这是可行的，我在AppJet中测试了它（你可以复制粘贴代码到它的IDE上，然后按“reload”查看它的工作情况，不需要创建帐户）

/* appjet:version 0.1 */
function Joes_remove(someArray) {
    var newArray = [];
    var element;
    for( element in someArray){
        if(someArray[element]!=undefined ) {
            newArray.push(someArray[element]);
        }
    }
    return newArray;
}

var myArray2 = [1,2,,3,,3,,,0,,,4,,4,,5,,6,,,,];

print("Original array:", myArray2);
print("Clenased array:", Joes_remove(myArray2) );
/*
Returns: [1,2,3,3,0,4,4,5,6]
*/

几个简单的方法：

var arr = [1,2,,3,,-3,null,,0,,undefined,4,,4,,5,,6,,,,];

arr.filter(n => n)
// [1, 2, 3, -3, 4, 4, 5, 6]

arr.filter(Number) 
// [1, 2, 3, -3, 4, 4, 5, 6]

arr.filter(Boolean) 
// [1, 2, 3, -3, 4, 4, 5, 6]

或-（仅适用于“text”类型的单个数组项）

['','1','2',3,,'4',,undefined,,,'5'].join('').split(''); 
// output:  ["1","2","3","4","5"]

或-经典方式：简单迭代

var arr = [1,2,null, undefined,3,,3,,,0,,,[],,{},,5,,6,,,,],
    len = arr.length, i;

for(i = 0; i < len; i++ )
    arr[i] && arr.push(arr[i]);  // copy non-empty values to the end of the array

arr.splice(0 , len);  // cut the array and leave only the non-empty values
// [1,2,3,3,[],Object{},5,6]

jQuery：

var arr = [1,2,,3,,3,,,0,,,4,,4,,5,,6,,,,];
    
arr = $.grep(arr, n => n == 0 || n);
// [1, 2, 3, 3, 0, 4, 4, 5, 6]

干净的方法。

var arr = [0,1,2,"Thomas","false",false,true,null,3,4,undefined,5,"end"];
arr = arr.filter(Boolean);
// [1, 2, "Thomas", "false", true, 3, 4, 5, "end"]

使用正则表达式筛选出无效条目

array = array.filter(/\w/);
filter + regexp

就地解决方案：

function pack(arr) { // remove undefined values
  let p = -1
  for (let i = 0, len = arr.length; i < len; i++) {
    if (arr[i] !== undefined) { if (p >= 0) { arr[p] = arr[i]; p++ } }
    else if (p < 0) p = i
  }
  if (p >= 0) arr.length = p
  return arr
}

let a = [1, 2, 3, undefined, undefined, 4, 5, undefined, null]
console.log(JSON.stringify(a))
pack(a)
console.log(JSON.stringify(a))

简单ES6

['a','b','',,,'w','b'].filter(v => v);