typescript sort

1// Numbers
2// When sorting numbers, you can use the compact comparison:
3
4var numericArray: number[] = [2, 3, 4, 1, 5, 8, 11];
5
6var sortedArray: number[] = numericArray.sort((n1,n2) => n1 - n2);
7// i.e. - rather than <.
8
9// Other Types
10// If you are comparing anything else, you'll need to convert the comparison into a number.
11
12var stringArray: string[] = ['AB', 'Z', 'A', 'AC'];
13
14var sortedArray: string[] = stringArray.sort((n1,n2) => {
15    if (n1 > n2) {
16        return 1;
17    }
18
19    if (n1 < n2) {
20        return -1;
21    }
22
23    return 0;
24});
25// Objects
26// For objects, you can sort based on a property, bear in mind the above information about being able to short-hand number types. The below example works irrespective of the type.
27
28var objectArray: { age: number; }[] = [{ age: 10}, { age: 1 }, {age: 5}];
29
30var sortedArray: { age: number; }[] = objectArray.sort((n1,n2) => {
31    if (n1.age > n2.age) {
32        return 1;
33    }
34
35    if (n1.age < n2.age) {
36        return -1;
37    }
38
39    return 0;
40});

1let arr:{key:number}[] = [{key : 2}, {key : 3}, {key : 4}, {key : 1}, {key : 5}, {key : 8}, {key : 11}];
2
3let sortFn2 = (obj1 , obj2) => {key:number} { return obj1.key - obj2.key; }
4
5const sortedArray2:{key:number}[] = arr.sort(sortFn2);
6

1var numericArray: number[] = [2, 3, 4, 1, 5, 8, 11];
2
3var sortedArray: number[] = numericArray.sort((n1,n2) => n1 - n2);

1var stringArray: string[] = ['AB', 'Z', 'A', 'AC'];
2
3var sortedArray: string[] = stringArray.sort((n1,n2) => {
4    if (n1 > n2) {
5        return 1;
6    }
7
8    if (n1 < n2) {
9        return -1;
10    }
11
12    return 0;
13});