从这个最初的问题,我将如何在多个字段应用排序?

使用这种稍作调整的结构,我将如何排序城市(上升)和价格(下降)?

var homes = [
    {"h_id":"3",
     "city":"Dallas",
     "state":"TX",
     "zip":"75201",
     "price":"162500"},
    {"h_id":"4",
     "city":"Bevery Hills",
     "state":"CA",
     "zip":"90210",
     "price":"319250"},
    {"h_id":"6",
     "city":"Dallas",
     "state":"TX",
     "zip":"75000",
     "price":"556699"},
    {"h_id":"5",
     "city":"New York",
     "state":"NY",
     "zip":"00010",
     "price":"962500"}
    ];

我喜欢的事实是,给出的答案提供了一个一般的方法。在我计划使用这段代码的地方,我将不得不对日期以及其他东西进行排序。“启动”对象的能力似乎很方便,如果不是有点麻烦的话。

我试图把这个答案构建成一个很好的通用示例,但我运气不太好。


当前回答

这是另一个可能更接近您对语法的想法的例子

function sortObjects(objArray, properties /*, primers*/) {
    var primers = arguments[2] || {}; // primers are optional

    properties = properties.map(function(prop) {
        if( !(prop instanceof Array) ) {
            prop = [prop, 'asc']
        }
        if( prop[1].toLowerCase() == 'desc' ) {
            prop[1] = -1;
        } else {
            prop[1] = 1;
        }
        return prop;
    });

    function valueCmp(x, y) {
        return x > y ? 1 : x < y ? -1 : 0; 
    }

    function arrayCmp(a, b) {
        var arr1 = [], arr2 = [];
        properties.forEach(function(prop) {
            var aValue = a[prop[0]],
                bValue = b[prop[0]];
            if( typeof primers[prop[0]] != 'undefined' ) {
                aValue = primers[prop[0]](aValue);
                bValue = primers[prop[0]](bValue);
            }
            arr1.push( prop[1] * valueCmp(aValue, bValue) );
            arr2.push( prop[1] * valueCmp(bValue, aValue) );
        });
        return arr1 < arr2 ? -1 : 1;
    }

    objArray.sort(function(a, b) {
        return arrayCmp(a, b);
    });
}

// just for fun use this to reverse the city name when sorting
function demoPrimer(str) {
    return str.split('').reverse().join('');
}

// Example
sortObjects(homes, ['city', ['price', 'desc']], {city: demoPrimer});

演示:http://jsfiddle.net/Nq4dk/2/


编辑:只是为了好玩,这里有一个变化,只需要一个类似sql的字符串,所以你可以做sortObjects(房屋,“城市,价格desc”)

function sortObjects(objArray, properties /*, primers*/) {
    var primers = arguments[2] || {};

    properties = properties.split(/\s*,\s*/).map(function(prop) {
        prop = prop.match(/^([^\s]+)(\s*desc)?/i);
        if( prop[2] && prop[2].toLowerCase() === 'desc' ) {
            return [prop[1] , -1];
        } else {
            return [prop[1] , 1];
        }
    });

    function valueCmp(x, y) {
        return x > y ? 1 : x < y ? -1 : 0; 
    }

    function arrayCmp(a, b) {
        var arr1 = [], arr2 = [];
        properties.forEach(function(prop) {
            var aValue = a[prop[0]],
                bValue = b[prop[0]];
            if( typeof primers[prop[0]] != 'undefined' ) {
                aValue = primers[prop[0]](aValue);
                bValue = primers[prop[0]](bValue);
            }
            arr1.push( prop[1] * valueCmp(aValue, bValue) );
            arr2.push( prop[1] * valueCmp(bValue, aValue) );
        });
        return arr1 < arr2 ? -1 : 1;
    }

    objArray.sort(function(a, b) {
        return arrayCmp(a, b);
    });
}

其他回答

改编自@chriskelly的回答。


大多数答案都忽略了,如果价值在1万美元以下或超过100万美元,价格将无法正确排序。原因是JS按字母顺序排序。这里回答得很好,为什么JavaScript不能对“5,10,1”排序,这里如何正确地对整数数组排序。

最后,如果我们要排序的字段或节点是一个数字,我们必须做一些计算。我并不是说在这种情况下使用parseInt()是正确的答案,排序结果更重要。

var homes = [{ "h_id": "2", "city": "Dallas", "state": "TX", "zip": "75201", "price": "62500" }, { "h_id": "1", "city": "Dallas", "state": "TX", "zip": "75201", "price": "62510" }, { "h_id": "3", "city": "Dallas", "state": "TX", "zip": "75201", "price": "162500" }, { "h_id": "4", "city": "Bevery Hills", "state": "CA", "zip": "90210", "price": "319250" }, { "h_id": "6", "city": "Dallas", "state": "TX", "zip": "75000", "price": "556699" }, { "h_id": "5", "city": "New York", "state": "NY", "zip": "00010", "price": "962500" }]; homes.sort(fieldSorter(['price'])); // homes.sort(fieldSorter(['zip', '-state', 'price'])); // alternative function fieldSorter(fields) { return function(a, b) { return fields .map(function(o) { var dir = 1; if (o[0] === '-') { dir = -1; o = o.substring(1); } if (!parseInt(a[o]) && !parseInt(b[o])) { if (a[o] > b[o]) return dir; if (a[o] < b[o]) return -(dir); return 0; } else { return dir > 0 ? a[o] - b[o] : b[o] - a[o]; } }) .reduce(function firstNonZeroValue(p, n) { return p ? p : n; }, 0); }; } document.getElementById("output").innerHTML = '<pre>' + JSON.stringify(homes, null, '\t') + '</pre>'; <div id="output"> </div>


用来测试的小提琴

一个动态的方法来做到这与多个键:

从排序的每个col/key中过滤唯一的值 按顺序排列或颠倒 根据indexOf(value)键值为每个对象添加weights width zeropad 使用计算的权重进行排序

Object.defineProperty(Array.prototype, 'orderBy', {
value: function(sorts) { 
    sorts.map(sort => {            
        sort.uniques = Array.from(
            new Set(this.map(obj => obj[sort.key]))
        );
        
        sort.uniques = sort.uniques.sort((a, b) => {
            if (typeof a == 'string') {
                return sort.inverse ? b.localeCompare(a) : a.localeCompare(b);
            }
            else if (typeof a == 'number') {
                return sort.inverse ? b - a : a - b;
            }
            else if (typeof a == 'boolean') {
                let x = sort.inverse ? (a === b) ? 0 : a? -1 : 1 : (a === b) ? 0 : a? 1 : -1;
                return x;
            }
            return 0;
        });
    });

    const weightOfObject = (obj) => {
        let weight = "";
        sorts.map(sort => {
            let zeropad = `${sort.uniques.length}`.length;
            weight += sort.uniques.indexOf(obj[sort.key]).toString().padStart(zeropad, '0');
        });
        //obj.weight = weight; // if you need to see weights
        return weight;
    }

    this.sort((a, b) => {
        return weightOfObject(a).localeCompare( weightOfObject(b) );
    });
    
    return this;
}
});

Use:

// works with string, number and boolean
let sortered = your_array.orderBy([
    {key: "type", inverse: false}, 
    {key: "title", inverse: false},
    {key: "spot", inverse: false},
    {key: "internal", inverse: true}
]);

对于你的具体问题,一个非通用的,简单的解决方案:

homes.sort(
   function(a, b) {          
      if (a.city === b.city) {
         // Price is only important when cities are the same
         return b.price - a.price;
      }
      return a.city > b.city ? 1 : -1;
   });

简单的一个:

var someArray = [...];

function generateSortFn(props) {
    return function (a, b) {
        for (var i = 0; i < props.length; i++) {
            var prop = props[i];
            var name = prop.name;
            var reverse = prop.reverse;
            if (a[name] < b[name])
                return reverse ? 1 : -1;
            if (a[name] > b[name])
                return reverse ? -1 : 1;
        }
        return 0;
    };
};

someArray.sort(generateSortFn([{name: 'prop1', reverse: true}, {name: 'prop2'}]));

通过添加3个相对简单的帮助程序,可以构建一个非常直观的功能解决方案。在深入研究之前,我们先来了解一下用法:

function usage(homes, { asc, desc, fallback }) { homes.sort(fallback( asc(home => home.city), desc(home => parseInt(home.price, 10)), )); console.log(homes); } var homes = [{ h_id: "3", city: "Dallas", state: "TX", zip: "75201", price: "162500", }, { h_id: "4", city: "Bevery Hills", state: "CA", zip: "90210", price: "319250", }, { h_id: "6", city: "Dallas", state: "TX", zip: "75000", price: "556699", }, { h_id: "5", city: "New York", state: "NY", zip: "00010", price: "962500", }]; const SortHelpers = (function () { const asc = (fn) => (a, b) => (a = fn(a), b = fn(b), -(a < b) || +(a > b)); const desc = (fn) => (a, b) => asc(fn)(b, a); const fallback = (...fns) => (a, b) => fns.reduce((diff, fn) => diff || fn(a, b), 0); return { asc, desc, fallback }; })(); usage(homes, SortHelpers);

如果你向下滚动代码片段,你可能已经看到了helper:

const asc  = (fn) => (a, b) => (a = fn(a), b = fn(b), -(a < b) || +(a > b));
const desc = (fn) => (a, b) => asc(fn)(b, a);
const fallback = (...fns) => (a, b) => fns.reduce((diff, fn) => diff || fn(a, b), 0);

让我快速解释一下这些函数的作用。

asc creates a comparator function. The provided function fn is called for both the comparator arguments a and b. The results of the two function calls are then compared. -1 is returned if resultA < resultB, 1 is returned if resultA > resultB, or 0 otherwise. These return values correspond with an ascending order direction. It could also be written like this: function asc(fn) { return function (a, b) { // apply `fn` to both `a` and `b` a = fn(a); b = fn(b); if (a < b) return -1; if (a > b) return 1; return 0; // or `return -(a < b) || +(a > b)` for short }; } desc is super simple, since it just calls asc but swaps the a and b arguments, resulting in descending order instead of ascending. fallback (there might be a better name for this) allows us to use multiple comparator functions with a single sort. Both asc and desc can be passed to sort by themself. homes.sort(asc(home => home.city)) There is however an issue if you want to combine multiple comparator functions. sort only accepts a single comparator function. fallback combines multiple comparator functions into a single comparator. The first comparator is called with arguments a and b, if the comparator returns the value 0 (meaning that the values are equal) then we fall back to the next comparator. This continues until a non-0 value is found, or until all comparators are called, in which case the return value is 0.

也可以为fallback()提供自定义比较器函数。假设您想使用localeCompare()而不是比较字符串与<和>。在这种情况下,您可以将asc(home => home.city)替换为(a, b) => a.city.localeCompare(b.city)。

homes.sort(fallback(
  (a, b) => a.city.localeCompare(b.city),
  desc(home => parseInt(home.price, 10)),
));

需要注意的一点是,在与<和>比较时,未定义的值总是返回false。因此,如果一个值可能缺失,您可能希望首先根据它的存在进行排序。

homes.sort(fallback(
  // homes with optionalProperty first, true (1) > false (0) so we use desc
  desc(home => home.optionalProperty != null), // checks for both null and undefined
  asc(home => home.optionalProperty),
  // ...
))

因为用localeCompare()比较字符串是一件很常见的事情,所以可以将其作为asc()的一部分。

function hasMethod(item, methodName) {
  return item != null && typeof item[methodName] === "function";
}

function asc(fn) {
  return function (a, b) {
    a = fn(a);
    b = fn(b);

    const areLocaleComparable =
      hasMethod(a, "localeCompare") && hasMethod(b, "localeCompare");

    if (areLocaleComparable) return a.localeCompare(b);

    return -(a < b) || +(a > b);
  };
}