Javascript: Ignore destructured object parameter if the object is not passed to the function

Question

In my express server I have a function that if a request body is passed to it, it destructures the properties from it as so:

export function createProxiedRequestBody({ dateRange, dateFrom, filters },
dateDefault) {
}

Its a common function where the object passed to it could be void but is obviously causing errors as such:

Cannot read property 'dateRange' of null.

Is there a fancy JavaScript way I can ignore these in the parameters rather than writing an if block statement in the function ?

Read about default values

TeaNyan
– TeaNyan

2019-01-17 13:51:43 +00:00
Commented Jan 17, 2019 at 13:51 — TeaNyan
– TeaNyan, Commented Jan 17, 2019 at 13:51

Beniamin H · Accepted Answer · 2019-01-17 13:56:19Z

1

You can use destructured parameter with default value assignment: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/Default_parameters#Destructured_parameter_with_default_value_assignment

function createProxiedRequestBody(
   { dateRange, dateFrom, filters } = {dateRange: null, dateFrom: null, filters: null},
   dateDefault) { 
       console.log('dateRange:' + dateRange);
}

createProxiedRequestBody();  // no error, dateRange === null
createProxiedRequestBody({dateRange: 2});

edited Jan 17, 2019 at 13:56

answered Jan 17, 2019 at 13:50

Beniamin H

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Comments

Prasad Telkikar · Accepted Answer · 2019-01-17 13:51:13Z

0

You can pass default value to parameter which you want to ignore

Something like below

    function foo (x, y = 10) {
        return x + y;
    }
    
    console.log("Ignore y. Sum = "+ foo(10)); //It will ignore y while calling foo function
    console.log("Without ignoring y. Sum = "+ foo(10, 20)); //It assign value to y i.e 20

answered Jan 17, 2019 at 13:51

Prasad Telkikar

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