You can do this using String#split and Array#sort, defining a custom compareFunction function in this scheme:

MDN Web Docs:

• If compareFunction(a, b) returns a value > than 0, sort b before a.
• If compareFunction(a, b) returns a value < than 0, sort a before b.
• If compareFunction(a, b) returns 0, a and b are considered equal.

Like this:

let arr = ["UserA | 3", "UserB | 0", "UserC | 2", "UserD | 1", "UserE | 2020"];

arr.sort((a, b) => a.split(" | ")[1] - b.split(" | ")[1]);

console.log(arr);

You can create a function to extract the last number id using a RegExp combined with String.prototype.match():

Regular expression: /\d+(?=\D*$)/

• \d matches a digit (equivalent to [0-9])
• + matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)
• Positive Lookahead (?=\D*$) Assert that the Regex below matches \D matches any character that's not a digit (equivalent to [^0-9])
• * matches the previous token between zero and unlimited times, as many times as possible, giving back as needed (greedy)
• $ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)

And finally use Array.prototype.sort()

Code:

const arr = ["UserE | 2020", "UserA | 3", "UserB | 0", "UserC | 2", "UserD | 1"]
const getId = s => +s.match(/\d+(?=\D*$)/)[0] 

arr.sort((a, b) => getId(a) - getId(b))

console.log(arr)