I have a function which has two default parameters and one regular parameter.
var Fields = function ($content, result = failedObject, job = true)
{
...
}
When I call Fields($content,job), job here could be true or false, the result parameter takes in the job value rather than the third parameter. Works good in the case if Fields($content,result).
Any good way to tackle this situation.
you would call Fields($content, undefined, job) to use the default value for the parameter result
if I understood your question correctly, javascript does not support named parameters natively, so you can call by Fields($content, undefined, job)
or using an object as parameter for your function instead of 3 params.
You could use a destructured argument :
var Fields = function ($content, { result = failedObject, job = true } = { }) {
...
}
Then, when you call the function :
const data = { result: anotherObject, job = false };
Fields('toto', data);
Or, in your specific case :
const data = { job = false };
Fields('toto', data); // result still is a failedObject
undefined as param valuePass empty/undefined/null:
Fields($content, , job);
Fields($content, , job) is syntax error. Passing null will set param value to null. Only undefined is proper way. developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…
You must switch the order(like below), or using an array and treat inside function.
var Fields = function ($content, job = true, result = failedObject, )
{
}
