Consider the following two javascript objects:
Object A:
{
specialPropertyA: string,
partTime: boolean,
role: { key: string, label: string }
}
Object B:
{
specialPropertyB: string,
partTime: boolean,
role: { key: number, label: string }
}
Now, consider an arrayA: ObjectA[]
and an arrayB: ObjectB[]
.
I am looking for a concise way to determine:
If any of the ObjectA's have partTime === true
AND role.key === 1
, are there any ObjectBs, which fulfill the same requirements? I want to check the same for role.key === 2
.
I know I could do sth like:
if(
arrayA.filter(objectA => objectA.partTime === true && objectA.role.key === 1)
&& arrayB.filter(objectB => objectB.partTime === true && objectB.role.key === 1)
|| arrayA.filter(objectA => objectA.partTime === true && objectA.role.key === 2)
&& arrayB.filter(objectB => objectB.partTime === true && objectB.role.key === 2)
)
But I don't like this solution at all, especially since I am repeating the same code for key 2. Any suggestions as on how to make this more concise?
CodePudding user response:
One way would be to make a higher-order function that you can pass the key number you want to find.
Also, because partTime
is a boolean, obj.partTime === true
simplifies to obj.partTime
.
const makeCb = keyNum => obj => obj.partTime && obj.role.key === keyNum;
if (
(arrayA.some(makeCb(1)) && arrayB.some(makeCb(1))) ||
(arrayA.some(makeCb(2)) && arrayB.some(makeCb(2)))
) {
You'll want .some
(or .find
, if you want to put the result into a variable). .filter
won't work in an if
because even if there are no results, empty arrays are truthy.
CodePudding user response:
If your arrays have many possible key values, which could match, then maybe first collect the found key values in the first array before scanning the second:
const keysA = new Set(arrayA.map(({partTime, key}) => partTime && key));
if (arrayB.some(({partTime, key}) => partTime && keysA.has(key))) {
/* ... */
}
This assumes that false
is not a possible value of key
.