This question already has answers here:
Why does non-equality check of one variable against many values always return true?
(3 answers)
Closed 8 years ago.
#!/bin/bash
(...)
if [ "$1" == "" ] || [ "$2" == "" ] || [ "$3" == "" ] || [ "$4" == "" ] ; then # WORKS!
echo "Erro #1: Nao inseriu todos os argumentos necessarios! Tente novamente!"
echo "Ajuda #1: Lembre-se, ep1.sh [diretoria] [modo] [informação] <nome_do_ficheiro>"
elif [[ "$2" != "contar" || "$2" != "hist" ]] ; then **# THE PROBLEM!!!**
echo "Erro #2: O segundo argumento está incorrecto! Tente novamente!"
echo "Ajuda #2: Use contar ou hist."
elif [[ "$3" != "palavras" || "$3" != "bytes" ]] ; then # CAN'T TEST BECAUSE OF FIRST ELIF
echo "Erro #3: O terceiro argumento está incorrecto! Tente novamente!"
echo "Ajuda #3: Use palavras ou bytes."
(...)
fi
So, my problem is in the first elif, when the elif is False, the program should not enter in the elif, but enters.
I do not see what could be the problem.
Can anyone help me?
Use && not ||.
elif [[ "$2" != "contar" && "$2" != "hist" ]] ; then
The reason for switching the operator is De Morgan's law. If you negate
if [[ $a == "foo" || $b == "bar" || $c == "baz" ]]
you get
# negated
if ! [[ $a == "foo" || $b == "bar" || $c == "baz" ]]
De Morgan's law says you can distribute the ! by switching all the =='s to !='s and changing || to &&.
# `!` distributed per De Morgan's law
if [[ $a != "foo" && $b != "bar" && $c != "baz" ]]
Related
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
I'm trying to be smart but it doesn't work. Can anybody help me to do this a bit simpler?
if [[ "${DATUM}" == "${today}" && "${STUNDE}" == "${HH24}" ]] || [[ "${DATUM}" == "${today}" && "${STUNDE}" == "20" ]] ||
[[ "${DATUM}" == "${today}" && "${STUNDE}" == "" && "20" == "${HH24}" ]]; then
Is there a way to combine it?
Your code can be translated to:
(C1 and C2) or (C1 and C3) or (C1 and C4 and C5)
Applying boolean arithmetics you can simplify it as:
C1 and (C2 or C3 or (C4 and C5))
This said, you can add a nested if statement to, first, check the C1 condition and, second, check the other conditions. It does not simplify the code a lot but here it is:
if [ "${DATUM}" = "${today}" ]; then
if [ "${STUNDE}" = "${HH24}" ] || [ "${STUNDE}" = "20" ] || { [ "${STUNDE}" = "" ] && [ "${HH24}" = "20" ]; }; then
# Insert the code to execute when the conditions are satisfied
fi
fi
As others have noted, your boolean expression can be simplified applying the law of distributivity of conjunction (⋀, *, AND) over disjunction (⋁, +, OR):
(a ⋀ b) ⋁ (a ⋀ c) = a ⋀ (b ⋁ c)
But to simplify it further, note you can, in bash, use && and || inside the (bash-specific) [[ .. ]] command:
[[ $a == 1 && $b == 2 ]]
Also, when using [[ .. ]] compound command (over POSIX [ .. ]) you don't have to quote variables. And to test for null-strings, you can use the shorter -z $var form over $var == "".
All this together yields:
if [[ $DATUM == $today ]] && [[ $STUNDE == $HH24 || $STUNDE == 20 || -z $STUNDE && $HH24 == 20 ]]; then
# ...
fi
To further simplify it, we would need to have more details on your application logic, possible values, etc.
I have just started learning to write bash scripts. This a simplified form of what I want to write.
The problem is despite of the input, it prints only "YES".
#! /usr/bin/bash
read input
if (("$input"== "y" || "$input" == "Y"))
then
echo "YES";
elif (("$input" == "n" || "$input" == "N"))
then
echo "NO";
else
echo "Not a valid input!!"
fi
Use [[ instead of (( like,
if [[ "$input" == "y" || "$input" == "Y" ]]
and also there must be a space exists before == operator.
ie,
input="n"
if [[ "$input" == "y" || "$input" == "Y" ]]
then
echo "YES";
elif [[ "$input" == "n" || "$input" == "N" ]]
then
echo "NO";
else
echo "Not a valid input!!"
fi
You could use regular expression also for condition checking purpose.
if [[ "$input" =~ ^[yY]$ ]]
then
echo "YES";
elif [[ "$input" =~ ^[nN]$ ]]
then
echo "NO";
else
echo "Not a valid input!!"
fi
When you automaticly convert the input to lowercase (using typeset), you do not have to bother with the the uppercases.
When you use an elif, always think 10 seconds for another solution. In this case you might want to use a "switch", in shell written as a case-statement:
#!/usr/bin/bash
typeset -l input
read input
case ${input} in
"y") echo "Yes";;
"n") echo "NO";;
*) echo "Not a valid input!!";;
esac
I have a shellscript that tells me missing ] in the line
if [ $status != "2" && $status != "3" && `echo "$temp1 > $upperLimit" | bc` = "1" ]
and also missing ] in the line
if [ $status = "2" && `cat motionsensordate` \> `date +%s` ]
Why is that?
The single [ doesn't support logical operators inside the brackets. You have to use them outside
if [ "$status" != 2 ] && [ "$status" != 3 ] ...
Use double quotes for variables in single brackets to prevent unary operator expected error when the variable is empty.
Or, switch to double brackets:
if [[ $status != 2 && $status != 3 ... ]]
Also, status different to 2 and 3 can be expressed by a pattern:
if [[ $status != [23] && ... ]]
And if you would like to (in addition to the answers here) group together conditions:
if [[ ( COND1 || COND2 ) && COND3 ]]
then
echo "$cmd"
break
fi
I'm having an issue getting a a simple y/n question to work. Consider the following code:
echo "Hi there"
read ans
if [[ $ans != "y" || $ans != "Y" || $ans != "YES" || $ans != "yes" ]]; then
echo "Foo"
exit 0
fi
I've looked at – I would argue – some of the more informative answers on StackOverflow for advice: Simple logical operators in Bash
I've tried all different types of variations such as:
if [[ ($ans != "y" || $ans != "Y" || $ans != "YES" || $ans != "yes") ]]; then
echo "Foo"
exit 0
fi
if [[ ($ans != "y*" || $ans != "Y*" || $ans != "YES*" || $ans != "yes*") ]]; then
echo "Foo"
exit 0
fi
if [[ ($ans != "y") || ($ans != "Y") || ($ans != "YES") || ($ans != "yes") ]]; then
echo "Foo"
exit 0
fi
Regardless of why I type in any of these cases, it automatically fails and I'm not sure why. If anyone has a better way to handle y/n answers then please let me know! Ideally I would like to use pattern matching (like I might do with Perl) but I'm not entirely sure the best way/most efficient way to accomplish a simple y/n question.
You need to use && instead of ||. As it stands you're saying if it's not equal to any of those possibilities, then execute the "then" block. You mean to say if it's not equal to all of them, then execute the "then" block. That requires &&.
You can use:
echo "Hi there"
read ans
case "$ans" in
y|Y|YES|yes)
;;
*)
echo "Foo"
exit 0
;;
esac
The logic needs to be adjusted:
echo "Hi there"
read ans
if ! [[ "$ans" == "y" || "$ans" == "Y" || "$ans" == "YES" || "$ans" == "yes" ]]; then
echo "Foo" # User answered no
exit 0
fi
The will echo "Foo" only if the answer was not one of "y", "Y" or "YES". By contrast, consider the original logic:
[[ $ans != "y" || $ans != "Y" || $ans != "YES" || $ans != "yes" ]]
At least one of these tests will be true regardless of what the user's answer is.
Using the case statement
You might consider using the case statement to analyze the user's answer:
read -p "Hi there: " ans
case "$ans" in
[yY]*) ;;
[nN]*) echo "Foo" ; exit 0 ;;
*) echo "You were supposed to answer yes or no" ;;
esac
Try read ans, not read $ans.