搜索对象数组的JavaScript函数返回undefined而不是object

Question

I'm working on a function that searches an array filled with objects (each one has strings, numbers, etc...).

My function

function findPlayer(array, playerToFind){
$.each(array, function(index, el) {
    console.log(el);
    if(el.user_id === playerToFind){
        return el;
    }
}); }

I'm calling it like this (when a button is clicked) I've already checked that $(this).text contains the string value I want. That string is somewhere in the arrayWithData and I want the full object.

var player = findPlayer(arrayWithData,$(this).text());
    console.log(player);

But "undefined" is returned. Why? Thanks in advance.

Answer 1

Because findPlayer is calling .each() and the return statement is in the callback you pass to .each() , you're only returning from the callback, not the original function.

Remember, you're passing the callback to jQuery, and jQuery is invoking it for you on every iteration, which means jQuery internally is receiving the value from your return statement.

Since you're abstracting the search operation behind the findPlayer anyway, I'd ditch .each() and just use a regular loop.

function findPlayer(array, playerToFind){
    for (var i=0; i < array.length; i++)
        if(array[i].user_id === playerToFind)
            return array[i];
}

Will be faster this way too.

---

Note that in ECMAScript 6, you'll be able to use the .find() method.

function findPlayer(array, playerToFind){
    return array.find(function(el) {
        return el.user_id === playerToFind;
    });
}

Both of these solutions have the advantage of both being native and short circuiting the loop.

Answer 2

findPlayer returns undefined because you don't have it returning any value.

You could modify the function to be closer to this:

function findPlayer(array, playerToFind) {
  var foundPlayer;

  $.each(array, function(index, el) {
      if (el.user_id === playerToFind) {
        foundPlayer = el;
        return false;
      }
  }); 

  return foundPlayer;
}

Personally I probably wouldn't do this, instead I'd either just simply iterate with a for loop, or use something like lodash/underscore's find method.