I'm learning BASH through HackerRank.There's an exercise in which the lengths of the triangle is given and then you need to find whether the triangle is isosceles,scalene or equilateral.I wrote the following code:
read a
read b
read c
if [ [ "$a" -eq "$b" ] && [ "$b" -eq "$c" ] ]
then
echo "EQUILATERAL"
elif [ [ "$a" -eq "$b" ] || [ "$b" -eq "$c" ] ]
then
echo "ISOSCELES"
else
echo "SCALENE"
fi
But then I get the following error
solution.sh: line 4: [: too many arguments
solution.sh: line 7: [: too many arguments
solution.sh: line 7: [: too many arguments
Why is this happening? I tried long and hard to rectify it but nothing worked out
You can combine conditions either ommiting the surrounding brackets like this
if [ "$a" -eq "$b" ] && [ "$b" -eq "$c" ]
or by combining the conditions with -a/-o like this
if [ "$a" -eq "$b" -a "$b" -eq "$c" ]
see http://wiki.bash-hackers.org/commands/classictest#and_and_or
&& and || are Bash list operators. In a chain of commands, the next command is executed only if the previous command returned 0 (&&) or nonzero (||).
[ is an alias for the Bash internal test command and has arguments such as -eq or -ne. ] ends its command line. Type help test for more information.
So if you write a conditional expression, you do not put the list operators inside brackets.
Try, for example, this instead of the respective line in your code:
if [ "$a" -eq "$b" ] && [ "$b" -eq "$c" ]
then
[ isn't a grouping operator in bash, you can't use it to group tests.
there are a number of different ways to express the tests you want to make, numeric evaluation mode is probably easiest to read
if (( a == b && b == c ))
if (( a == b || b == c || c == a ))
This is going to break if you have decimal fractions, but will work fine for integers.
[ is a conditional command, like an alias for sh's test built-in command.
[[ is the same for bash which has more test options.
So make a choice between [ and [[ but not [ [ which means two command.
Example:
# [ [ -n 'test' ] ]
bash: [: too many arguments
# [ -n 'test' ] && echo $?
0
# [[ -n 'test' ]] && echo $?
0
Related
This question already has answers here:
Meaning of "[: too many arguments" error from if [] (square brackets)
(6 answers)
Closed 4 years ago.
I want to check that of two variables both or neither are set. I've tried multiple options, and this is the most clean solution I've come up with:
if [ ! -z $A -a -z $B ] || [ -z $A -a ! -z $B ]; then
#error
fi
#success
When I run the script with both A and B set - it runs fine. But when I run it with A missing I get:
./test.sh: line 3: [: too many arguments
./test.sh: line 3: [: too many arguments
line 3 being the condition statement.
When I run it with B missing, I get:
./test.sh: line 3: [: argument expected
./test.sh: line 3: [: argument expected
Is it my condition that has wrong syntax or am I missing smth else?
You should try to avoid -a; it's non-standard and considered obsolete by the POSIX standard. Since || and && have equal precedence, you need to use { ... } to properly group the individual tests.
(This is in addition to the immediate need to quote your parameter expansions.)
if { [ ! -z "$A" ] && [ -z "$B" ]; } || { [ -z "$A" ] && [ ! -z "$B" ]; }; then
However, a simpler expression might be
if [ -z "$A$B" ] || { [ "$A" ] && [ "$B" ]; }; then
The concatenation of two strings is empty if and only if both strings are also empty.
[ "$A" ] is short for [ -n "$A" ], which is equivalent to [ ! -z "$A" ].
Using bash's [[ ... ]] command, you can write the more natural
if [[ -z $A && -n $B || -n $A && -z $B ]];
Quotes are optional in this case, and || and && are usable inside [[ ... ]] with the precedence you expect.
Quote your variables:
if [ ! -z "$A" -a -z "$B" ] || [ -z "$A" -a ! -z "$B" ]; then
If the variables are unquoted and unset, they are replaced with nothing, meaning that the command essentially becomes:
if [ ! -z -a -z ] || [ -z -a ! -z ]; then
resulting in the error you see.
You forgot to use quotation marks around your vars:
if [ ! -z "$A" -a -z "$B" ] || [ -z "$A" -a ! -z "$B" ]; then
echo "error"
fi
Bash will replace your vars in your script with the values, so when A=5 and B is unset, your version will read:
if [ ! -z 5 -a -z ] || [ -z 5 -a ! -z ]; then
You see that the syntax is wrong, as -z expects an argument. When using quotes, is reads:
if [ ! -z "5" -a -z "" ] || [ -z "5" -a ! -z "" ]; then
AS you can see, now the argument for B is an empty string, which is valid.
Also your version would have failed when setting A="string with spaces" when unquoted.
Suppose I am writing the following in a bash script:
if [ -z $a ] || [ -z $b ] ; then
usage
fi
It works but I would like to write it with short-circuiting as follows:
[ -z $a ] || [ -z $b ] || usage
Unfortunately it does not work. What am I missing ?
You want to execute usage in case either 1st or 2nd condition are accomplished. For that, you can do:
[ -z $a ] || [ -z $b ] && usage
Test:
$ [ -z "$a" ] || [ -z "$b" ] && echo "yes"
yes
$ b="a"
$ [ -z "$a" ] || [ -z "$b" ] && echo "yes"
yes
$ a="a"
$ [ -z "$a" ] || [ -z "$b" ] && echo "yes"
$
You could make use of the following form:
[[ expression ]]
and say:
[[ -z "$a" || -z "$b" ]] && usage
This would execute usage if either a or b is empty.
Always quote your variables. Saying
[ -z $a ]
if the variable a is set to foo bar would return an error:
bash: [: foo: binary operator expected
When I run this bash script :
if [ [$EUID -ne 0] ]; then
echo "This script must be run as root" 1>&2
exit 1
else
printf " whathever "
exit 0
fi
I have this error :
./myScript: 15: [: Illegal number: [
Do you see any problem ?
You have syntax error in your if condition, use this if condition:
if [ "$EUID" -ne 0 ];
OR using [[ and ]]
if [[ "$EUID" -ne 0 ]];
You have syntax error in your if condition, use this if condition:
if [ "$EUID" -ne 0 ];
OR using [[ and ]]
if [[ "$EUID" -ne 0 ]];
If you use the KSH88+/Bash 3+ internal instruction [[, it's not necessary to use doubles quotes around the variables operands :
[ ~/test]$ [[ $var2 = "string with spaces" ]] && echo "OK" || echo "KO"
OK
Instead of the external command test or his fork [ :
[ ~/test]$ [ $var2 = "string with spaces" ] && echo "OK" || echo "KO"
bash: [: too many arguments
KO
[ ~/test]$ [ "$var2" = "string with spaces" ] && echo "OK" || echo "KO"
OK
Of course, you also have to choose the operators according to the type of operands :
[ ~/test]$ var1="01"
[ ~/test]$ [ "$var1" = "1" ] && echo "OK" || echo "KO"
KO
[ ~/test]$ [ "$var1" -eq "1" ] && echo "OK" || echo "KO"
OK
two suggestions apart from what everyone else has pointed out already.
rather than doing else [bunch of code because we are root] fi, just replace the else with fi. once you've tested for the failure condition you are concerned about and taken appropriate action, no need to continue to be within the body of the conditional.
$EUID is a bashism, if you would like to make this portable to shells such as ksh, replacing it with:
if [ $(id -u) -ne 0 ]; then echo "ur not root bro"; exit 1; fi
would be a good way to do it.
using
sudo bash shell_script.sh
instead of
sudo sh shell_script.sh
solved in my case.
I've seen two ways in tutorials to do syntax for if statements in BASH shell:
This one wouldn't work unless I put quotes around the variable and added additional [ and ]:
if [[ "$step" -eq 0 ]]
This one worked without putting quotes around the variable and the additional [ and ] weren't needed:
if [ $step -ge 1 ] && [ $step -le 52 ]
Which is correct and best practice? What are the differences? Thanks!
"When referencing a variable, it is generally advisable to enclose its name in double quotes" -- http://tldp.org/LDP/abs/html/quotingvar.html
if [ $step -ge 1 ] && [ $step -le 52 ] can be replaced as
if [ "$step" -ge 1 -a "$step" -le 52 ]
if [[ "$step" -eq 0 ]] can be replaced as if [ "$step" -eq 0 ]
Also, suppose you have the following script:
#!/bin/bash
if [ $x -eq 0 ]
then
echo "hello"
fi
You get this error when you run the script -- example.sh: line 2: [: -eq: unary operator expected
But using if [ "$x" -eq 0 ]
You get a different error when you run the script -- example.sh: line 2: [: : integer expression expected
Thus, it is always best to put variables inside quotes...
if [[ .... ]] syntax is particularly useful when you have regex in the condition statement -- http://honglus.blogspot.com/2010/03/regular-expression-in-condition.html
EDIT: When we deal with strings --
#!/bin/bash
if [ $x = "name" ]
then
echo "hello"
fi
You get this error when you run the script -- example.sh: line 2: [: =: unary operator expected
But, if you use if [ "$x" = "name" ] it runs fine (i.e. no errors ) and if statement is evaluated as false, as value of x is null which does not match name.
I'm trying to write a script that will check two error flags, and in case one flag (or both) are changed it'll echo-- error happened. My script:
my_error_flag=0
my_error_flag_o=0
do something.....
if [[ "$my_error_flag"=="1" || "$my_error_flag_o"=="2" ] || [ "$my_error_flag"="1" && "$my_error_flag_o"="2" ]]; then
echo "$my_error_flag"
else
echo "no flag"
fi
Basically, it should be, something along:
if ((a=1 or b=2) or (a=1 and b=2))
then
display error
else
no error
fi
The error I get is:
line 26: conditional binary operator expected
line 26: syntax error near `]'
line 26: `if [[ "$my_error_flag"=="1" || "$my_error_flag_o"=="2" ] || [ "$my_error_flag"="1" && "$my_error_flag_o"="2" ]]; then'
Are my brackets messed up?
Use -a (for and) and -o (for or) operations.
tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html
Update
Actually you could still use && and || with the -eq operation. So your script would be like this:
my_error_flag=1
my_error_flag_o=1
if [ $my_error_flag -eq 1 ] || [ $my_error_flag_o -eq 2 ] || ([ $my_error_flag -eq 1 ] && [ $my_error_flag_o -eq 2 ]); then
echo "$my_error_flag"
else
echo "no flag"
fi
Although in your case you can discard the last two expressions and just stick with one or operation like this:
my_error_flag=1
my_error_flag_o=1
if [ $my_error_flag -eq 1 ] || [ $my_error_flag_o -eq 2 ]; then
echo "$my_error_flag"
else
echo "no flag"
fi
You can use either [[ or (( keyword. When you use [[ keyword, you have to use string operators such as -eq, -lt. I think, (( is most preferred for arithmetic, because you can directly use operators such as ==, < and >.
Using [[ operator
a=$1
b=$2
if [[ a -eq 1 || b -eq 2 ]] || [[ a -eq 3 && b -eq 4 ]]
then
echo "Error"
else
echo "No Error"
fi
Using (( operator
a=$1
b=$2
if (( a == 1 || b == 2 )) || (( a == 3 && b == 4 ))
then
echo "Error"
else
echo "No Error"
fi
Do not use -a or -o operators Since it is not Portable.
Please try following
if ([ $dateR -ge 234 ] && [ $dateR -lt 238 ]) || ([ $dateR -ge 834 ] && [ $dateR -lt 838 ]) || ([ $dateR -ge 1434 ] && [ $dateR -lt 1438 ]) || ([ $dateR -ge 2034 ] && [ $dateR -lt 2038 ]) ;
then
echo "WORKING"
else
echo "Out of range!"
You can get some inspiration by reading an entrypoint.sh script written by the contributors from MySQL that checks whether the specified variables were set.
As the script shows, you can pipe them with -a, e.g.:
if [ -z "$MYSQL_ROOT_PASSWORD" -a -z "$MYSQL_ALLOW_EMPTY_PASSWORD" -a -z "$MYSQL_RANDOM_ROOT_PASSWORD" ]; then
...
fi