javascript - Cara menggunakan break untuk mengawal gelung supaya nilai console.log(sum) adalah berbeza

Question

var a = [1, 2, 3, 4, 5, 6, 7, 8];
sum = [];
array = parseInt(10 * Math.random());

for (var j = 0; j < 3; j++) {
    for (var i = 0; i < a.length; i++) {

       
        sum[i] = a[i];


        if (i == array) {
            break
        };

    }

    console.log(sum);
}

 

Answer 1

pecah boleh ditanggalkan...

let arr = [1, 2, 3, 4, 5, 6, 7, 8]
let result = []

for (let i = 0; i < arr.length - 1; i++) {
  // 每次循环重置 result
  result = []
  for (let j = 0; j < arr.length; j++) {
    result.push(arr[j])
    if (j >= i) break
  }
  console.log(result)
}

Answer 2

Juli[b] dalam gelung tidak boleh ditambah pada tatasusunan dengan betul setiap kali

Jika syarat dipenuhi, saya akan menggunakan rehat untuk mengganggu

Jika subskrip yang diperolehi oleh acc mencapai 5, jumlahkan 1234 yang pertama dan seterusnya

                for (var a = 0; a < cresult.length; a++) {

                    for (var i = 1, j = 0; i < arr.length; i++, j++) {

                        result[j] = [countDisPoToLine(cresult[a]["2"], cresult[a]["1"], arr[j]["2"], arr[j]["1"], arr[i]["2"], arr[i]["1"]), arr[i]["1"], arr[i]["2"], arr[j]["1"], arr[j]["2"], arr[j]["0"], lineDis(arr[i]["1"], arr[i]["2"], arr[j]["1"], arr[j]["2"])];

                        arrjoin[j] = result[j][0]; //循环数组中第一组关于距离的值

                        juli[j] = result[j][6];
                        
                    }
                    //console.log(juli); 
                    mindis = Math.min.apply(Math, arrjoin); //
                    //console.log(arrmin.length, "点距线最短距离");
                    console.log(mindis, "点距线最短距离");
//result = [];
                    for (var b = 0; b < result.length; b++) {
                        ///sum = lineDis(result[b][1],result[b][2],result[b][3],result[b][4]); //计算两点之间的长度

                        acc = isHasElementTwo(arrjoin, mindis); //查找item 在array中的下标位置
                        
                        
                       sum.push(juli[b]);        
                    //console.log(juli[b]);     
   //console.log(sumb(acc,juli));
                        if (b == acc) //满足最近点下标的条件，连续相加产生连续里程。
                        {

                            minxy = getPointOnPolyline(cresult[a]["2"], cresult[a]["1"], result[b][4], result[b][2], result[b][3], result[b][1]);

                            ctt = lineDis(result[b][3], result[b][4], minxy[1], minxy[0]);

                            console.log(minxy, "垂足坐标");

                            console.log(ctt, "距离最近点的距离");

                            break
                        };

                    }
        
                    //abc.push(sum );
                    //console.log(eval(abc.join("+")));
                    console.log(sum);

                }
                
  

**1.0374510479565615 "Jarak terpendek antara titik dan garis" (indeks):84
[736260.8899769672, 1411419.6883407037] "Koordinat menegak" (indeks):107.38 "(indeks):107.38 titik" (indeks): 104
[13.999781212443958, 54.4198045292843 70.38912223496114 880968758, 646.61679543939, 300622202552, 417.420483917605, 64.576888332512566, 216.4881564657956, 316.1135609002259, 81. 465181 56, 29.40872826215661, 22.45736237849337, 19.645029625828716, 28.46256518660203, 41.71105215644795, 18.333020182101933, 16.940 88291682246, 32.23066924532746, 14.250265576472243, 204.50641699722792, 20.740160172962312, 24.035660943608878, 13.999781212 443958, 54.4198045292843, 70.38912223496114, 10.536416088967684, 685.858750934864, 255.56736787796788, 62 .58699880968758, 646.6167989543939, 313.57803581242166, 1067.5293031990727, 958.2300622502552, 417.420483917605, 142.39781 082935613 A 6924532746, 14.250265576472243, 204.50641699722792, 20.740160172962312, 13.999781212443958, 646.6167989543939, 313.57803581242166, .1135609002259, 81.2835112492013, 56.39033259343886, 561.7188548465188, 142.39781082935613, 893.5690 953081938, 92.02929710155902, 16.993059288952495 85963614054475]**