I am trying to use the simple fuinction below. But i get error sayin unary oprator expected and the output is always one. Can any1 help me correct it.
#!/bin/bash
checkit ()
{
if [ $1 = "none" ]
then
echo "none"
else
echo "one"
fi
}
checkit
$1 is an argument to the entire script and not to the function checkit(). So send the same argument to the function too.
#!/bin/bash
checkit ()
{
if [ $1 = "none" ]
then
echo "none"
else
echo "one"
fi
}
checkit $1
This has to work.
[ "$1" = "none" ] ... checkit "$1"
[ $var == "" ] ... if the variable $var is empty the test becomes, [ == "" ] and ends up throwing an error. It's always better to quote your variables as @glenn said.
= and ==: POSIX standard defines only = operator (see opengroup.org/onlinepubs/009695399/utilities/test.html), == is an extension supported by some shells and some implementations of test. You should always try to make your scripts portable and hence avoid ==.Use the [[ keyword instead of the [ builtin:
#!/bin/bash
checkit ()
{
if [[ $1 = none ]]
then
echo none
else
echo one
fi
}
checkit
Please read the Bash pitfalls and the Bash FAQ.
one because $1 is empty, not none. Please read up some basics about programming in bash.[ instead of [[ because [[ is non-standard and is not supported by some shells. Just remember to quote your variables properly.Surround $1 with quotes like this:
#!/bin/bash
checkit ()
{
if [ "$1" = "none" ]
then
echo "none"
else
echo "one"
fi
}
checkit
man [ there is no such thing as using == in the expression. = compares strings.