Below is my script, I am not getting the right output, I am typing alice I am getting bob as output, but when i am typing bob I should get alice as output, but i am getting bob as output. Kindly let me know what's the error in my script
#!/bin/bash
echo "Enter your name"
read a
if [ $a==alice ];
then
echo "bob"
elif [ $a==bob ];
then
echo "alice"
else
echo "STDERR"
fi
Wrong syntax, should be:
[[ $a == alice ]]
Check: http://tldp.org/LDP/abs/html/comparison-ops.html
You should put spaces around your == operator, so that it's a three-argument form of [.
Without the spaces, it's using the one-argument form which gives true if the string length is non-zero (as it is with the string bob==alice).
Spaces are required surrounding your == expression. When using the [ test operator or using the keyword test, you will want to use the = for posix compliance and portability:
[ "$a" = alice ] # remember to quote your variable when comparing strings
While == is permissive with [ or test, you will lose portability. Alternatively, if you are using [[ operator, then == is correct:
[[ "$a" == alice ]] # quoting is not required, but is good practice.
Related
So I'm trying to write a simple script that takes two stdin variables then tests to see if the string value of those two variables are equal to a set string. Here is my code:
echo "First Name:"
read fN
echo "Last Name"
read lN
if (( $fN + $lN=="louis smith" ))
then
echo "You are root, you may continue."
else
echo "Access denied, you are not root."
first
When I run the script and enter the first name as "louis" and the last name as "smith" I get this error:
((: louis + smith==louis smith : syntax error in expression (error token is "smith ")
Then it tells me that I am not root when I clearly am. xD
Any input helps :D
(( ... )) is for arithmetic expressions, you can't use it to compare strings. And + is for adding numbers, not concatenating strings. Use [[ ... ]] for conditional expressions.
if [[ "$fN $lN" = "louis smith" ]]
In addition to the other answer, you need to make sure you familiarize yourself with each of the ways the conditional can be properly written. There is no magic, just read the appropriate section of man bash. The [[ test is actually a Reserved Word in bash and is a bashism (it works in bash, but not in POSIX shell). It is the most flexible and forgiving test construct for bash and should be used if portability is not a concern. Make sure you understand the difference in quoting requirements, word splitting and pathname expansion, between the [[ test clause and the [ and test builtins.
In addition to [[, you can also make use of [ and test for testing. ([ and test are equivalent). The following are also correct and portable to POSIX shell:
if [ "$fN $lN" = "louis smith" ]
if test "$fN $lN" = "louis smith"
To answer your original question directly, you test for more than one condition by either using the -a (and) or -o (or) within the test expression itself (older syntax) or by separating multiple test expressions with && (and) or || (or). For instance to check for multiple conditions using your example you could do:
if [ "$fN" = "louis" -a "$LN" = "smith" ]
or
if test "$fN" = "louis" -a "$LN" = "smith"
Written using && or ||:
if [ "$fN" = "louis" ] && [ "$LN" = "smith" ]
or
if test "$fN" = "louis" && test "$LN" = "smith"
(note: no matter which you use you must always leave a space between the test expression and the [ and ] and a space or a newline (or line break indicator ;) between if test "$a" = b ; then...)
I am trying to make a bash script with the output based on the input.
My code looks like this:
#!/bin/bash
echo "Letter:"
read a
if a=3
then
echo "LOL"
fi
if a=4
then
echo "ROFL"
fi
But when I enter 3 or 4, I get both LOL and ROFL.
Is there a way for me to get LOL for 3 and ROFL for 4?
Sorry if I'm using incorrect terms and stuff, I'm new to bash scripting.
In bash, a=3 is an assignment, not a test. Use, e.g.:
if [ "$a" = 3 ]
Inside [...], the equal sign tests for string (character) equality. If you want to test for numeric value instead, then use '-eq` as in:
if [ "$a" -eq 3 ]
The quotes around "$a" above are necessary to avoid an "operator" error when a is empty.
bash also offers a conditional expressions that begin with [[ and have a different format. Many like the [[ format better (it avoids, for example, the quote issue mentioned above) but the cost is loss of compatibility with other shells. In particular, note that dash, which is the default shell (/bin/sh) for scripts under Debian-derived distributions, does not have [[.
Bash thinks you're trying to assign a variable by saying a=3. You can do the following to fix this:
Use the = operator whilst referencing the variable with a $, like so: if [[ $a = 3 ]]
Use the -eq operator, which is special and doesn't require you to reference the variable with a $, but may not be compatible with all sh-derived shells: if [[ a -eq 3 ]]. If you wish to use -eq without Bash reference the variable: if [[ $a -eq 3 ]]
Note:
The double square brackets [[ ... ]] are a preferred format with specifically Bash conditionals. [ ... ] is good with any sh-derived shell (zsh, tcsh, etc).
if a=3 will assign value 3 to variable a
unless a is readonly variable, if a=3 always returns TRUE
same for if a=4
To compare variable a with a value, you can do this if [ $a = 3 ]
so the script should change to
#!/bin/bash
echo "Letter:"
read a
if [ $a = 3 ]
then
echo "LOL"
fi
if [ $a = 4 ]
then
echo "ROFL"
fi
Since a is read from user input, there is possibility user key in:
non numeric value
a string with empty space
nothing, user may just press Enter key
so a safer way to check is:
if [ "x$a" = "x3" ]
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.