Question

这是我正在使用的JSON对象：

var object = {
    "units": [
        { "unitNum": "1", "fuels": [{ "fuelDesc": "Coal" }] },
        { "unitNum": "2", "fuels": [{ "fuelDesc": "Pipeline Natural Gas" }] }
    ]
}

将所有fuelDesc个字符串连接成一个字符串，以,分隔的最具表现力的方法是什么？

注意：我在这里简化了对象。实际的对象要大得多，所以我关心的是性能。此外，燃料阵列可以有多个项目。

结果将是“煤炭，管道天然气”。

我这样做了：

var fuelsStr = "";
for (i = 0; i < units.length; i += 1) {
    var mergedFuels = [].concat.apply([], units[i].fuels)
    var unitFuelStr = mergedFuels.map(function (mergedFuels) {
        return mergedFuels["fuelDesc"];
    }).join(",");
    fuelsStr += unitFuelStr;
}

Answer 1

对于JSON字符串，可以使用JSON.parse reviver：

a = [], j = '{"units":[{"unitNum":"1","fuels":[{"fuelDesc":"Coal"}]},{"unitNum":"2","fuels":[{"fuelDesc":"Pipeline Natural Gas"}]}]}'

JSON.parse(j, (k, v) => k === "fuelDesc" ? a.push(v) : 0)

console.log(a + '')

对于JavaScript对象，JSON.stringify replacer：

a = [], o = {"units":[{"unitNum":"1","fuels":[{"fuelDesc":"Coal"}]},{"unitNum":"2","fuels":[{"fuelDesc":"Pipeline Natural Gas"}]}]}

JSON.stringify(o, (k, v) => k === "fuelDesc" ? a.push(v) : v)

console.log(a + '')

Answer 2

基于假设您在燃料数组中始终有一个元素。您可以一起使用 .map 和 .join

＆＃13;

var myobject = {"units":[{"unitNum": "1","fuels":[{"fuelDesc":"Coal"}]},
{"unitNum": "2","fuels":[{"fuelDesc":"Pipeline Natural Gas"}]}]};

var result = myobject.units.map(function(elem){
    return elem.fuels[0].fuelDesc;
}).join(",");

console.log(result);

＆＃13;

Answer 3

我会结合使用javascript的reduce和map函数

var object = {
  "units": [{
    "unitNum": "1",
    "fuels": [{
      "fuelDesc": "Coal"
    }]
  }, {
    "unitNum": "2",
    "fuels": [{
      "fuelDesc": "Pipeline Natural Gas"
    }, {
      "fuelDesc": "some fuel"
    }]
  }]
};

var myStr = object.units.reduce(
  (first, second) =>
  first.fuels.map(item =>
    item.fuelDesc)
  .concat(second.fuels.map(item => item.fuelDesc))).join()
  
  console.log(myStr);

我相信这和你能得到的效率差不多，但是因为我已经完成了涉及大O的数学，所以如果有人可以证明我错了，请告诉我！

如何在嵌套在另一个对象数组中的对象数组中连接字符串？

3 个答案: