I've seen questions regarding the same issue, but all of them are about strings. How about integers? Why am I getting the "unary operator expected" error?
if [ $(date +%k%M) -ge ${!BLOCK1FRAN} ] ; then whatever ; fi
You are using indirection. If the variable ${BLOCK1FRAN} points to an empty variable, you'll get the error message. Make sure that the variable pointed by ${BLOCK1FRAN} contains a valid numeric value.
If you want an empty string and nonnumeric values to be evaluated as zero (0), use the following syntax.
if [[ $(date +%k%M) -ge ${!BLOCK1FRAN} ]]; then whatever ; fi
It looks good to me. Are you sure you've set BLOCK1FRAN correctly?
whatever() { echo "it works"; }
foo=42
BLOCK1FRAN=foo
if [ $(date +%k%M) -ge ${!BLOCK1FRAN} ] ; then whatever ; fi
it works
Related
This question already has answers here:
What is the proper way to test a Bash function's return value?
(5 answers)
Closed 2 years ago.
I would like to do the following in shell scripting:
if [ (is_number $arg1) -ne 0 ] || [ (is_number $arg2) -ne 0 ] ; then
printf "Exit. Valid input must be positive integer.\n"
return 1
fi
But it gave me syntax error. Do you have better solution? Thanks!
I want a solution without accessing $? or storing return value into variable, unless such answer doesn't exist.
I see you guys taking advantage of 0 and not 0. It make sense since 0 signify whether succeed or not.
Now, speaking it in a pure syntax way, I want to do the following:
if [ (is_number $arg1) -ne int ] || [ (is_number $arg2) -ne int] ; then
printf "Exit. Valid input must be positive integer.\n"
return 1
fi
Please provide a way, thanks.
Do you have better solution?
Call the actual functions.
is_number() { [[ "$1" =~ ^[0-9]+$ ]]; }
if ! is_number "$arg1" || ! is_number "$arg2"; then
I'm pretty new with Bash scripting and am having trouble getting my 'while' loop to run. When I echo keywords, a whole list of words prints and then when I echo length, it prints 124. I believe I'm using the while loop and condition correctly, so I can't figure out what I'm doing wrong. Any thoughts?
keywords=$1
length=${#keywords}
echo "$keywords"
echo "$length"
if [ -z "$keywords" ]; then
while [ $length -gt 100 ]; do
echo "$keywords"
echo "$length"
keywords="${keywords%,*}"
length=${#keywords}
done
fi
echo $keywords
The problem is here:
[ -z "$keywords" ]
-z is true if its argument is an empty string. Something of length 124 is definitely far from empty. You probably meant -n.
Next time, please also include the input in the question so we can reproduce the problem.
I have two variables
var=""
var1=abcd
Here is my shell script code
if [ $var == $var1 ]; then
do something
else
do something
fi
If I run this code it will prompt a warning
[: ==: unary operator expected
How can I solve this?
Since the value of $var is the empty string, this:
if [ $var == $var1 ]; then
expands to this:
if [ == abcd ]; then
which is a syntax error.
You need to quote the arguments:
if [ "$var" == "$var1" ]; then
You can also use = rather than ==; that's the original syntax, and it's a bit more portable.
If you're using bash, you can use the [[ syntax, which doesn't require the quotes:
if [[ $var = $var1 ]]; then
Even then, it doesn't hurt to quote the variable reference, and adding quotes:
if [[ "$var" = "$var1" ]]; then
might save a future reader a moment trying to remember whether [[ ... ]] requires them.
Why all people want to use '==' instead of simple '=' ? It is bad habit! It used only in [[ ]] expression. And in (( )) too. But you may use just = too! It work well in any case. If you use numbers, not strings use not parcing to strings and then compare like strings but compare numbers. like that
let -i i=5 # garantee that i is nubmber
test $i -eq 5 && echo "$i is equal 5" || echo "$i not equal 5"
It's match better and quicker. I'm expert in C/C++, Java, JavaScript. But if I use bash i never use '==' instead '='. Why you do so?
I couldn't find any one simple straightforward resource spelling out the meaning of and fix for the following BASH shell error, so I'm posting what I found after researching it.
The error:
-bash: [: too many arguments
Google-friendly version: bash open square bracket colon too many arguments.
Context: an if condition in single square brackets with a simple comparison operator like equals, greater than etc, for example:
VARIABLE=$(/some/command);
if [ $VARIABLE == 0 ]; then
# some action
fi
If your $VARIABLE is a string containing spaces or other special characters, and single square brackets are used (which is a shortcut for the test command), then the string may be split out into multiple words. Each of these is treated as a separate argument.
So that one variable is split out into many arguments:
VARIABLE=$(/some/command);
# returns "hello world"
if [ $VARIABLE == 0 ]; then
# fails as if you wrote:
# if [ hello world == 0 ]
fi
The same will be true for any function call that puts down a string containing spaces or other special characters.
Easy fix
Wrap the variable output in double quotes, forcing it to stay as one string (therefore one argument). For example,
VARIABLE=$(/some/command);
if [ "$VARIABLE" == 0 ]; then
# some action
fi
Simple as that. But skip to "Also beware..." below if you also can't guarantee your variable won't be an empty string, or a string that contains nothing but whitespace.
Or, an alternate fix is to use double square brackets (which is a shortcut for the new test command).
This exists only in bash (and apparently korn and zsh) however, and so may not be compatible with default shells called by /bin/sh etc.
This means on some systems, it might work from the console but not when called elsewhere, like from cron, depending on how everything is configured.
It would look like this:
VARIABLE=$(/some/command);
if [[ $VARIABLE == 0 ]]; then
# some action
fi
If your command contains double square brackets like this and you get errors in logs but it works from the console, try swapping out the [[ for an alternative suggested here, or, ensure that whatever runs your script uses a shell that supports [[ aka new test.
Also beware of the [: unary operator expected error
If you're seeing the "too many arguments" error, chances are you're getting a string from a function with unpredictable output. If it's also possible to get an empty string (or all whitespace string), this would be treated as zero arguments even with the above "quick fix", and would fail with [: unary operator expected
It's the same 'gotcha' if you're used to other languages - you don't expect the contents of a variable to be effectively printed into the code like this before it is evaluated.
Here's an example that prevents both the [: too many arguments and the [: unary operator expected errors: replacing the output with a default value if it is empty (in this example, 0), with double quotes wrapped around the whole thing:
VARIABLE=$(/some/command);
if [ "${VARIABLE:-0}" == 0 ]; then
# some action
fi
(here, the action will happen if $VARIABLE is 0, or empty. Naturally, you should change the 0 (the default value) to a different default value if different behaviour is wanted)
Final note: Since [ is a shortcut for test, all the above is also true for the error test: too many arguments (and also test: unary operator expected)
Just bumped into this post, by getting the same error, trying to test if two variables are both empty (or non-empty). That turns out to be a compound comparison - 7.3. Other Comparison Operators - Advanced Bash-Scripting Guide; and I thought I should note the following:
I used -e thinking it means "empty" at first; but that means "file exists" - use -z for testing empty variable (string)
String variables need to be quoted
For compound logical AND comparison, either:
use two tests and && them: [ ... ] && [ ... ]
or use the -a operator in a single test: [ ... -a ... ]
Here is a working command (searching through all txt files in a directory, and dumping those that grep finds contain both of two words):
find /usr/share/doc -name '*.txt' | while read file; do \
a1=$(grep -H "description" $file); \
a2=$(grep -H "changes" $file); \
[ ! -z "$a1" -a ! -z "$a2" ] && echo -e "$a1 \n $a2" ; \
done
Edit 12 Aug 2013: related problem note:
Note that when checking string equality with classic test (single square bracket [), you MUST have a space between the "is equal" operator, which in this case is a single "equals" = sign (although two equals' signs == seem to be accepted as equality operator too). Thus, this fails (silently):
$ if [ "1"=="" ] ; then echo A; else echo B; fi
A
$ if [ "1"="" ] ; then echo A; else echo B; fi
A
$ if [ "1"="" ] && [ "1"="1" ] ; then echo A; else echo B; fi
A
$ if [ "1"=="" ] && [ "1"=="1" ] ; then echo A; else echo B; fi
A
... but add the space - and all looks good:
$ if [ "1" = "" ] ; then echo A; else echo B; fi
B
$ if [ "1" == "" ] ; then echo A; else echo B; fi
B
$ if [ "1" = "" -a "1" = "1" ] ; then echo A; else echo B; fi
B
$ if [ "1" == "" -a "1" == "1" ] ; then echo A; else echo B; fi
B
Another scenario that you can get the [: too many arguments or [: a: binary operator expected errors is if you try to test for all arguments "$#"
if [ -z "$#" ]
then
echo "Argument required."
fi
It works correctly if you call foo.sh or foo.sh arg1. But if you pass multiple args like foo.sh arg1 arg2, you will get errors. This is because it's being expanded to [ -z arg1 arg2 ], which is not a valid syntax.
The correct way to check for existence of arguments is [ "$#" -eq 0 ]. ($# is the number of arguments).
I also faced same problem. #sdaau answer helped me in logical way. Here what I was doing which seems syntactically correct to me but getting too many arguments error.
Wrong Syntax:
if [ $Name != '' ] && [ $age != '' ] && [ $sex != '' ] && [ $birthyear != '' ] && [ $gender != '' ]
then
echo "$Name"
echo "$age"
echo "$sex"
echo "$birthyear"
echo "$gender"
else
echo "Enter all the values"
fi
in above if statement, if I pass the values of variable as mentioned below then also I was getting syntax error
export "Name"="John"
export "age"="31"
export "birthyear"="1990"
export "gender"="M"
With below syntax I am getting expected output.
Correct syntax:
if [ "$Name" != "" -a "$age" != "" -a "$sex" != "" -a "$birthyear" != "" -a "$gender" != "" ]
then
echo "$Name"
echo "$age"
echo "$sex"
echo "$birthyear"
echo "$gender"
else
echo "it failed"
fi
There are few points which we need to keep in mind
use "" instead of ''
use -a instead of &&
put space before and after operator sign like [ a = b], don't use as [ a=b ] in if condition
Hence above solution worked for me !!!
Some times If you touch the keyboard accidentally and removed a space.
if [ "$myvar" = "something"]; then
do something
fi
Will trigger this error message. Note the space before ']' is required.
I have had same problem with my scripts. But when I did some modifications it worked for me. I did like this :-
export k=$(date "+%k");
if [ $k -ge 16 ]
then exit 0;
else
echo "good job for nothing";
fi;
that way I resolved my problem. Hope that will help for you too.
i am testing with the shell script below:
#!/bin/ksh -x
instance=`echo $1 | cut -d= -f2`
if [ $instance == "ALL" ]
then
echo "strings matched \n"
fi
It's giving this error in the if condition:
: ==: unknown test operator
is == really not the correct syntax to use?
I am running on the command line as below
test_lsn_2 INSTANCE=ALL
Could anybody please suggest a solution.
Thanks.
To compare strings you need a single =, not a double. And you should put it in double quotes in case the string is empty:
if [ "$instance" = "ALL" ]
then
echo "strings matched \n"
fi
I see that you are using ksh, but you added bash as a tag, do you accept a bash-related answer?
Using bash you can do it in these ways:
if [[ "$instance" == "ALL" ]]
if [ "$instance" = "ALL" ]
if [[ "$instance" -eq "ALL" ]]
See here for more on that.
Try
if [ "$instance" = "ALL" ]; then
There were several mistakes:
You need double quotes around the variable to protect against the (unlikely) case that it's empty. In this case, the shell would see if [ = "ALL" ]; then which isn't valid.
Equals in the shell uses a single = (there is no way to assign a value in an if in the shell).
totest=$1
case "$totest" in
"ALL" ) echo "ok" ;;
* ) echo "not ok" ;;
esac
I'va already answered a similar question. Basically the operator you need is = (not ==) and the syntax breaks if your variable is empty (i.e. it becomes if [ = ALL]). Have a look at the other answer for details.