Question

我在数组中有这个结构：

[
    {
        "playerName": "Stack",
        "leaguePoints": 100,
        "wins": 280,
        "miniSeries": {
            "progress": "WNN",
            "losses": 0,
            "wins": 1
        }
    },
    {
        "playerName": "Overflow",
        "leaguePoints": 90,
        "wins": 280
    }
    {
        "playerName": "StackOverflow",
        "leaguePoints": 0,
        "wins": 100
    }
]

我需要像这样对数组元素进行排序：

如果leaguePoints = 100则排序（miniSeries.wins + miniSeries.losses（更多游戏，排名更高））或miniSeries.progress中留下的N（中性）数量（减去Ns，排名更高）< / p>

如果leaguePoints＆lt; 100然后按leaguePoints排序

如果leaguePoints = 0则按胜利排序

我一直在使用Ege Özcan's multiple parameters sort，它按照排名和胜利排序，但我无法使用miniSeries。

它应该如何结束：

Name    leaguePoints    Wins    MiniSeries
Stack   100             10      LWN
Stack   100             25      WNN
Stack   100             5       NNN
Stack   99              50      ---
Stack   70              250     ---
Stack   0               300     ---
Stack   0               200     ---
Stack   0               100     ---

Answer 1

我认为你最好做2遍处理首先将数据分为3组，leaguePoints100，leaguePoints0，其余为。

您知道如何单独对每个组进行排序，然后使用Array.concat()合并三个排序的数组，请参阅示例here。

Lo-Dash可以轻松完成任务，groupBy() leaguePoints：

groups = _.groupBy(data, function(item) {
  switch (item.leaguePoints) {
    case 100:
        return 'leaguePoints100';
    case 0:
        return 'leaguePoints0';
    default:
        return 'other';
  }
);

然后sortBy()三组：

groups['leaguePoints100'] = _.sortBy(group['leaguePoints100'], sortLeaguePoints100);
groups['leaguePoints0'] = _.sortBy(group['leaguePoints0'], 'wins');
groups['other'] = _.sortBy(group['other'], 'leaguePoints');

这使代码更清晰，但机械确实 KISS ，伙计。

Answer 2

我想你想要

function byProgress(a, b) {
    // less "N"s in the value
    return b.split("N").length - a.split("N").length;
}
function byMiniSeries(a, b) {
    // more games played
    return a.wins+a.losses - b.wins-b.losses || byProgress(a.progress, b.progress);
}
function byRank(a, b) {
    // highest league
    return a.leaguePoints - b.leaguePoints || (a.leaguePoints == 0
      ? a.wins - b.wins
      : (a.leaguePoints == 100 
        ? byMiniSeries(a.miniSeries, b.miniSeries)
        : 0 ) );
}
playerArray.sort(byRank);

||使用短路评估，如果左边的操作数是假的，则返回右操作数（此处：0，即值相等）。

基于3个字段对json对象的数组进行排序

2 个答案: