Question

我有json对象，我想获得最低价格。以下是回复。

[
  {
    "room": {
      "price": 217,
      "available": true
    }
  },
  {
    "room": {
      "price": 302,
      "available": true,
    }
  },
  {
    "room": {
      "price": 427,
      "available": true,
    }
  }
]

我已尝试过Stackoverflow的解决方案，但无论如何都不会有效。

  var arr = Object.keys( response ).map(function ( key ) { return response[key]; });
  var min = Math.min.apply( null, arr );

请帮忙

Answer 1

你可以试试这个：

＆＃13;

let response = [
  {
    "room": {
      "price": 217,
      "available": true
    }
  },
  {
    "room": {
      "price": 302,
      "available": true,
    }
  },
  {
    "room": {
      "price": 427,
      "available": true,
    }
  }
];
let values  = response.map(function(v) {
  return v.room.price;
});
var min = Math.min.apply( null, values );
console.log(min)

＆＃13;

使用ES2015，您也可以将它放在一行：

var min = Math.min.apply( null, response.map((v) => v.room.price));

Answer 2

您有数组不是对象，因此您无法使用Object.keys()。您也可以使用这样的扩展语法。

＆＃13;

var data = [{
  "room": {
    "price": 217,
    "available": true
  }
}, {
  "room": {
    "price": 302,
    "available": true,
  }
}, {
  "room": {
    "price": 427,
    "available": true,
  }
}]

var min = Math.min(...data.map(e => e.room.price))
console.log(min)

＆＃13;

Answer 3

您可以使用Array.protype.reduce()

var rooms = [{
    "room": {
      "price": 217,
      "available": true
    }
  },
  {
    "room": {
      "price": 302,
      "available": true,
    }
  },
  {
    "room": {
      "price": 427,
      "available": true,
    }
  }
];
console.log(rooms.reduce((prev, curr) => prev.price > curr.price ? curr : prev).room.price);

Answer 4

        var response = [
            {
                "room": {
                    "price": 217,
                    "available": true
                }
            },
            {
                "room": {
                    "price": 302,
                    "available": true,
                }
            },
            {
                "room": {
                    "price": 427,
                    "available": true,
                }
            }
        ];

        debugger;
        if (response && response.length > 0) 
        {
            var min = response[0].room.price;

            for (var i = 0; i < response.length; i++)           
                if (response[i].room.price < min)
                    min = response[i].room.price;

            console.log(min);
        }

Answer 5

原生JS解决方案：

var t =[ { "room": { "price": 217, "available": true } }, { "room": { "price": 302, "available": true, } }, { "room": { "price": 427, "available": true, } } ]
var min = t.map(function(el){return el.room.price}).reduce(function(el){return Math.min(el)});

working fiddle

Answer 6

lbrutty代码有点不对，函数总是返回第一个元素。所以有一点修复。

var t = [ { "room": { "price": 300, "available": true } }, { "room": { "price": 102, "available": true, } }, { "room": { "price": 427, "available": true, } } ];
var min = t.map(function(el){return el.room.price}).reduce(function(prevEl, el){return Math.min(prevEl, el)});

https://jsfiddle.net/xe0dcoym/17/

从嵌套的json对象获取最小值

6 个答案: