javascript - 过滤数组键

Question

我有一个对象主体，我想从中获取键的值。如何以最少的计算量获取 membershipId、serviceId、userId？

我只想要数据，只是我不想使用过滤器 3 次

var body = {
  "custom_fields": [{
    "label": "membership_id",
    "data": "1",
  }, {
    "label": "service_id",
    "data": "1",

  }, {
    "label": "user_id",
    "data": "26134",
  }]
};

var membershipInfo = body.custom_fields.filter(n=>{
  return n.label == "membership_id";
});

var serviceInfo = body.custom_fields.filter(n=>{
  return n.label == "service_id";
})
var userInfo = body.custom_fields.filter(n=>{
  return n.label == "user_id";
})

var membershipId = membershipInfo[0].data;
var serviceId    = serviceInfo[0].data;
var userId       = userInfo[0].data;

console.log(membershipId,serviceId,userId);

Answer 1

...in minimum computations

嗯，重复循环数组并不是最小的。您将使用一个循环，可能是一个相当无聊的循环:

var membershipId, serviceId, userId;
var fields = body.custom_fields;
for (var i = 0, len = fields.length; i < len; ++i) {
    var entry = fields[i];
    switch (entry && entry.label) {
        case "membership_id":
            membershipId = entry.data;
            break;
        case "service_id":
            serviceId = entry.data;
            break;
        case "user_id":
            userId = entry.data;
            break;
    }
}

实例:

var body = {
    "custom_fields": [{
        "label": "membership_id",
        "data": "1",
    }, {
        "label": "service_id",
        "data": "1",

    }, {
        "label": "user_id",
        "data": "26134",
    }]
};

var membershipId, serviceId, userId;
var fields = body.custom_fields;
for (var i = 0; i < fields.length; ++i) {
    var entry = fields[i];
    switch (entry && entry.label) {
        case "membership_id":
            membershipId = entry.data;
            break;
        case "service_id":
            serviceId = entry.data;
            break;
        case "user_id":
            userId = entry.data;
            break;
    }
}

console.log(membershipId, serviceId, userId);

并不令人兴奋，但你说“以最少的计算量”。 :-)

这不是最简洁的方式，但它可能是运行时最高效的方式。您可以进行的调整:

• 如果你知道永远不会有 null或undefined条目，您可以删除 entry &&参与switch

• 如果有很多条目与任何情况都不匹配，如果您已获得所需的全部内容，您可能会提前停止