Question

有一个例子

此javascript：

读取并显示数据

let dataString = "Worl, Seymour|Jones, Jim|Smith, Paul|Jolly, Roger|";
let splitString = dataString.split("|");

for (let i = 0; i < splitString.length;) {
    $("#I" + i).val(splitString[i-1]);  i++;
}

和支持json：

var standing = [
    {"code" : "A", "rank" :  "1"},
    {"code" : "B", "rank" :  "2"},
    {"code" : "C", "rank" :  "3"},
    {"code" : "D", "rank" :  "4"}
];

var members = [
    {"Class": "A", "Name": "Jolly, Roger"},
    {"Class": "B", "Name": "Smith, Paul"},
    {"Class": "C","Name": "Jones, Jim"},
    {"Class": "D","Name": "Worl, Seymour"}
];

我想要：“Worl，Seymour | Jones，Jim | Smith，Paul | Jolly，Roger |”

成为“Jolly，Roger | Smith，Paul | Jones，Jim | Worl，Seymour |”;

Answer 1

根据您对代码运行方式的评论，我认为这将为您完成工作。我假设code数组中的standing与class数组中的members相同。在成员列表中找不到的名称将排在最底层。

let dataString = "Worl, Seymour|Jones, Jim|Smith, Paul|Jolly, Roger|";
let splitString = dataString.split("|");

var standing = [
{"code" : "A", "rank" :  "1"},
{"code" : "B", "rank" :  "2"},
{"code" : "C", "rank" :  "3"},
{"code" : "D", "rank" :  "4"}
];

var members = [
{"Class": "A", "Name": "Jolly, Roger"},
{"Class": "B", "Name": "Smith, Paul"},
{"Class": "C","Name": "Jones, Jim"},
{"Class": "D","Name": "Worl, Seymour"}
];

function getRank(name) {
  var i;
  for (i = 0; i < members.length; i++)
    if (members[i].Name == name) break;
  if (i == members.length) return 9999;
  let code = members[i].Class;
  for (i = 0; i < standing.length; i++)
    if (standing[i].code == code) break;
  if (i == standing.length) return 9999;
  return parseInt(standing[i].rank);
}

function rank(name1, name2) {
  return getRank(name1) - getRank(name2);
}

console.log(splitString.sort(rank).join('|'));
Jolly, Roger|Smith, Paul|Jones, Jim|Worl, Seymour|

Answer 2

以下代码符合您的订购标准。我没有把琴弦放回原处;你的问题中没有足够的细节来解决潜在的角落案件。但我假设你真正需要的只是订购，你可以自己重新组装字符串。

let dataString = "Worl, Seymour|Jones, Jim|Smith, Paul|Jolly, Roger|";

var standing = [
    {"code" : "A", "rank" :  "1"},
    {"code" : "B", "rank" :  "2"},
    {"code" : "C", "rank" :  "3"},
    {"code" : "D", "rank" :  "4"}
];

var members = [
    {"Class": "A", "Name": "Jolly, Roger"},
    {"Class": "B", "Name": "Smith, Paul"},
    {"Class": "C","Name": "Jones, Jim"},
    {"Class": "D","Name": "Worl, Seymour"}
];

function getOrderFromClass(lhsClass, rhsClass) {
    return getOrderFromCode(lhsClass, rhsClass);
}

function getOrderFromCode(lhsCode, rhsCode) {
  var lhsRank, rhsRank;

  for(let s of standing) {
      if(s["code"] === lhsCode) { lhsRank = s["rank"]; }
      if(s["code"] === rhsCode) { rhsRank = s["rank"]; }
  }

  if(lhsRank < rhsRank) { return -1; }
  else if(lhsRank > rhsRank) { return 1; }
  else { return 0; }
}

let splitString = dataString.split("|");
console.log(splitString);

splitString.sort(function(lhsName, rhsName) {
    var lhsClass, rhsClass;

    for(let m of members) {
        if(m["Name"] === lhsName) { lhsClass = m["Class"]; }
        if(m["Name"] === rhsName) { rhsClass = m["Class"]; }
    }

    return getOrderFromClass(lhsClass, rhsClass);
});

// All done
console.log(splitString);

Answer 3

我将假设所有名称都有效，即存在于members数组中，以保持代码简单。

var standing = [
  { "code": "A", "rank": "1" },
  { "code": "B", "rank": "2" },
  { "code": "C", "rank": "3" },
  { "code": "D", "rank": "4" }
];
var members = [

  { "Class": "A", "Name": "Jolly, Roger" },
  { "Class": "B", "Name": "Smith, Paul" },
  { "Class": "C", "Name": "Jones, Jim" },
  { "Class": "D", "Name": "Worl, Seymour" }
];

const dataString = "Worl, Seymour|Jones, Jim|Smith, Paul|Jolly, Roger|";
// Filter out the empty string that split() produces at the end
const names = dataString.split("|").filter(name => name.length > 0);

// Map the names to members
const ms = names.map(name => members.find(m => m.Name === name));
// Map the members to standings
const ss = ms.map(m => standing.find(s => s.code === m.Class));
// Now fill the result with the names in the order of their ranking
const result = [];
names.forEach((name, i) => result[ss[i].rank - 1] = name);

for (let i = 0; i < result.length; i++) {
  $("#I" + (i+1)).val(result[i]);
}

在更新DOM元素的最后一部分中，您似乎遇到了某种索引问题，因此我最好地猜测您要在那里做什么。但是，由于您尚未发布HTML，我不知道这是否完全正确。

用json字段对分隔的字符串进行排序

3 个答案: