1

I want to define a custom bash function, which gets an argument as a part of a dir path.

I'm new to bash scripts. The codes provided online are somehow confusing for me or don't work properly.

For example, the expected bash script looks like:

function my_copy() {
    sudo cp ~/workspace/{$1} ~/tmp/{$2} 
}

If I type my_copy a b,

then I expect the function executes sudo cp ~/workspace/a ~/tmp/b in the terminal.

Thanks in advance.

TrW236
  • 367
  • 1
  • 5
  • 14

2 Answers2

3

If you have the below function in say copy.sh file and if you source it ( source copy.sh or . copy.sh) then the function call my_copy will work as expected.

$1 and $2 are positional parameters.

i.e. when you call my_copy a b, $1 will have the first command line argument as its value which is a in your case and $2 which is second command line argument, will have the value b. The function will work as expected.

Also you have a logical error in the function, you have given {$1} instead of ${1}. It will expand to {a} instead of a in your function and it will throw an error that says cp: cannot stat '~/workspace/{a}': No such file or directory when you run it.

Additionally, if the number of positional parameters are greater than 10, only then it is required to use {} in between otherwise you can avoid it. eg: ${11} instead of $11. 

   function my_copy() {
        sudo cp ~/workspace/$1 ~/tmp/$2 
    }

So above function will execute the statement sudo cp ~/workspace/a ~/tmp/b as expected.

To understand the concept, you can try echo $1, echo ${1}, echo {$1}, echo {$2}, echo ${2} and echo $2 inside the script to see the resulting values. For more special $ sign shell variables

j23
  • 3,139
  • 1
  • 6
  • 13
1

There is a syntax error in your code. You don't call a variable like {$foo}. If 1=a and 2=b then you execute

sudo cp ~/workspace/{$1} ~/tmp/{$2}

BASH is going to replace $1 with a and $2 with b, so, BASH is going to execute

sudo cp ~/workspace/{a} ~/tmp/{b}

That means tha cp is going to fail because there is no file with a name like {a}

There is some ways to call a variable

echo $foo

echo ${foo}

echo "$foo"

echo "${foo}"

Otherwise, your code looks good, should work.

Take a look a this links first and second, it's really important to quoting your variables. If you want more information about BASH or can't sleep at night try with the Official Manual, it have everything you must know about BASH and it's a good somniferous too ;)

PS: I know $1, $2, etc are positional parameters, I called it variables because you treat it as a variable, and my anwser can be applied for both.

karu
  • 11
  • 1