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.
Related
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 am trying to run while loop in shell
NODESTATE="0"
LOOPC="1"
while [ "$NODESTATE" -ne "UP" ]; do
echo "node is up "
but it is throwing me an error with [: UP: integer expression expected
or shoud i use != instead of -ne
The problem is in your condition:
while [ "$NODESTATE" -ne "UP" ]; do
The -ne option is used to determine whether one comparator is numerically equal to another. But you're doing a string comparison, not a numeric one. Instead, try the following:
while [ "$NODESTATE" != "UP" ]; do
You can read man test to see how the various options to [ work.
$i="500,600"
$j="600"
if[$i -ne $j]; then
#some line
else
#some line
fi
This if condition is not going inside.
this if condition fails. else is pass.
how is this possible
can someone help me on this
The trick is to consider the "[" sign as a command (in fact it is one indeed), whose last argument must be a "]". So you must ensure that there is a space after [ and all of its arguments go to the proper place. In your case:
if [ "$i" -ne "$j" ]
then
# some code
else
# some code
fi
Since [ is a command, you might want to omit the if structure and use logical operators, taking advantage of lazy evaluation. The following means the same:
[ "$i" -ne "$j" ] && {
echo "hello" ;
echo "world" ;
} || {
echo "bye bye" ;
echo "world" ;
}
I couldn't find any one simple straightforward resource spelling out the meaning of and fix for the following BASH shell error, so I'm posting what I found after researching it.
The error:
-bash: [: too many arguments
Google-friendly version: bash open square bracket colon too many arguments.
Context: an if condition in single square brackets with a simple comparison operator like equals, greater than etc, for example:
VARIABLE=$(/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 [ $k -ge 16 ]
then exit 0;
else
echo "good job for nothing";
fi;
that way I resolved my problem. Hope that will help for you too.
this is a bash script. im getting the error in the if statement line
# 2. read a line of input from the keyboard
read answer
if [-z "$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).