如何过滤除与Javascript中特定模式匹配的非数字字符之外的非数字字符？

Question

I have an array containing strings that represent numbers. The only two valid types of elements in the array are:

1. All-numeric strings (Eg: "123","123344")
2. Strings that match a specific pattern of xxx-xxxxx where all x are numbers (Eg:"123-12345", "233-30000").

This pattern can be added to more values in the future ie xx-xxxxx and x-xx-xxx where all x are still numbers:

I have the below code which solves 1 but not 2 :

const arr = ["ab's-test#s", "ab-c", "124", "123-12345"];
var arr2 = arr.filter(function(el) {
    return el.length && el==+el && el.match(new RegExp("^\\d{3}(-\\d{5})?$"));
});
console.log(arr2)

This prints out ["124"] However - I want it to print out ["124","123-12345"]

Is there a way I can allow numbers and numeric patterns as a part of the same filter function? Thanks in advance.

Answer 1

You can use conditional OR with regex to test your string is either number or number with pattern xxx-xxxxx .

 const arr = ["ab's-test#s", "ab-c", "124", "123-12345"], arr2 = arr.filter((str) => /^\\d{3}-\\d{5}$/.test(str) || /^\\d+$/.test(str)); console.log(arr2)

Answer 2

You can use a single pattern placing the or | in the regex itself, using a non capture group (?:....) to match either of the alternatives between the anchors to assert the start and the end of the string:

The updated pattern will become:

^(?:\d{3}-\d{5}|\d+)$

 const arr = ["ab's-test#s", "ab-c", "124", "123-12345"]; const arr2 = arr.filter(str => /^(?:\\d{3}-\\d{5}|\\d+)$/.test(str)); console.log(arr2)