具有未定义值的排序函数

Question

我有一个对象数组，如下所示：

 var stores = [
      {
        "name": "store3",
        "ditance": 8
      },
      {
        "name": "Store5",
        "distance": 7,
        "web": {
          "validateAttributes": {
            "isSelectedDispostionSupported": true,
            "isFutureOrderPossible": false
          }
        }
      },
      {
        "name": "Store1",
        "distance": 12,
        "web": {
          "validateAttributes": {
            "isSelectedDispostionSupported": true,
            "isOpen": true
          }
        }
      },
      {
        "name": "store2",
        "distance": 13,
        "web": {
          "validateAttributes": {
            "isSelectedDispostionSupported": true,
            "isOpen": true
          }
        }
      }
    ]

根据 isopen 和距离排序的预期结果

[
  {
    "name": "Store1",
    "distance": 12,
    "web": {
      "validateAttributes": {
        "isSelectedDispostionSupported": true,
        "isOpen": true
      }
    }
  },
  {
    "name": "store2",
    "distance": 13,
    "web": {
      "validateAttributes": {
        "isSelectedDispostionSupported": true,
        "isOpen": true
      }
    }
  },
  {
    "name": "Store5",
    "distance": 7,
    "web": {
      "validateAttributes": {
        "isSelectedDispostionSupported": true,
        "isFutureOrderPossible": false
      }
    }
  },
  {
    "name": "store3",
    "ditance": 8
  }
]

需要根据 isOpen 和 distance 对数组进行排序。挑战是一些对象有网络属性，而一些对象没有。即使网络有时可用 isOpen 也不会。我已经尝试了以下方法它不起作用

const sorter = (a, b) => {
                if (a.web) {
                  if (a.web.validateAttributes) {
                    if (a.web.validateAttributes.isOpen) {
                      return 1;
                    } else if (b.web.validateAttributes.isOpen) {
                      return -1;
                    } else {
                      return 1;
                    };
                  } else {
                    return 1;
                  }
                 
                } else {
                  return 1;
                }
              };
  stores.sort(sorter);

Answer 1

在这里修正了一些错别字之后是另一种选择。

如果未设置，则默认每个 isOpen 为 false 并存储。 如果相同，则按distance 排序，否则按isOpen 排序。

这使用了Optional Chaining operator 和Nullish Coalescing operator 的组合

如果链上的任何属性不存在，可选链运算符将返回null。因此，在.web 或.isOpen 不存在的情况下，会生成null。

后面是 Nullish Caolescing 运算符，因此如果如上所述生成 null，则默认值为 false，因此排序顺序基于该值（true 将放置在 @987654334 之前@。

let stores = [{"name":"Store1","distance":13,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isOpen":true}}},{"name":"store2","distance":12,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isOpen":true}}},{"name":"Store5","distance":8,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isFutureOrderPossible":false}}},{"name":"store3","distance":7}];

function comparator(a,b) {
  let aResult = a?.web?.validateAttributes?.isOpen ?? false;
  let bResult = b?.web?.validateAttributes?.isOpen ?? false;
  return (aResult === bResult)? a.distance - b.distance : bResult - aResult;
}
console.log(stores.sort(comparator));

Answer 2

试试这个

const stores = [{"name":"store3","ditance":8},{"name":"Store5","distance":7,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isFutureOrderPossible":false}}},{"name":"Store1","distance":12,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isOpen":true}}},{"name":"store2","distance":13,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isOpen":true}}}]

   const sorter = (a,b)=> {
      if(!b.web) {
          return -1;
      }
      else if(!b.web.validateAttributes) {
        return -1;
      }
      else if(!b.web.validateAttributes.isOpen) {
        return -1;
      }
      return (a.distance - b.distance)
   }
   console.log(stores.sort(sorter)) 

Answer 3

试试这个。

const stores = [{"name":"store3","ditance":8},{"name":"Store5","distance":7,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isFutureOrderPossible":false}}},{"name":"Store1","distance":12,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isOpen":true}}},{"name":"store2","distance":13,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isOpen":true}}}]

const sorted = stores.slice();

sorted.sort((a, b) => {
    if (!a.web || !b.web) {
        return -1;
    }
    if (!a.web.validateAttributes || !b.web.validateAttributes) {
        return -1;
    }
    const x = a.web.validateAttributes.isOpen || false;
    const y = b.web.validateAttributes.isOpen || false;

    return (x === y) ? a.distance - b.distance : -1;
})
console.log(sorted)

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Answer 4

我发现了 2 个问题。

1. 第一项“距离”的错字：8
2. 您的排序功能。 这是我的逻辑。首先，检查两者是Opening 还是Closing，然后比较距离。 否则只是简单地比较状态打开/关闭

var stores = [{"name":"store3","distance":8},{"name":"Store5","distance":7,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isFutureOrderPossible":false}}},{"name":"Store1","distance":12,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isOpen":true}}},{"name":"store2","distance":13,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isOpen":true}}}];
    
stores.sort((a, b) => {
  if (a.web?.validateAttributes?.isOpen && b.web?.validateAttributes?.isOpen || !a.web?.validateAttributes?.isOpen && !b.web?.validateAttributes?.isOpen) {
    return a.distance - b.distance;
  }
  if (a.web?.validateAttributes?.isOpen) {
    return -1;
  }
  return 1;
})

console.log(stores);

Answer 5

您可以使用一些optional chaining 将isOpen 属性转换为-1 如果真实 或1 如果虚假 来确定它们的位置。

如果这些属性相等，则回退到距离比较。

// fixed "ditance" to "distance" in "store3"
const stores = [{"name":"store3","distance":8},{"name":"Store5","distance":7,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isFutureOrderPossible":false}}},{"name":"Store1","distance":12,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isOpen":true}}},{"name":"store2","distance":13,"web":{"validateAttributes":{"isSelectedDispostionSupported":true,"isOpen":true}}}]

const sorted = [...stores].sort((a, b) => 
  (a.web?.validateAttributes?.isOpen ? -1 : 1) -
  (b.web?.validateAttributes?.isOpen ? -1 : 1) ||
  (a.distance - b.distance))
  
console.log(sorted)

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