this is a bash script. im getting the error in the if statement line
# 2. read a line of input from the keyboard
read answer
if [-z "$answer"]
then
$answer=$default
else
$default=$answer
fi
i don't do much bash anymore, i can't see the error, ive tried
if [-z "$answer"]; then
and that failed as well with the same error. can anyone else see the error?
EDIT UPDATE
i changed it to this
read answer
if [ -z "$answer" ]
then
$answer=$defaultEntry
else
$defaultEntry=$answer
fi
and the same error occures
simply leave spaces between the brackets and condition:
if [ -z "$answer" ]
There should be a space between [ and -z, and between " and ].
[ is an alias to the test program.
[ is a program, which bash can execute.
[-z isn't, thus an error.
You need spaces around [ and ]. E.g.:
if [ -z "$answer" ]
then
$answer=$default
else
$default=$answer
fi
The reason for this is that [ is actually a shell builtin command.
[me#home]$ type [
[ is a shell builtin
[me#home]$ which [
/usr/bin/[
If you omit the space after [, you're mangling the command name into something that does not exist.
[me#home]$ [ -z "something" ] # OK. calling command [ with some args
[me#home]$ [-z "something" ] # fail. calling command [-z with some args
-bash: [-z: command not found
The [ command also checks that the last argument is ], therefore if you don't have spaces around that it becomes part of the previous argument and the [ command will complain:
[me#home]$ [ -z "something" ] # this runs fine becuase last arg is ]
[me#home]$ [ -z "something"] # will fail. last arg is "something"]
-bash: [: missing `]'
The space around the [ and ] is important, as others have pointed out.
Also, you should not be using $ when assigning to a variable.
Try this:
read answer
if [ -z "$answer" ]
then
answer=$defaultEntry
else
defaultEntry=$answer
fi
read answer
if [ -z "$answer" ] #Spaces!
then
answer=$default # Use "answer" and not "$answer"
else
default=$answer # Use "default" and not "$default"
fi
This now works except that I don't have a value for $default. I suspect you've done that elsewhere (as well as set a prompt).
Related
I can understand if [ -f /etc/bashrc ] which just checks the existence of the file "/etc/bashrc". but what is "if [ ! -f ]". This line I came across at grub.cfg.
This may be a simple syntax but I couldn't get the answer by simply googling, as its not an usual syntax.
Thanks
The documentation for [ can be found at https://pubs.opengroup.org/onlinepubs/9699919799/utilities/test.html; the relevant section being:
If $1 is '!', exit true if $2 is null, false if $2 is not null.
Since the first argument is ! and the second argument is not the null string, the command returns false (non-zero).
Usually, this is an error, probably caused by a failure to properly quote a variable. At some point, the intent was probably to do something like if [ ! -f $filename ], and somewhere in the build chain the variable was the null string and that test expanded to if [ ! -f ].
I am missing something fundamental concerning either the bash's if construct/operators or string comparison.
Consider the following script:
#!/bin/bash
baseSystem="testdir1"
testme="NA"
if [ "$baseSystem"=="$testme" ]; then
echo "In error case"
fi
if [ "$baseSystem"!="$testme" ]; then
echo "In error case"
fi
I get:
In error case
In error case
So it enters each case even though they should be mututally exclusive.
Any help is appreciated.
bash happens to be somewhat particular about spaces.
Add spaces around the operators:
if [ "$baseSystem" == "$testme" ]; then
...
if [ "$baseSystem" != "$testme" ]; then
The following are not equivalent:
[ "$a"="$b" ]
[ "$a" = "$b" ]
Your first test is essentially the same as saying if [ "testdir1==NA" ]; then which would always be true.
I am trying to run the below logic
if [-f $file] && [$variable -lt 1] ; then
some logic
else
print "This is wrong"
fi
It fails with the following error
MyScipt.ksh[10]: [-f: not found
Where 10th line is the if condition , I have put in .
I have also tried
if [-f $file && $variable -lt 1] ; then
which gives the same error.
I know this is a syntax mistake somehwere , but I am not sure , what is the correct syntax when I am using multiple conditions with && in a if block
[ is not an operator, it's the name of a program (or a builtin, sometimes). Use type [ to check. Regardless, you need to put a space after it so that the command line parser knows what to do:
if [ -f $file ]
The && operator might not do what you want in this case, either. You should probably read the bash(1) documentation. In this specific case, it seems like what you want is:
if [ -f $file -a $variable -lt 1 ]
Or in more modern bash syntax:
if [[ -f $file && $variable -lt 1 ]]
The [ syntax is secretly a program!
$ type [
[ is a shell builtin
$ ls -l $(which [)
-rwxr-xr-x 1 root root 35264 Nov 19 16:25 /usr/bin/[
Because of the way the shell parses (technically "lexes") your command line, it sees this:
if - keyword
[-f - the program [-f
$file] - A string argument to the [-f program, made by the value of $file and ]. If $file was "asdf", then this would be asdf]
And so forth, down your command. What you need to do is include spaces, which the shell uses to separate the different parts (tokens) of your command:
if [ -f "$file" ]; then
Now [ stands on its own, and can be recognized as a command/program. Also, ] stands on its own as an argument to [, otherwise [ will complain. A couple more notes about this:
You don't need to put a space before or after ;, because that is a special separator that the shell recognizes.
You should always "double quote" $variables because they get expanded before the shell does the lexing. This means that if an unquoted variable contains a space, the shell will see the value as separate tokens, instead of one string.
Using && in an if-test like that isn't the usual way to do it. [ (also known as test) understands -a to mean "and," so this does what you intended:
if [ -f "$file" -a "$variable" -lt 1 ]; then
Use -a in an if block to represent AND.
Note the space preceding the -f option.
if [ -f $file -a $variable -lt 1] ; then
some logic
else
print "This is wrong"
fi
I couldn't find any one simple straightforward resource spelling out the meaning of and fix for the following BASH shell error, so I'm posting what I found after researching it.
The error:
-bash: [: too many arguments
Google-friendly version: bash open square bracket colon too many arguments.
Context: an if condition in single square brackets with a simple comparison operator like equals, greater than etc, for example:
VARIABLE=$(/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.