之前写过同样的原因（适用于任意数量的数组）：

/**
 * Returns with the union of the given arrays.
 *
 * @param Any amount of arrays to be united.
 * @returns {array} The union array.
 */
function uniteArrays()
{
    var union = [];
    for (var argumentIndex = 0; argumentIndex < arguments.length; argumentIndex++)
    {
        eachArgument = arguments[argumentIndex];
        if (typeof eachArgument !== 'array')
        {
            eachArray = eachArgument;
            for (var index = 0; index < eachArray.length; index++)
            {
                eachValue = eachArray[index];
                if (arrayHasValue(union, eachValue) == false)
                union.push(eachValue);
            }
        }
    }

    return union;
}    

function arrayHasValue(array, value)
{ return array.indexOf(value) != -1; }

最佳解决方案。。。

您可以直接在浏览器控制台中点击。。。

无重复项

a = [1, 2, 3];
b = [3, 2, 1, "prince"];

a.concat(b.filter(function(el) {
    return a.indexOf(el) === -1;
}));

具有重复项

["prince", "asish", 5].concat(["ravi", 4])

如果你想要没有重复，你可以从这里尝试一个更好的解决方案-大喊代码。

[1, 2, 3].concat([3, 2, 1, "prince"].filter(function(el) {
    return [1, 2, 3].indexOf(el) === -1;
}));

在Chrome浏览器控制台上试用

 f12 > console

输出：

["prince", "asish", 5, "ravi", 4]

[1, 2, 3, "prince"]

在Dojo 1.6中+

var unique = []; 
var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];
var array3 = array1.concat(array2); // Merged both arrays

dojo.forEach(array3, function(item) {
    if (dojo.indexOf(unique, item) > -1) return;
    unique.push(item); 
});

使现代化

参见工作代码。

http://jsfiddle.net/UAxJa/1/

//Array.indexOf was introduced in javascript 1.6 (ECMA-262) 
//We need to implement it explicitly for other browsers, 
if (!Array.prototype.indexOf)
{
  Array.prototype.indexOf = function(elt, from)
  {
    var len = this.length >>> 0;

    for (; from < len; from++)
    {
      if (from in this &&
          this[from] === elt)
        return from;
    }
    return -1;
  };
}
//now, on to the problem

var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];

var merged = array1.concat(array2);
var t;
for(i = 0; i < merged.length; i++)
  if((t = merged.indexOf(i + 1, merged[i])) != -1)
  {
    merged.splice(t, 1);
    i--;//in case of multiple occurrences
  }

其他浏览器的indexOf方法的实现取自MDC

合并无限数量的数组或非数组并保持其唯一性：

function flatMerge() {
    return Array.prototype.reduce.call(arguments, function (result, current) {
        if (!(current instanceof Array)) {
            if (result.indexOf(current) === -1) {
                result.push(current);
            }
        } else {
            current.forEach(function (value) {
                console.log(value);
                if (result.indexOf(value) === -1) {
                    result.push(value);
                }
            });
        }
        return result;
    }, []);
}

flatMerge([1,2,3], 4, 4, [3, 2, 1, 5], [7, 6, 8, 9], 5, [4], 2, [3, 2, 5]);
// [1, 2, 3, 4, 5, 7, 6, 8, 9]

flatMerge([1,2,3], [3, 2, 1, 5], [7, 6, 8, 9]);
// [1, 2, 3, 5, 7, 6, 8, 9]

flatMerge(1, 3, 5, 7);
// [1, 3, 5, 7]

使用reduce func查看的另一种方法：

function mergeDistinct(arResult, candidate){
  if (-1 == arResult.indexOf(candidate)) {
    arResult.push(candidate);
  }
  return arResult;
}

var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];

var arMerge = [];
arMerge = array1.reduce(mergeDistinct, arMerge);
arMerge = array2.reduce(mergeDistinct, arMerge);//["Vijendra","Singh","Shakya"];