JavaScript通过正则表达式拆分字符串

Question

I will have a string never long than 8 characters in length, eg:

// represented as array to demonstrate multiple examples
var strs = [
    '11111111',
    '1RBN4',
    '12B5'
]    

When ran through a function, I would like all digit characters to be summed to return a final string:

var strsAfterFunction = [
    '8',
    '1RBN4',
    '3B5'
]

Where you can see all of the 8 single 1 characters in the first string end up as a single 8 character string, the second string remains unchanged as at no point are there adjacent digit characters and the third string changes as the 1 and 2 characters become a 3 and the rest of the string is unchanged.

I believe the best way to do this, in pseudo-code, would be:

1. split the array by regex to find multiple digit characters that are adjacent
2. if an item in the split array contains digits, add them together
3. join the split array items

What would be the .split regex to split by multiple adajcent digit characters, eg:

var str = '12RB1N1'
  => ['12', 'R', 'B', '1', 'N', '1']

EDIT:

question: What about the string "999" should the result be "27", or "9"

If it was clear, always SUM the digits, 999 => 27 , 234 => 9

Answer 1

You can do this for the whole transformation :

var results = strs.map(function(s){
    return s.replace(/\d+/g, function(n){
       return n.split('').reduce(function(s,i){ return +i+s }, 0)
    })
});

For your strs array, it returns ["8", "1RBN4", "3B5"] .

Answer 2

var results = string.match(/(\d+|\D+)/g);

Testing:

"aoueoe34243euouoe34432euooue34243".match(/(\d+|\D+)/g)

Returns

["aoueoe", "34243", "euouoe", "34432", "euooue", "34243"]

Answer 3

George... My answer was originally similar to dystroy's, but when I got home tonight and found your comment I couldn't pass up a challenge

:)

Here it is without regexp. fwiw it might be faster, it would be an interesting benchmark since the iterations are native.

function p(s){
  var str = "", num = 0;
  s.split("").forEach(function(v){
    if(!isNaN(v)){
        (num = (num||0) + +v);
    } else if(num!==undefined){
        (str += num + v,num = undefined);
    } else {
        str += v;
    }
  });
  return str+(num||"");
};

// TESTING
console.log(p("345abc567"));
// 12abc18
console.log(p("35abc2134mb1234mnbmn-135"));
// 8abc10mb10mnbmn-9
console.log(p("1 d0n't kn0w wh@t 3153 t0 thr0w @t th15 th1n6"));
// 1d0n't0kn0w0wh@t12t0thr0w0@t0th6th1n6

// EXTRY CREDIT
function fn(s){
    var a = p(s);
    return a === s ? a : fn(a);
}

console.log(fn("9599999gh999999999999999h999999999999345"));
// 5gh9h3

and here is the Fiddle & a new Fiddle without overly clever ternary