多条件排序函数

Posted 2023-02-14

【中文标题】多条件排序函数

【英文标题】：Sort function with multiple condition

【发布时间】：2022-01-18 11:03:24

【问题描述】：

我有一个对象数组，如下所示：

var data = [
  
    name": "Name2",
    "webOrderingEnabled": true,
    "distance": 1.6989125091571928
  ,
  
    "name": "Name3",
    "webOrderingEnabled": false,
    "distance": 1.9178283920396098
  ,
  
    "name": "Name4",
    "webOrderingEnabled": false,
    "shutdown": 
      "message": "",
      "status": true
    ,
    "distance": 6.94478210395609
  ,
  
    "name": "Name1",
    "webOrderingEnabled": true,
    "shutdown": 
      "message": "",
      "status": false
    ,
    "distance": 0.5368834377514055
  
]

我想对这个对象数组进行排序 1.webOrderingEnabled 2.shutdown.status = false 3. 距离 挑战是某些对象没有关闭键，如果它不存在则考虑存储打开 我尝试了以下方法，但对我不起作用

  data.sort((a, b) => 
   if (a.shutdown?.status && b.shutdown?.starus || !a.shutdown?.status && !b.shutdown?.status && 
       a.webOrderingEnabled || b.webOrderingEnabled) 
              return a.distance - b.distance;
            
            if (a.shutdown?.status) 
              return -1;
            
            return 1;
          );

预期：

 [
  "name": "Name1",
  "webOrderingEnabled": true,
  "shutdown": 
  "message": "",
  "status": false
  ,
  "distance": 0.5368834377514055
  ,
  
  "name": "Name2",
  "webOrderingEnabled": true
  "distance": 1.6989125091571928
  ,
  
  "name": "Name3",
  "webOrderingEnabled": false,
  "distance": 1.9178283920396098
  ,
  
  "name": "Name4",
  "webOrderingEnabled": false,
  "shutdown": 
  "message": "",
  "status": true
  ,
  "distance": 6.94478210395609
  ]

【参考方案1】：

一般来说，如果您想根据字段 A、B、C（按此顺序）对对象数组进行排序，您只需这样做：

if (a.A < b.A) return -1;
else if (a.A > b.A) return 1;
// otherwise a.A == b.A, so let's proceed with considering B now...

if (a.B < b.B) return -1;
else if (a.B > b.B) return 1;
// otherwise a.A == b.A && a.B == b.B, so let's proceed with considering C now...
if (a.C < b.C) return -1;
else if (a.C > b.C) return 1;
return 0;

在这里，您可以将a.A < b.A 视为用于比较单个属性值的伪代码，具体执行方式取决于类型。

所以你的代码应该看起来像

// I assume by "consider to be store open" you mean shutdown.status === false
function getShutdownStatus(a) 
    return a?.shutdown?.status || false;

data.sort((a, b) => 
  // Use ! here because true > false but it seems you want true to come first
  if (!a.webOrderingEnabled < !b.webOrderingEnabled) return -1;
  else if (!a.webOrderingEnabled > !b.webOrderingEnabled) return 1;
  const aShutdown = getShutdownStatus(a);
  const bShutdown = getShutdownStatus(b);
  if (aShutdown < bShutdown) return -1;
  else if (aShutdown > bShutdown) return 1;
  // otherwise compare by distance
  if (a.distance < b.distance) return -1;
  else if (a.distance > b.distance) return 1
  return 0;
); 

完成此工作后，您可以简化代码：

data.sort((a, b) => 
  if (a.webOrderingEnabled != b.webOrderingEnabled) return b.webOrderingEnabled - a.webOrderingEnabled;
  const aShutdown = getShutdownStatus(a);
  const bShutdown = getShutdownStatus(b);
  if (aShutdown != bShutdown) return aShutdown - bShutdown;
  return a.distance - b.distance;
);