There are already a lot of cool features in ES6/ES7 for defining Javascript objects. However, the following pattern is common in Javascript:
const obj = {
requiredKey1: ...,
requiredKey2: ...
};
if (someCondition) {
obj.optionalKey1 = ...;
}
Is there a way to define the object all at once with both optional and required keys?
You can use object spread to have an optional property:
let flag1 = true;
let flag2 = false;
// extra cases added by Abdull
let optionalKey8 = 8;
let optionalKey9 = undefined;
let optionalKey10 = false;
let optionalKey11 = null;
let optionalKey12 = "twelve";
const obj = {
requiredKey1: 1,
requiredKey2: 2,
...(flag1 && { optionalKey3: 3 }),
...(flag2 && { optionalKey4: 4, optionalKey5: 5 }), // ignored
...(flag1 && { optionalKey6: 6, optionalKey7: 7 }),
...(optionalKey8 && { optionalKey8 }),
...(optionalKey9 && { optionalKey9 }), // ignored
...(optionalKey10 && { optionalKey10 }), // ignored
...(optionalKey11 && { optionalKey11 }), // ignored
...(optionalKey12 && { optionalKey12 })
};
console.log(obj);
...flag1 && { optionalKey1: 5 }, is fine too.
To indicate optional key, you can assign to it null, if the condition is false
const someCondition = true;
const obj = {
requiredKey1: 1,
requiredKey2: 2,
optionalKey1: someCondition ? 'optional' : null
};
console.log(obj);
optionalKey1 still appears as one of the keys of the object if the condition is false (and has the value null), whereas OPs original snippet will create an object that lacks the key entirely if the condition is false.
null to the property is more understandable that it is optional and does not have value than checking for the existence
null value doesn't make it optional. It means that the object has an actual member whose assigned value is null which actually is a value. You may thing that null is not a value, but it is. Therefore this doesn't make a member optional. An optional member would be one, that wouldn't be returned by the Object.entries(obj) function. If at least you'd assign it a value of undefined which would mean that the member is undefined and doesn't have a value assigned to it. Object.entries(obj) would still see the member though.
null would be removing it in the next line: Object.keys(obj).forEach((k) => obj[k] == null && delete obj[k]); @SurenSrapyan If you add this line, that will make this answer also one of the options.You can declare the variable by passing it through the below function
function removeOptionalKeys(x){
let obj = {};
Object.keys(x).forEach((a) => {
if(x[a]){
obj[a] = x[a]
}
})
return(obj)
}
Then declare the variable as follows
let obj = removeOptionalKeys({
requiredKey1: 1,
requiredKey2: 2,
optionalKey1: someCondition ? 'optional' : null
})
the following pattern is common in Javascript
It should not. Having many objects of different shapes can incur a performance penalty. Records should always contain the same keys. So just use
const obj = {
requiredKey1: …,
requiredKey2: …,
optionalKey1: someCondition ? … : undefined,
};
optionKey1: someCondition ? value : undefined?{ key?: optionalValue }or with property shorthand:{ optionalValue? }