Question

我有这个JSON

{
"people": [{
        "games": [
            "destiny",
            "horizon",
            "fifa",
            "cuphead"
        ],
        "name": "Cartman"
    },
    {
        "games": [
            "fifa",
            "pes"
        ],
        "name": "Kyle"
    },
    {
        "games": [
            "cuphead",
            "destiny"
        ],
        "name": "Stan"
    },

    {
        "games": [
            "pokemon",
            "metroid"
        ],
        "name": "Clyde"
    },
    {
        "games": [
            "fifa",
            "metroid",
            "pes"
        ],
        "name": "Butters"
    }

]

}

我需要让至少有两个游戏共享的用户，所以对于这个JSON，结果将是：

卡特曼和斯坦扮演命运和杯头

Kyle和Butters扮演FIFA和Pes

注意：Clyde将不在结果中，因为只有一场比赛与其他球员匹配

我想过两个解决方法

1-有一个字典使用两个标签（涵盖所有可能性）作为一个密钥并填写它如果它不存在和其他密钥匹配后，我会知道已经有两个匹配

2-让第一个用户将其与其他用户进行比较，如果匹配至少两个，则将其打印，然后继续下一个并与其他人进行比较等等

感谢任何帮助

Answer 1

方法通过检查是否只有一个存在来减少游戏/人员，然后删除另一个对象的响应部分。

它返回的项目多于一项，但超过两项。

要获得结果，您可以删除包含两个部分的结果并调整其余部分。

＆＃13;

var object = { people: [{ games: ["destiny", "horizon", "fifa", "cuphead"], name: "Cartman" }, { games: ["fifa", "pes"], name: "Kyle" }, { games: ["cuphead", "destiny"], name: "Stan" }, { games: ["pokemon", "metroid"], name: "Clyde" }, { games: ["fifa", "metroid", "pes"], name: "Butters" }] },
    peoples = {},
    games = {},
    change = false;

object.people.forEach(function (person) {
    person.games.forEach(function (game) {
        games[game] = games[game] || {};
        games[game][person.name] = true;

        peoples[person.name] = peoples[person.name] || {};
        peoples[person.name][game] = true;
    });
});

do {
    change = false;
    Object.keys(games).forEach(function (k) {
        var keys = Object.keys(games[k]);
        if (keys.length === 1) {
            delete peoples[keys[0]][k];
            delete games[k];
            change = true;
        }
    });
    Object.keys(peoples).forEach(function (k) {
        var keys = Object.keys(peoples[k]);
        if (keys.length === 1) {
            delete games[keys[0]][k];
            delete peoples[k];
            change = true;
            return;
        }
    });
} while (change);

console.log(peoples);
console.log(games);

＆＃13;

.as-console-wrapper { max-height: 100% !important; top: 0; }

＆＃13;

从json获得至少两场比赛

1 个答案: