Xcode tableview不显示firebase对象

Question

我目前正在尝试在tableview中显示firebase对象。但是，我只是获得空的原型细胞。谢谢，如果你能看一眼就可以了！

if (!Array.prototype.everyCheckAll) {
  Array.prototype.everyCheckAll = function(callbackfn, thisArg) {
    'use strict';
    var T, k;

    if (this == null) {
      throw new TypeError('this is null or not defined');
    }

    // 1. Let O be the result of calling ToObject passing the this 
    //    value as the argument.
    var O = Object(this);

    // 2. Let lenValue be the result of calling the Get internal method
    //    of O with the argument "length".
    // 3. Let len be ToUint32(lenValue).
    var len = O.length >>> 0;

    // 4. If IsCallable(callbackfn) is false, throw a TypeError exception.
    if (typeof callbackfn !== 'function') {
      throw new TypeError();
    }

    // 5. If thisArg was supplied, let T be thisArg; else let T be undefined.
    if (arguments.length > 1) {
      T = thisArg;
    }

    // 6. Let k be 0.
    k = 0;

    // 7. Repeat, while k < len
    var oneFalse = false;
    while (k < len) {

      var kValue;

      // a. Let Pk be ToString(k).
      //   This is implicit for LHS operands of the in operator
      // b. Let kPresent be the result of calling the HasProperty internal 
      //    method of O with argument Pk.
      //   This step can be combined with c
      // c. If kPresent is true, then
      if (k in O) {

        // i. Let kValue be the result of calling the Get internal method
        //    of O with argument Pk.
        kValue = O[k];

        // ii. Let testResult be the result of calling the Call internal method
        //     of callbackfn with T as the this value and argument list 
        //     containing kValue, k, and O.
        var testResult = callbackfn.call(T, kValue, k, O);

        // iii. If ToBoolean(testResult) is false, return false.
        if (!testResult) {
          oneFalse = true;
        }
      }
      k++;
    }
    return !oneFalse;
  };
}

function gt2(element, index, array) {
  return element > 2;
}
document.write([1,2,3].everyCheckAll(gt2));   // false
document.write([4,5,6].everyCheckAll(gt2));   // true

再次感谢愿意提供帮助的人！我目前正在使用Swift 4和Google的Firebase API。

P.S。我知道我已经连接了delegate和dataSource。我也确定我已正确标题我的标识符。谢谢！