CODEWITHSUNDEEP
bashshell
Per Lundberg
Per Lundberg

Reputation: 4220

How can I assign the output of a function *with parameters* to a variable using bash?

Similar to How can I assign the output of a function to a variable using bash?, but slightly different.

If I have a function like this:

function scan {
  echo "output"
}

...I can easily assign this to a variable like this:

VAR=$(scan)

Now, what if my function takes one or more parameters, how can I pass them to the function using the "shell expansion" syntax? E.g. this:

function greet {
  echo "Hello, $1"
}

# Does not work
VAR=$(greet("John Doe"))

The above produces an error like this with my bash (version 5.0.3(1)-release):

$ ./foo.sh 
./foo.sh: command substitution: line 8: syntax error near unexpected token `"John Doe"'
./foo.sh: command substitution: line 8: `greet("John Doe"))'

Upvotes: 0

Views: 1702

Answers (2)

Zig Razor
Zig Razor

Reputation: 3535

in this code :

function greet {
  echo "Hello, $1"
}

# Does not work
VAR=$(greet("John Doe"))

the error is the parenthesis passing parameters. try to do this:

function greet {
  echo "Hello, $1"
}

# works
VAR=$(greet "John Doe")

it should work.

Explanation: when you use the $ an the parenthesis you have to write inside the parenthesis command as in a shell so the parameters are passed without parenthesis.

Upvotes: 2

Per Lundberg
Per Lundberg

Reputation: 4220

I have obviously been writing to much Java code lately. Drop the parentheses when calling the function and everything works flawlessly:

function greet {
  echo "Hello, $1"
}

VAR=$(greet "John Doe")

Upvotes: 0

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