如何在javascript中对嵌套的对象数组进行排序？
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【中文标题】如何在javascript中对嵌套的对象数组进行排序？【英文标题】：How to sort a nested array of objects in javascript? 【发布时间】：2019-09-30 00:21:26 【问题描述】：我有一个对象数组：
entities: [
name: "zBroomsticks PTY",
id: 34098365,
entityType: "personal",
facilities: [
type: "Home loan",
account: "032654 987465",
existing: true,
modified: "04/12/2018",
limit: 100000
,
type: "Credt card",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 200000
,
type: "Credt card",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 10000
,
type: "Credt card",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 10000
]
,
name: "Mr John Doe -3409865, Mrs Jane Doe -34098365",
id: 34098365,
entityType: "business",
facilities: [
type: "Overdraft",
account: "032654 987465",
existing: false,
modified: "04/12/2018" ,
limit: 10000
]
,
name: "Mr Jack",
id: 34098365,
entityType: "mixed",
facilities: [
type: "Overdraft",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 10000
]
]
我想按特定顺序排序：
entity.name：按字母升序排列。
entity.entityType: 1. 个人 2. 商务 3. 混合
entity.facilities.limit：降序
这是我目前得到的代码：
sortData(entities)
var order =
entityType: personal: 2, business: 1
;
return entities.map(ent =>
entities.sort(
(a, b) =>
a.name - b.name ||
order.entityType[a.facilities.entityType] - order.entityType[b.facilities.entityType]
)
);
我知道如何执行名称排序，但找不到 2 和 3 的方法？
Link to code
【问题讨论】：
limit 只是为了对facilities 数组进行排序，还是你希望它也会影响到外部数组entities？
只是设施数组。
【参考方案1】：
首先，要通过names 订购，您可以使用localeCompare()。其次，facilities 数组中没有属性 entityType，但您正在尝试访问它。现在，一种解决方案是首先使用Array.map() 来获取一个新数组，其中facilities 数组按limit 属性排序，然后您可以将map() 返回的新数组首先按names 排序，然后再排序通过entityType 属性，如下所示：
const input = [name:"zBroomsticks PTY",id:34098365,entityType:"personal",facilities:[type:"Home loan",account:"032654 987465",existing:true,modified:"04/12/2018",limit:100000,type:"Credt card",account:"032654 987465",existing:false,modified:"04/12/2018",limit:200000,type:"Credt card",account:"032654 987465",existing:false,modified:"04/12/2018",limit:10000,type:"Credt card",account:"032654 987465",existing:false,modified:"04/12/2018",limit:10000],name:"Mr John Doe -3409865, Mrs Jane Doe -34098365",id:34098365,entityType:"business",facilities:[type:"Overdraft",account:"032654 987465",existing:false,modified:"04/12/2018",limit:10000],name:"Mr Jack",id:34098365,entityType:"mixed",facilities:[type:"Overdraft",account:"032654 987465",existing:false,modified:"04/12/2018",limit:10000],name:"Mr Jack",id:34098365,entityType:"personal",facilities:[type:"Overdraft",account:"032654 987465",existing:false,modified:"04/12/2018",limit:10000]];
let order =
entityType: personal:1, business:2, mixed:3
;
function sortData(entities)
let limitOrder = entities.map(e =>
e.facilities.sort((a, b) => b.limit - a.limit);
return e;
);
return limitOrder.sort((a, b) =>
return a.name.localeCompare(b.name) ||
order.entityType[a.entityType] - order.entityType[b.entityType];
);
console.log(sortData(input));.as-console background-color:black !important; color:lime;
.as-console-wrapper max-height:100% !important; top:0;注意，我使用不同的entityType 复制了与name: "Mr Jack" 相关的对象，因此您可以看到当有两个等于names 的对象时算法的执行情况。
【讨论】：
【参考方案2】：按照1、2 和最后3 的优先顺序对数组进行排序
2 条件，您可以使用index 对象来映射entityType 及其权重。
3 因为要按descending order 排序，所以在facilities 列表中找到最小的limit 值并与其他项进行比较。
let entities = [
name: "zBroomsticks PTY",
id: 34098365,
entityType: "personal",
facilities: [
type: "Home loan",
account: "032654 987465",
existing: true,
modified: "04/12/2018",
limit: 100000
,
type: "Credt card",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 200000
,
type: "Credt card",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 10000
,
type: "Credt card",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 10000
]
,
name: "Mr John Doe -3409865, Mrs Jane Doe -34098365",
id: 34098365,
entityType: "business",
facilities: [
type: "Overdraft",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 10000
]
,
name: "Mr Jack",
id: 34098365,
entityType: "mixed",
facilities: [
type: "Overdraft",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 10000
]
];
const entityIndex = personal: 1, business: 2, mixed: 3 ;
let result = entities.sort((a, b) =>
if (a.name > b.name) return -1;
if (a.name < b.name) return 1;
let et = entityIndex[a.entityType] - entityIndex[b.entityType];
if (et != 0) return et;
const aMinLimit = Math.min(...a.facilities.map(i => i.limit));
const bMinLimit = Math.min(...b.facilities.map(i => i.limit));
return bMinLimit - aMinLimit;
)
console.log(JSON.stringify(result));
【讨论】：
【参考方案3】：试试
eType = personal:1, business:2, mixed:3;
entities.sort((a,b) =>
if(a.name>b.name) return 1;
if(a.name<b.name) return -1;
let et=eType[a.entityType]-eType[b.entityType];
return et;
)
entities.forEach(e=> e.facilities.sort((a,b)=> b.limit - a.limit ))
let entities = [
name: "zBroomsticks PTY",
id: 34098365,
entityType: "personal",
facilities: [
type: "Home loan",
account: "032654 987465",
existing: true,
modified: "04/12/2018",
limit: 100000
,
type: "Credt card",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 200000
,
type: "Credt card",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 10000
,
type: "Credt card",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 10000
]
,
name: "Mr John Doe -3409865, Mrs Jane Doe -34098365",
id: 34098365,
entityType: "business",
facilities: [
type: "Overdraft",
account: "032654 987465",
existing: false,
modified: "04/12/2018" ,
limit: 10000
]
,
name: "Mr Jack",
id: 34098365,
entityType: "mixed",
facilities: [
type: "Overdraft",
account: "032654 987465",
existing: false,
modified: "04/12/2018",
limit: 10000
]
];
eType = personal:1, business:2, mixed:3;
entities.sort((a,b) =>
if(a.name>b.name) return 1;
if(a.name<b.name) return -1;
let et=eType[a.entityType]-eType[b.entityType];
return et;
)
entities.forEach(e=> e.facilities.sort((a,b)=> b.limit - a.limit ))
console.log(entities);【讨论】：
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