Hi I have the script as below.
if [[ 'whoami' -eq "test" ]]; then
echo "test user"
else
echo "Not a test user"
fi
This is returning test user even if my user is not text. It would be great if someone can pin point my silly mistake.
-eq is for integer testing.
e.g.:
kent$ [[ "a" -eq "b" ]] && echo "y" || echo "n"
y
kent$ [[ 34 -eq 35 ]] && echo "y" || echo "n"
n
try:
kent$ [[ "a" == "b" ]] && echo "y" || echo "n"
n
You are using single quotes in place of backticks:
if [[ `whoami` -eq "test" ]]; then # Not 'whoami'
For improved readability (and easier nesting, if applicable), use
if [[ $(whoami) -eq "test" ]]; then
instead.
Just like this:
if [[ `whoami` == "test" ]]; then # Not 'whoami' -eq
echo "test user"
else
echo "Not a test user"
fi
don't use single quote and '-eq'
Related
$var=1545334345
The $var is dynamic. How do I check if it starts with 15? I tried:
if [[ $var == 15* ]]; then
echo "Yes"
fi
and
if (( $var -eq 15* )); then
echo "Yes"
fi
Neither works.
Remove the $ from the assignment.
var=1545334345
The first if statement you wrote is correct.
if [[ $var == 15* ]]; then
echo "Yes"
fi
Here is my code
#! /bin/bash
read var
if [ $var="Y" -o $var="y" ]
then
echo "YES"
else
echo "NO"
fi
I want to print YES if the user presses y or Y, otherwise I want to print NO. Why doesn't this code work?
Basically, your Condition is wrong. Quote your variables and leave spaces between operators (like shellter wrote). So it should look like:
#! /bin/bash
read var
if [ "$var" = "Y" ] || [ "$var" = "y" ]
then
echo "YES"
else
echo "NO"
fi
Edit: for POSIX ccompatibility
Replaced == with = - see comments
Replaced -o syntax with || syntax - see comments
With Bash, you can also use regular expression in your test with the =~ operator:
read var
[[ "$var" =~ [Yy] ]] && echo "YES" || echo "NO"
Or as Benjamin W. mentionned, simply use character range with the == operator:
read var
[[ "$var" == [Yy] ]] && echo "YES" || echo "NO"
There is minor syntax error in your code.
Correction : There should be a white space between operators and variables
read var
if [ $var = "Y" -o $var = "y" ]
then
echo "YES"
else
echo "NO"
fi
Try the above bash script.
Hope it would work fine.
Happy Coding!
If all you require is a upper/lowercase comparison, use the ,, operator on the variable being compared ( note the ${var,,} ):
#!/bin/bash
read var
if [ ${var,,} = "y" ]
then
echo "YES"
else
echo "NO"
fi
or more succinctly:
#!/bin/bash
read var
[ ${var,,} = 'y' ] && echo 'YES' || echo 'NO'
or the way I might actually do it:
#!/bin/bash
read var
[[ "${var,,}" == 'y' ]] && echo 'YES' || echo 'NO'
Below is the code that I tried.
#! /bin/bash
read -p "Are you Sure?(Y/N) " answer
if [ "$answer" = "y" ] || [ "$answer" = "Y" ]; then
echo "Do your stuff."
else
echo "Do your other stuff"
fi
Add whitespace around '=' and your code will run fine.
#! /bin/bash
read var
if [ $var = "Y" -o $var = "y" ]
then
echo "YES"
else
echo "NO"
fi
Try this code.
#! /bin/bash
read var
echo -e "YES\nNO\n" | grep -i $var
I have just started learning to write bash scripts. This a simplified form of what I want to write.
The problem is despite of the input, it prints only "YES".
#! /usr/bin/bash
read input
if (("$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 am attempting to run a block of code if one flag is set to true and the other is set to false. ie
var1=true
var2=false
if [[ $var1 && ! $var2 ]]; then var2="something"; fi
Since that did not evaluate the way that I expected I wrote several other test cases and I am having a hard time understanding how they are being evaluated.
aa=true
bb=false
cc="python"
if [[ "$aa" ]]; then echo "Test0" ; fi
if [[ "$bb" ]]; then echo "Test0.1" ; fi
if [[ !"$aa" ]]; then echo "Test0.2" ; fi
if [[ ! "$aa" ]]; then echo "Test0.3" ; fi
if [[ "$aa" && ! "$bb" ]]; then echo "Test1" ; fi
if [[ "$aa" && ! "$aa" ]]; then echo "Test2" ; fi
if [[ "$aa" ]] && ! [[ "$bb" ]]; then echo "test3" ; fi
if [[ "$aa" ]] && ! [[ "$cc" ]]; then echo "test4" ; fi
if [[ $aa && ! $bb ]]; then echo "Test5" ; fi
if [[ $aa && ! $aa ]]; then echo "Test6" ; fi
if [[ $aa ]] && ! [[ $bb ]]; then echo "test7" ; fi
if [[ $aa ]] && ! [[ $cc ]]; then echo "test8" ; fi
When I run the preceding codeblock the only output I get is
Test0
Test0.1
Test0.2
however, my expectation is that I would get
Test0
Test1
Test3
Test5
Test7
I have tried to understand the best way to run similar tests, however most examples I have found are set up in the format of
if [[ "$aa" == true ]];
which is not quite what I want to do. So my question is what is the best way to make comparisons like this, and why do several of the test cases that I would expect to pass simply not?
Thank you!
Without any operators, [[ only checks if the variable is empty. If it is, then it is considered false, otherwise it is considered true. The contents of the variables do not matter.
Your understanding of booleans in shell context is incorrect.
var1=true
var2=false
Both the above variables are true since those are non-empty strings.
You could instead make use of arithmetic context:
$ a=1
$ b=0
$ ((a==1 && b==0)) && echo y
y
$ ((a==0 && b==0)) && echo y
$
$ ((a && !(b))) && echo y; # This seems to be analogous to what you were attempting
y
The shell does not have Boolean variables, per se. However, there are commands named true and false whose exit statuses are 0 and 1, respectively, and so can be used similarly to Boolean values.
var1=true
var2=false
if $var1 && ! $var2; then var2="something"; fi
The difference is that instead of testing if var1 is set to a true value, you expand it to the name of a command, which runs and succeeds. Likewise, var2 is expanded to a command name which runs and fails, but because it is prefixed with ! the exit status is inverted to indicate success.
(Note that unlike most programming languages, an exit status of 0 indicates success because while most commands have 1 way to succeed, there are many different ways they could fail, so different non-zero values can be assigned different meanings.)
true and false are evaluated as strings ;)
[[ $var ]] is an equivalent of [[ -n $var ]] that check if $var is empty or not.
Then, no need to quote your variables inside [[. See this reminder.
Finally, here is an explication of the difference between && inside brackets and outside.
The closest you can come seems to be use functions instead of variables because you can use their return status in conditionals.
$ var1() { return 0; }
$ var2() { return 1; } # !0 = failure ~ false
and we can test this way
$ var1 && echo "it's true" || echo "it's false"
it's true
$ var2 && echo "it's true" || echo "it's false"
it's false
or this way
$ if var1; then echo "it's true"; else echo "it's false"; fi
it's true
$ if var2; then echo "it's true"; else echo "it's false"; fi
it's false
Hope this helps.
I found an interesting Bash script that will test if a variable is numeric/integer. I like it, but I do not understand why the "0" is not recognized as a number? I can not ask the author, hi/shi is an anonymous.
#!/bin/bash
n="$1"
echo "Test numeric '$n' "
if ((n)) 2>/dev/null; then
n=$((n))
echo "Yes: $n"
else
echo "No: $n"
fi
Thank you!
UPDATE - Apr 27, 2012.
This is my final code (short version):
#!/bin/bash
ANSWER=0
DEFAULT=5
INDEX=86
read -p 'Not choosing / Wrong typing is equivalent to default (#5): ' ANSWER;
shopt -s extglob
if [[ $ANSWER == ?(-)+([0-9]) ]]
then ANSWER=$((ANSWER));
else ANSWER=$DEFAULT;
fi
if [ $ANSWER -lt 1 ] || [ $ANSWER -gt $INDEX ]
then ANSWER=$DEFAULT;
fi
It doesn't test if it is a numeric/integer. It tests if n evaluates to true or false, if 0 it is false, else (numeric or other character string) it is true.
use pattern matching to test:
if [[ $n == *[^0-9]* ]]; then echo "not numeric"; else echo numeric; fi
That won't match a negative integer though, and it will falsely match an empty string as numeric. For a more precise pattern, enable the shell's extended globbing:
shopt -s extglob
if [[ $n == ?(-)+([0-9]) ]]; then echo numeric; else echo "not numeric"; fi
And to match a fractional number
[[ $n == #(?(-)+([0-9])?(.*(0-9))|?(-)*([0-9]).+([0-9])) ]]