I want to pass the value of \'undefined\' on a multiple parameter function but without omitting the parameter.
What do I mean with \"without omitting the paramet
Just to give an examples on what @dbrin was explaining, and @alphapilgrim might have wished to check.
simpleFunction({params1:12, params2:"abc"}); //use expected arguments
simpleFunction({params2:"abc"}); //use only one
simpleFunction({params2:"abc", params1:12, iDoNotExist:"I will not be included"}); //use expected arguments in any order
simpleFunction(); //ignore arguments and uses defaults
simpleFunction(2123); //ignores non object arguments
function simpleFunction(someParams){
var myParams = {
params1:null,
params2:"cde" //default value for second parameter
};
simpleExtend(myParams, someParams);
console.log(myParams);
}
//to include in your utilities
function simpleExtend(currentParams, newParams){
if(typeof currentParams !== "undefined" && typeof newParams !== "undefined"){
Object.keys(newParams).forEach(function(key){
if(typeof currentParams[key] !== "undefined"){
currentParams[key] = newParams[key];
}
});
}
}