在bash中匹配变量的正确语法是什么？

Question

What's the correct syntax for matching a variable in bash?

#!/bin/bash
# test1.sh

say_hello()
{
    if [ "$1" = true ]; then
        echo "hello"
    else
        echo "goodbye"
    fi
}

say_hello


$ sh test1.sh true
> goodbye

If I remove the function, then it works as I expect it to.

#!/bin/bash

# test2.sh
if [ "$1" = true ]; then
    echo "hello"
else
    echo "goodbye"
fi


$ sh test2.sh true
> hello

I'm not sure what's wrong with the first script's syntax.

Solved and explained by you all, thank you for teaching me.

Answer 1

A function in bash has its own list of positional parameters - the arguments, you passed to the function.

As in other programming languages, you need to pass the parameter from the main script to the function:

say_hello()
{
    if [ "$1" = "true" ]; then
        echo "hello"
    else
        echo "goodbye"
    fi
}

say_hello "$1"

Btw, "true" is just a string in this context, not a boolean value like in other programming languages. I would quote it, simply to express that.

---

Another thing, you are using #!/bin/bash in the shebang line, but you call the script using sh script.sh . While /bin/sh is a link to /bin/bash in some systems, it does not necessarily need to be. Even then, bash will recognize that is has started as /bin/sh and will run in compatibility mode. If you want to call a script with bash , use bash script.sh .

Answer 2

$1 is the first input argument to the script, but inside the scope of a function it is the first argument to that function. So, this would work:

#!/bin/bash
# test1.sh

say_hello()
{
    if [ "$1" = true ]; then
        echo "hello"
    else
        echo "goodbye"
    fi
}

say_hello "$1"


$ sh test1.sh true
> goodbye

Note that here we are passing the $1 argument along to the function.