我有一个这样定义的对象：{ _id: "5d406a171ed43384972f04b5", index: 0, age: 28, eyeColor: "brown", name: { first: "Myra", last: "Navarro" }, company: "SUSTENZA", email: "myra.navarro@sustenza.net" }我需要能够对所有键进行搜索，即使是像name.例如，我有一个搜索字符串name.first等，它可能是更深层次的复合对象。我设法做到了这一点，但例如name.first我无法让它发挥作用。你能帮我个忙吗？
3 回答
梦里花落0921
TA贡献1772条经验 获得超5个赞
第一个问题是 javascript 不会自动使用点访问子项。
接下来，一旦你有了想要比较的值。
如果值是字符串，则需要使用特殊的字符串比较函数（使用 typeof 进行检查）。或者使用 valA - valB 以便获得 1、0 和 -1 作为用于排序的返回值。
Qyouu
TA贡献1786条经验 获得超11个赞
我希望这就是您所需要的。该child功能可以满足您的需求。你在评论里有解释。
let datatable = [
{
_id: "5d406a171ed43384972f04b5",
index: 0,
age: 28,
eyeColor: "brown",
name: {
first: "Myra",
last: "Navarro"
},
company: "SUSTENZA",
email: "myra.navarro@sustenza.net"
},
{
_id: "5d406a170db0f4b04d9a9acf",
index: 1,
age: 23,
eyeColor: "blue",
name: {
first: "Harriett",
last: "Tanner"
},
company: "VALPREAL",
email: "harriett.tanner@valpreal.com"
},
{
_id: "5d406a17e95da8ff80a759c5",
index: 2,
age: 39,
eyeColor: "blue",
name: {
first: "Vega",
last: "Hanson"
},
company: "BEDLAM",
email: "vega.hanson@bedlam.tv"
},
{
_id: "5d406a175505da190e6875ec",
index: 3,
age: 31,
eyeColor: "blue",
name: {
first: "Rosemary",
last: "Fields"
},
company: "QUAILCOM",
email: "rosemary.fields@quailcom.me"
},
{
_id: "5d406a17ea96044c027f4e50",
index: 4,
age: 27,
eyeColor: "brown",
name: {
first: "Dale",
last: "Wilkinson"
},
company: "QIAO",
email: "dale.wilkinson@qiao.org"
},
{
_id: "5d406a17c5fff1ff6653a555",
index: 5,
age: 25,
eyeColor: "blue",
name: {
first: "Beatrice",
last: "Contreras"
},
company: "ZENOLUX",
email: "beatrice.contreras@zenolux.us"
},
{
_id: "5d406a17a199efcba25e1f26",
index: 6,
age: 34,
eyeColor: "blue",
name: {
first: "Hancock",
last: "Wynn"
},
company: "PLASMOS",
email: "hancock.wynn@plasmos.co.uk"
},
{
_id: "5d406a17019a2a4544a4f134",
index: 7,
age: 40,
eyeColor: "blue",
name: {
first: "Brown",
last: "Stanton"
},
company: "SNACKTION",
email: "brown.stanton@snacktion.name"
},
{
_id: "5d406a17e516dd71af8210d4",
index: 8,
age: 39,
eyeColor: "blue",
name: {
first: "Barnes",
last: "Dunn"
},
company: "PORTALINE",
email: "barnes.dunn@portaline.ca"
},
{
_id: "5d406a17516936a025b73c33",
index: 9,
age: 34,
eyeColor: "green",
name: {
first: "Blanche",
last: "Cherry"
},
company: "ISOSWITCH",
email: "blanche.cherry@isoswitch.io"
},
{
_id: "5d406a17527a4d2c6a7897dd",
index: 10,
age: 33,
eyeColor: "blue",
name: {
first: "Gilliam",
last: "Farley"
},
company: "AMTAS",
email: "gilliam.farley@amtas.biz"
},
{
_id: "5d406a175ff11478c416c30b",
index: 11,
age: 26,
eyeColor: "brown",
name: {
first: "Laura",
last: "Short"
},
company: "FISHLAND",
email: "laura.short@fishland.info"
},
{
_id: "5d406a1738181b471847339a",
index: 12,
age: 20,
eyeColor: "brown",
name: {
first: "Moreno",
last: "Barber"
},
company: "KEENGEN",
email: "moreno.barber@keengen.net"
},
{
_id: "5d406a17a6bcae6fe3ad1735",
index: 13,
age: 30,
eyeColor: "brown",
name: {
first: "Fischer",
last: "French"
},
company: "INCUBUS",
email: "fischer.french@incubus.com"
},
{
_id: "5d406a17600ca53e8f63f263",
index: 14,
age: 30,
eyeColor: "brown",
name: {
first: "Donaldson",
last: "Carr"
},
company: "SUNCLIPSE",
email: "donaldson.carr@sunclipse.tv"
},
{
_id: "5d406a17530655789a27174f",
index: 15,
age: 35,
eyeColor: "green",
name: {
first: "Sophia",
last: "Payne"
},
company: "PRISMATIC",
email: "sophia.payne@prismatic.me"
},
{
_id: "5d406a175dbc687b4c7669d8",
index: 16,
age: 34,
eyeColor: "green",
name: {
first: "Simone",
last: "Pollard"
},
company: "DIGIGEN",
email: "simone.pollard@digigen.org"
},
{
_id: "5d406a179f35ed326a6a5567",
index: 17,
age: 28,
eyeColor: "green",
name: {
first: "Yvette",
last: "Daugherty"
},
company: "CHILLIUM",
email: "yvette.daugherty@chillium.us"
}
];
const sortAsc = true;
const sortField = "name.first";//index,eyeColor
var result = datatable.sort((a, b) => {
let [x, z] = sortAsc ? [a, b] : [b, a];
//console.log(x[sortField], z[sortField], x[sortField] > z[sortField] ? 1 : -1);
//return x[sortField] > z[sortField] ? 1 : -1;
return child(x, sortField) > child(z, sortField) ? 1: -1;
});
console.log(result);
function child(obj, str){
//take dot-separated chain of properties' names and split them into an array
const props = str.split('.');
let child = obj;
//looop through all properties' names and get the access to the following [Object] children sequentially
props.forEach((prop)=> child = child[prop]);
return child;
}
30秒到达战场
TA贡献1828条经验 获得超6个赞
选项1：
const objRoute = sortField.split(".");
const getValue = obj => objRoute.reduce((v, c) => v[c], obj);
const result = datatable.sort((a, b) =>
sortAsc
? getValue(a).localeCompare() - getValue(b).localeCompare()
: getValue(b).localeCompare() - getValue(a).localeCompare()
);
选项2：
使用 eval，eval() 计算表达式。建议谨慎使用该功能：
const result = datatable.sort((a, b, i) => eval(`a.${sortField}`) - eval(`b.${sortField}`));
添加回答
举报
0/150
提交
取消