Check JS function when its stored in a variable with arguments

Question

I currently store a function in a variable, and then call it:

var function_name = 'test_function';


// some other JS here    


if (typeof app[function_name] != 'undefined') {
     app[function_name];
}


app.test_function = function() {
    alert('woo');
};

This works fine, but what if I store arguments in the function name variable? Like this:

var function_name = 'test_function("mike")';


// some other JS here    


if (typeof app[function_name] != 'undefined') {
     app[function_name];
}


app.test_function = function(name) {
    alert(name);
};

This doesn't work, because typeof app[function_name] is undefined. How can I do this?

Why not pack the arguments separately from the function definition? You could then use the same logic to lookup the function and delegate calling the function with the supplied arguments, much like lodash.curry — AL the X
– AL the X, Commented Jun 16, 2016 at 12:23

gurvinder372 · Accepted Answer · 2016-06-16 12:09:00Z

1

'test_function("mike")' is not a valid function name.

Your logic needs to change to

var function_name = 'test_function("mike")';
var app = {};
app.test_function = function(name) 
{
  console.log(name)
};
var split = function_name.split(/[()]/);
var funName = split[0];
var arguments = [];
if (funName != function_name) 
{
  var arguments = split[1].split(",");
}
if (typeof app[funName] != 'undefined') 
{
  app[funName].apply(null, arguments);
}

answered Jun 16, 2016 at 12:09

gurvinder372

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Comments

Rayon · Accepted Answer · 2016-06-16 11:51:26Z

1

function name should/will not change but the definition will!

var function_name = 'test_function';
var app = {};
if (typeof app[function_name] != 'undefined') {
  app[function_name];
}
app.test_function = function(name) {
  alert(name);
};
app.test_function('Hey');

answered Jun 16, 2016 at 11:51

Rayon

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