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

您可以使用链式排序方法,取值的增量,直到它达到不等于零的值。

var data = [{ 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" }]; data.sort(function (a, b) { return a.city.localeCompare(b.city) || b.price - a.price; }); console.log(data); .as-console-wrapper { max-height: 100% !important; top: 0; }

或者,使用es6,简单地:

data.sort((a, b) => a.city.localeCompare(b.city) || b.price - a.price);

其他回答

在这里,您可以尝试按多个字段进行排序的更小且方便的方法!

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" } ]; homes.sort((a, b)=> { if (a.city === b.city){ return a.price < b.price ? -1 : 1 } else { return a.city < b.city ? -1 : 1 } }) console.log(homes);

homes.sort(function(a,b) { return a.city - b.city } );
homes.sort(function(a,b){
    if (a.city==b.city){
        return parseFloat(b.price) - parseFloat(a.price);
    } else {
        return 0;
    }
});

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);
    });
}

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;
   });

这是一个通用的多维排序,允许在每个层次上进行反转和/或映射。

用Typescript编写。对于Javascript,请查看这个JSFiddle

的代码

type itemMap = (n: any) => any;

interface SortConfig<T> {
  key: keyof T;
  reverse?: boolean;
  map?: itemMap;
}

export function byObjectValues<T extends object>(keys: ((keyof T) | SortConfig<T>)[]): (a: T, b: T) => 0 | 1 | -1 {
  return function(a: T, b: T) {
    const firstKey: keyof T | SortConfig<T> = keys[0];
    const isSimple = typeof firstKey === 'string';
    const key: keyof T = isSimple ? (firstKey as keyof T) : (firstKey as SortConfig<T>).key;
    const reverse: boolean = isSimple ? false : !!(firstKey as SortConfig<T>).reverse;
    const map: itemMap | null = isSimple ? null : (firstKey as SortConfig<T>).map || null;

    const valA = map ? map(a[key]) : a[key];
    const valB = map ? map(b[key]) : b[key];
    if (valA === valB) {
      if (keys.length === 1) {
        return 0;
      }
      return byObjectValues<T>(keys.slice(1))(a, b);
    }
    if (reverse) {
      return valA > valB ? -1 : 1;
    }
    return valA > valB ? 1 : -1;
  };
}

用法示例

先按姓排序,再按名排序:

interface Person {
  firstName: string;
  lastName: string;
}

people.sort(byObjectValues<Person>(['lastName','firstName']));

按语言代码的名称排序,而不是按语言代码排序(见地图),然后按降序排序(见反向)。

interface Language {
  code: string;
  version: number;
}

// languageCodeToName(code) is defined elsewhere in code

languageCodes.sort(byObjectValues<Language>([
  {
    key: 'code',
    map(code:string) => languageCodeToName(code),
  },
  {
    key: 'version',
    reverse: true,
  }
]));