重复的玩家名称不应该在同一个数组中

Question

我正在制作一个匹配系统，将匹配 2 个相同级别的玩家。

我的目标是如何在同一个文件中删除重复的玩家名称 大批？删除该重复条目后，应将其存储在 另一个数组（等待列表）。

我在下面提供了一个示例和屏幕截图。任何帮助将不胜感激。谢谢。

匹配脚本：（匹配同级玩家）

const combine = (source) => {
  return source.reduce((acc, curr) => {
    if (acc[curr.level]) {
      const levelArr = acc[curr.level];
      const last = levelArr[levelArr.length - 1];
      if (last.length === 2) {
        levelArr.push([curr])
      } else {
        last.push(curr)
      }
    } else {
      acc[curr.level] = [
        [curr]
      ];
    }
    return acc;
  }, {})
};

脚本 ajax：（从数据库中获取数据）

let ajaxResult = []; // the pushed data will be saved here
let save_method;
let table;
let base_url = "<?php echo base_url();?>";

let result = [];
var html = "";

$(document).ready(function() {
    //datatables
    table = $("#entry_list1").DataTable({


        processing: false,
        serverSide: true,
        order: [],
        searching: false,
        paging: false,
        info: false,

        ajax: {
            url: "<?php echo site_url('controller/fetch')?>",
            type: "POST",
            async: true,
            dataType: "json",

             success: function(data) {
             result = combine(data.data2);
console.log(result)
var keys = Object.keys(result)
                
for (var i = 0; i < keys.length; i++) {
  result[keys[i]].forEach(function(val) {

    val.forEach(function(value, index) {


      var entryIDs = index == 0 ? "entryIDM[]" : "entryIDW[]"
      var players = index == 0 ? "playerM[]" : "playerW[]"
      var levels = index == 0 ? "levelM[]" : "levelW[]"
      html += `<input type="text" name="${entryIDs}" value="${value.entryID}"> 
                 <input type="text" name="${players}" value="${value.player}">
                 <input type="text" name="${levels}" value="${value.level}">
                 `
    })
  })
}
document.getElementById("result").innerHTML = html //add html to div

                
               
              
                

            },
        },

        "columnDefs": [{
                "targets": [0], //first column
                "orderable": false, //set not orderable
            },
            {
                "targets": [-1], //last column
                "orderable": false, //set not orderable
            },

        ],
    });
});

Answer 1

我建议在将重复的播放器传递给combine 函数之前删除它们。这样您就不会看到匹配的重复玩家。

我会创建一个removeDuplicates 函数来从玩家列表中删除重复的玩家。

在这个例子中，我们有 18 个玩家，每个玩家重复一次。

我们通过 removeDuplicates 函数运行这些，现在我们有 9 个不同的玩家。

然后我们将唯一的玩家传递给combine 函数，得到 3 组，每组 2 个匹配的玩家和 3 个不匹配的玩家。

const players = Array.from({ length: 18 }, (v,k) => ( { level: Math.floor(k / 6) + 1, player: `test-${Math.floor(k/2)+1}` }));

function removeDuplicates(players) {
   return Object.values(players.reduce((acc, curr) => { 
       acc[curr.player] = acc[curr.player] || curr;
       return acc;
   }, {}))
}

const combine = (source) => {
  return source.reduce((acc, curr) => {
    if (acc[curr.level]) {
      const levelArr = acc[curr.level];
      const last = levelArr[levelArr.length - 1];
      if (last.length === 2) {
        levelArr.push([curr])
      } else {
        last.push(curr)
      }
    } else {
      acc[curr.level] = [
        [curr]
      ];
    }
    return acc;
  }, {})
};

console.log("All players:", players);

const uniquePlayers = removeDuplicates(players);

console.log("Unique players:", uniquePlayers);

const matched = Object.values(combine(uniquePlayers)).flatMap(a => a.filter(x => x.length === 2));
const unmatched = Object.values(combine(uniquePlayers)).flatMap(a => a.filter(x => x.length === 1));
console.log("Matched players:", matched);
console.log("Unmatched players:", unmatched);