I am tying to check if a variable is null, It throws the below error. I have tried the below combinations but it says the same error. I am trying to find out the files in a folder
cmd="ls -tm"
output=$($cmd)
echo $output // doesn't print out anything as the folder is empty
if [ $output != "" ];
For the below one it goes to the loop
if[ -z $output ];
Error output:
line 25: [: !=: unary operator expected
Any suggestions for handling this?
Have you tried using if [ "$output" != "" ]; instead of if [ $output != "" ];?
To expand on #Quine's suggestion, if $output is null, your if statement expands to
if [ != "" ];
which makes bash think an argument is missing. Quoting the parameter name overcomes this.
Related
I have a very simple shell script which is failing at if-else. And I am not able to understand what is wrong. I checked my syntax and indentation.
echo_usr()
{
read -p "Do you want to preserve $EIP_HOME/conf and $EIP_HOME/installer/.datastore? [Y/N] " usr_in
kill_java()
if [ "$usr_in" == "y*" ] || [ "$usr_in" == "Y*" ]; then
preserve_conf()
else
if [ "$usr_in" == "n*" ] || [ "$usr_in" == "N*" ]; then
reset_home_eip()
else
echo "Invalid input"
fi
fi
reset_db()
}
It is giving me the following error:
syntax error near unexpected token `else'
To call a function with no arguments, just use the name of the function. Empty parentheses are not necessary, and in fact are a syntax error. (Or, more precisely, it's part of the syntax for a function definition, and if you try to use it as a function call you'll get a syntax error message later -- if you're lucky.)
The idea is that a call to a shell function looks and acts like an invocation of an external program.
For example, change:
preserve_conf()
to
preserve_conf
(Using elif rather than else followed by if is a good idea, but it's not the cause of the problem you're seeing.)
There's another problem that you won't see until you get past the syntax error (thanks to Sam for pointing it out in a comment). This:
if [ "$usr_in" == "y*" ] || [ "$usr_in" == "Y*" ]; then
isn't going to work correctly; the == operator used by the [ / test built-in doesn't do wildcard matching. If you're using bash, you can use [[ rather than [, and you can also combine the y and Y cases into a single pattern:
if [[ "$usr_in" == [yY]* ]] ; then
...
elif [[ "$usr_in" == [nN]* ]] ; then
...
fi
If you don't have a shell that supports that syntax, a case statement might be your next best bet -- in fact that's probably even cleaner than the if version:
case "$usr_in" in
[yY]*)
...
;;
[nN]*)
...
;;
*)
...
;;
esac
Okay, so within my script (this is my first time working with Bash) I am being met with two unary operator expected errors. The code itself is actually working fine, but it's presenting me with these errors at runtime:
[: !=: unary operator expected
For the line:
if [ ${netmask[1]} != "" ]; do
So for the first error, it's thrown when ${netmask[1]} is "" (null). I have tried multiple ideas and still can't get it to work without returning that error in the process.
I solved it by adding quotation marks (grrr)
if [ "${netmask[1]}" != "" ]; do
If you want to check for the null value for a variable, use the -z operator:
if [ -z "${netmask[1]}" ]; then
On example:
VAR=""
if [ -z "$VAR" ]; then
echo This will get printed
fi
Please note the parentheses around the variable: "$VAR".
I couldn't find any one simple straightforward resource spelling out the meaning of and fix for the following BASH shell error, so I'm posting what I found after researching it.
The error:
-bash: [: too many arguments
Google-friendly version: bash open square bracket colon too many arguments.
Context: an if condition in single square brackets with a simple comparison operator like equals, greater than etc, for example:
VARIABLE=$(/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.
if a script has
if [ $1 == "-?" ]; then #line 4
echo "usage: ...."
fi
when the script get runs without any parameter, it will complain that
./script.sh: line 4: [: ==: unary operator expected
but if instead
if [ "$1" == "-?" ]; then #line 4
echo "usage: ...."
fi
then everything's fine
why is that?
thanks
If the first argument is missing or empty, your first script evaluates to:
if [ == "-?" ] ; then
... which is a syntax error. As you noticed, to prevent that you need to make use of "", then it evaluates to:
if [ "" == "-?" ] ; then
AFAIK this is due to the way the original Bourne shell was working. You should make it a habit of enclosing variables in "" to also work correctly with arguments that have spaces in it. For example, if you would call your script like this:
./myScript "first argument has spaces"
Then your first script would evaluate to:
if [ first argument has spaces == "-?" ] ; then
which is also a syntax error. Also things like rm $1 will not do what you want if you pass filenames with spaces. Do rm "$1" instead.
Because [ replaces the values before executing. [[ doesn't, so will work as expected.
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.