo "yes";
else
echo "no";
fi
always giving no?
If you want to call out to the expr program, you have to actually call out to it:
if [ $(expr $count / 4) = 0 ]; then echo "yes"; else echo "no"; fi
However, bash can do it in-house:
if (( $count / 4 == 0 )); then echo "yes"; else echo "no"; fi
You need to use command substitution ($() or backticks) to evaluate the eval expression. Also, use -eq for integer comparison:
if [ $(expr $count / 4) -eq 0 ];
then
echo "yes";
else
echo "no";
fi
How about this
[[ count/4 -eq 0 ]] && echo 'yes' || echo 'no'
Related
I would like to write in one line this:
if [$SERVICESTATEID$ -eq 2]; then echo "CRITICAL"; else echo "OK"; fi
So to do a test in my shell I did:
if [2 -eq 3]; then echo "CRITICAL"; else echo "OK"; fi
The result is
-bash: [2: command not found
OK
So it doesn't work.
Space -- the final frontier. This works:
if [ $SERVICESTATEID -eq 2 ]; then echo "CRITICAL"; else echo "OK"; fi
Note spaces after [ and before ] -- [ is a command name! And I removed an extra $ at the end of $SERVICESTATEID.
An alternative is to spell out test. Then you don't need the final ], which is what I prefer:
if test $SERVICESTATEID -eq 2; then echo "CRITICAL"; else echo "OK"; fi
Write like this, space is required before and after [ and ] in shell
if [ 2 -eq 3 ]; then echo "CRITICAL"; else echo "OK"; fi
Shorter format.
( [ 2 -eq 3 ] && echo "CRITICAL" ) || echo "OK"
Regex pattern type numbers : 10,12.1,+3.33,-1,0004,-48.9
Oneliner attacks again!
( [ `echo $number 2>/dev/null | grep -E "^[ ]*(\+|\-){0,1}[0-9]+(\.[0-9]+)?$"` ] && echo "NUMBER" ) || echo "NOT NUMBER"
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
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 && $option -lt 4 ]]; then
echo "correct"
c
break
else
echo "Incorrect option selected,choose an option between [1-4]"
fi
done
I would like to write in one line this:
if [$SERVICESTATEID$ -eq 2]; then echo "CRITICAL"; else echo "OK"; fi
So to do a test in my shell I did:
if [2 -eq 3]; then echo "CRITICAL"; else echo "OK"; fi
The result is
-bash: [2: command not found
OK
So it doesn't work.
Space -- the final frontier. This works:
if [ $SERVICESTATEID -eq 2 ]; then echo "CRITICAL"; else echo "OK"; fi
Note spaces after [ and before ] -- [ is a command name! And I removed an extra $ at the end of $SERVICESTATEID.
An alternative is to spell out test. Then you don't need the final ], which is what I prefer:
if test $SERVICESTATEID -eq 2; then echo "CRITICAL"; else echo "OK"; fi
Write like this, space is required before and after [ and ] in shell
if [ 2 -eq 3 ]; then echo "CRITICAL"; else echo "OK"; fi
Shorter format.
( [ 2 -eq 3 ] && echo "CRITICAL" ) || echo "OK"
Regex pattern type numbers : 10,12.1,+3.33,-1,0004,-48.9
Oneliner attacks again!
( [ `echo $number 2>/dev/null | grep -E "^[ ]*(\+|\-){0,1}[0-9]+(\.[0-9]+)?$"` ] && echo "NUMBER" ) || echo "NOT NUMBER"
It seems that these two operators are pretty much the same - is there a difference? When should I use = and when ==?
You must use == in numeric comparisons in (( ... )):
$ if (( 3 == 3 )); then echo "yes"; fi
yes
$ if (( 3 = 3 )); then echo "yes"; fi
bash: ((: 3 = 3 : attempted assignment to non-variable (error token is "= 3 ")
You may use either for string comparisons in [[ ... ]] or [ ... ] or test:
$ if [[ 3 == 3 ]]; then echo "yes"; fi
yes
$ if [[ 3 = 3 ]]; then echo "yes"; fi
yes
$ if [ 3 == 3 ]; then echo "yes"; fi
yes
$ if [ 3 = 3 ]; then echo "yes"; fi
yes
$ if test 3 == 3; then echo "yes"; fi
yes
$ if test 3 = 3; then echo "yes"; fi
yes
"String comparisons?", you say?
$ if [[ 10 < 2 ]]; then echo "yes"; fi # string comparison
yes
$ if (( 10 < 2 )); then echo "yes"; else echo "no"; fi # numeric comparison
no
$ if [[ 10 -lt 2 ]]; then echo "yes"; else echo "no"; fi # numeric comparison
no
There's a subtle difference with regards to POSIX. Excerpt from the Bash reference:
string1 == string2
True if the strings are equal. = may be used in place of == for strict POSIX compliance.