This can be done through a recursion. For the convenience of demonstration, I changed it to generate a set of non-repeating positive integer random numbers within 10.
The main method it relies on is the indexOf() method, which is used to find the index of a certain value in the array. If it is not found in the array, -1 is returned.
The code is as follows:

var arr=[];
function getRandom(){
    var random=Math.floor(Math.random()*1000000);
    console.log(random);
    //判断生成的数在数组中是否存在，判断是否是6位数
    //如果不存在而且是6位数，放入数组
    if(random.toString().length==6&&arr.indexOf(random)==-1){     
        arr.push(random)
    }else{
    //如果存在或者不是6位数，接着调用这个函数，生成满足要求的随机数
        console.log("不符合要求的"+random)
        getRandom();
    }
}
for(var i=0;i<6;i++){
    getRandom();
}
console.log(arr);

Time will not repeat, and using timestamps to generate random numbers will not repeat either.

console.log((Math.random() * Date.now()).toFixed(0));
// 6位的
console.log((Math.random() * Date.now() / 1000000).toFixed(0));

And to a certain extent, there is no absolute non-repeating random number in this world, and all combinations are not infinite. Even if all the generated random numbers are stored, and then compared when generated, if they are found to already exist, they will be generated. New random numbers will eventually form an infinite loop when all combinations have been tried. What's more, it's only 6 digits, and there are only 472,392 combinations in total.

Let me give you a simple and easy-to-use one. The above cannot guarantee that it will never be repeated. If it is repeated, remember to buy a lottery ticket. Math.random().toString(36).slice(2,8)

Non-repetitive random number sequence generation algorithm
I saw an article that is very clever. The results and efficiency can be guaranteed. Use code + comments to implement it:

function getRandom(numCount) {
        var numList = [];
        var numMin = 100000;
        var numMax = 999999;
        var listLen = numMax - numMin + 1;

        var outPut = [];

        // 将所有数顺序装填到数字列表中
        for (var i = numMin; i < numMax + 1; i++) {
            numList.push(i);
        }

        var randNum;
        for (var j = 0; j < numCount; j++) {
            // 产生一个小于列表长度的随机数randNum
            randNum = Math.floor(Math.random() * listLen);
            // 将列表的第randNum项输出
            outPut.push(numList[randNum]);
            // 将列表的最后一项替换到刚被取出的数的位置
            numList[randNum] = numList[listLen - 1];
            // 列表长度减一,即列表最后一项不会再被取到;
            listLen--;
        }

        return outPut;
    }
    var arr = getRandom(30);
    console.log(arr);

JS is not well written (escape

Write a loop to generate a random number each time and throw it into the set. When the set is long enough, it will be converted into an array and returned

If the required number in the array is small, you can use the array method to determine it

console.time('time:');
function createRandomArr(l){
    var r = [];
    var o = {};
    var a;
    for (var i = 0;i < l;i++){
        a = Math.random().toString().slice(2,8);
        o[a] ? i-- : (r.push(a),o[a] = true);
    }
    return r;
}
var res = createRandomArr(10000);
console.log(res,res.length);
console.timeEnd("time:");