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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?
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The word unary is basically synonymous with “single.” In the context of mathematics, this could be a single number or other component of an equation. So, when Bash says that it is expecting a unary operator, it is just saying that you are missing a number in the script.
$? is the exit status of the most recently-executed command; by convention, 0 means success and anything else indicates failure. That line is testing whether the grep command succeeded. The grep manpage states: The exit status is 0 if selected lines are found, and 1 if not found.
Inside the function, $1 expands to the first positional parameter that was passed to the shell function, rather than to the script it's part of. Thus, like $# , the special parameters $1 , $2 , etc., as well as $@ and $* , also pertain to the arguments passed to a function, when they are expanded in the function.
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.
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