How to useType Assertions in Typescript
Using type assertions to sort an array with null properties involves asserting the property as a defined type and providing a default value if it is null. This ensures TypeScript treats the property as defined for comparison.
Example
interface Item {
name?: string | null;
value?: number | null;
}
const items: Item[] = [
{ name: 'item1', value: 10 },
{ name: 'item2', value: null },
{ name: null, value: 5 },
{ name: 'item3', value: 15 },
];
items.sort((a, b) => {
return (a.value ?? 0) - (b.value ?? 0);
});
console.log(items);
Output:
[
{ name: 'item2', value: null },
{ name: null, value: 5 },
{ name: 'item1', value: 10 },
{ name: 'item3', value: 15 }
]
How to Sort an Array of Objects with Null Properties in TypeScript ?
Sorting an array of objects in TypeScript involves arranging objects based on specified properties. When dealing with null properties, ensure consistency by placing them either at the start or end of the sorted array to maintain predictability in sorting outcomes.
Below are the approaches used to sort an array of objects with null properties in TypeScript:
Table of Content
- Using Array.prototype.sort with a ternary operator
- Using Array.prototype.reduce
- Using Array.prototype.filter() and concat()
- Using Array.prototype.sort with a custom comparison function
- Using Type Assertions
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