This question already has answers here:
How do I compare two string variables in an 'if' statement in Bash? [duplicate]
(12 answers)
Closed 6 years ago.
Here is a random number generator followed by a line clear timer and you are then to try and input the number that you saw. I cant seem to set up the comparison between the randNum variable and the input. I tried setting a defined value for the input as well and still receive an error "command not found" when checking input vs randNum variable
n=$RANDOM$RANDOM$RANDOM; let "n %= 10000000000";
echo $n
for i in {5..1}; do echo $i; sleep 1; tput cuu1; tput el; done
tput cuu1; tput el
echo "what was that number?"
#read input
input=999999999
if [$input == $n]
then
echo "you are correct"
else
echo "you are incorrect"
fi
Shells need spaces around brackets [ and ] when used for this purpose (replacement syntax for the test command), and you should better use -eq instead of == for numeric comparison:
if [ $input -eq $n ]
If you use the arithmetic comparison, rather than the old testcommand (a.k.a. [) then you can avoid these issues. Spaces here are less important inside the brackets:
if (( input == n ))
Note that the leading $ is not required (and using it can cause issues)
Related
This question already has answers here:
How do I do if statement arithmetic in bash?
(5 answers)
When are square brackets required in a Bash if statement?
(3 answers)
Closed 2 years ago.
The code I wrote to solve a challenge keeps giving me a response of "True" and I have no idea why. I've played with the syntax and keep getting "True", so there's probably something I'm not understanding about the way Bash works.
I apologize if this is too simple or if there are similar questions out there. I looked but don't know how to phrase the issue.
The test values input are n=3 x=3 y=4 (says true, should be false) and n=12 x=3 x=4 (does not reach this point of the test.
#!/bin/bash
read n
echo "n is " $n
read x
echo "x is " $x
read y
echo "y is " $y
if [[ ((n%x==0 && n%y==0)) ]]; then
echo "true"
else
echo "false"
fi
Get rid of the unnecessary [[]]. Your new if statement would look like if ((n%x==0 && n%y==0)).
This question already has answers here:
How do I test if a variable is a number in Bash?
(40 answers)
How to check if a variable is set in Bash
(38 answers)
Closed 4 years ago.
Given the following code:
Max=
if [[ something exists.. ]]; then
Max=2
// .. more code that can changes the value of Max
fi
// HERE
How can I check at "HERE" if Max is equal to some number (setted)?
if [ -z "$Max" ]
then
echo "Max: not set"
else if [ $Max -eq some_number ]
then
echo "Max is equal to some number"
else
echo "Max is set but not equal to some number"
fi
or
if [ -n "$Max" -a "$Max" = "some_number" ]
...
Notice that the second example is doing a string comparison which can solve some headaches but may hurt the sensibilities of purists.
Cool !
Max is set by default to an empty value, that when used in an arithmetic context, returns zero. For example:
Try running echo $(( definitelyNotAGivenName)) it comes out as zero, cool!
You can use the round brackets for arithmetic comparing - see more here and here
This question already has answers here:
How can I compare two floating point numbers in Bash?
(22 answers)
Closed 6 years ago.
I am new to bash scripting. My code is :
MK=M
SM=1.4
if [[ $MK == "M" ]]
then
if [ $SM -gt 1.3 ]
then
echo "Greater than 1.3M"
else
echo "Less than 1.3M"
fi
else
echo "Not yet M...."
fi
Reply:
/tmp/tmp.sh: line 6: [: 1.4: integer expression expected
Less than 1.3M
What am i doing wrong ?
Here's what man bash says:
arg1 OP arg2 ... Arg1 and arg2 may be positive or negative integers.
You seem to be trying to compare floating point numbers.
It's ultimately because bash is not terribly patient when it comes to floating point numbers. In very simple terms here, I would suggest you do one of two things:
It looks like you are trying to determine if something is larger than 1.3 Mb or not, is this correct? If this is the case, then leave everything the way you have it, and just use Kb for $sm and the compare
like this:
#/bin/bash
mk="p"
km="p"
sm="1400"
ms="1300"
if [[ $mk == $km ]]
then
if [ $sm > $ms ]
then
echo "Greater than 1.3M"
else
echo "Less than 1.3M"
fi
else
echo "Not yet M...."
fi
or
Use bc for the calculation of the floating point numbers...
# /bin/bash
mk="p"
km="p"
sm="1.4"
ms="1.3"
if [ $(echo "$mk == $km" | bc) ]
then
if [ $(echo "$sm > $ms" | bc) ]
then
echo "Greater than 1.3M"
else
echo "Less than 1.3M"
fi
else
echo "Not yet M...."
fi
One more thing to note here, is that, as you can see from my code, I have primed new variables with the data, rather than using raw letters and numbers in boolean and compare operations, which can have some genuinely unexpected results. Also, while they may work in some conditions, temporarily, bash prefers all variable names to be in lower-case. Let me know if you have any questions.. However, I have tested both code chunks, and they both work fine.
This question already has answers here:
Sequences expansion and variable in bash [duplicate]
(7 answers)
Closed 7 years ago.
Im trying a code in bash to generate prime nos as follows:
#!/bin/bash
echo "Enter till where u wish to generate"
read num
echo "Generating prime numbers from 2 to $num"
flag="prime"
for i in {2..$num}
do
for j in {2..$((${num}-1))}
do
[ $((${i}%${j})) -eq 0 ] && flag="nprime" || flag="prime"
break
done
[ "$flag" == "prime" ] && echo "$i"
done
Upon execution, it throws an error because the for loop takes the sequence mentioned in the curly braces as it is not as a sequence.
Could you guide me as to where am i going wrong ?
man bash in my version says:
A sequence expression takes the form {x..y[..incr]}, where x and y are either integers or single characters, and incr, an optional increment, is an integer.
You can't use variables in ranges. Try seq instead:
for i in $(seq 2 $num) ; do
Note that incr for seq goes between x and y.
Use:
for ((i=2; i<=$num; i++))
This question already has answers here:
How do I compare two string variables in an 'if' statement in Bash? [duplicate]
(12 answers)
Closed 8 years ago.
I don't know why this isn't working. Please help!
#!/bin/bash
clear
echo "Enter an option"
read $option
if ("$option" == 1) then
echo "Blah"
fi
I tried like this
if ("$option" -eq 1) then
I can't see why the if statement isn't being run. All I want to do is check what the user entered and do something depending on the value entered.
The syntax for an equality check is:
if [[ $option == 1 ]]; then
echo "Blah"
fi
Or, for compatibility with older non-bash shells:
if [ "$option" = 1 ]; then
echo "Blah"
fi
In either one, the whitespace is important. Do not delete the spaces around the square brackets.
That is not the syntax for an if statement in bash. Try this:
if [ "$option" = "1" ]; then
echo "Blah"
fi
I'm not sure where you got your syntax from...try this:
if [ "$option" -eq 1 ]; then
In the shell, [ is a command, not a syntactic construct.
Alternatively, you can use an arithmetic context in bash:
if (( option == 1 )); then
Bash if uses [ or [[ as test constructs, instead of parenthesis which are used in other languages.