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

以下是@ snowburn的解决方案的通用版本:

var sortarray = [{field:'city', direction:'asc'}, {field:'price', direction:'desc'}];
array.sort(function(a,b){
    for(var i=0; i<sortarray.length; i++){
        retval = a[sortarray[i].field] < b[sortarray[i].field] ? -1 : a[sortarray[i].field] > b[sortarray[i].field] ? 1 : 0;
        if (sortarray[i].direction == "desc") {
            retval = retval * -1;
        }
        if (retval !== 0) {
            return retval;
        }
    }
}


})

这是基于我正在使用的排序例程。我没有测试这个特定的代码,所以它可能有错误,但你知道的。其思想是基于第一个表示差异的字段进行排序,然后停止并转到下一个记录。因此,如果您按三个字段排序,并且compare中的第一个字段足以确定正在排序的两个记录的排序顺序,那么将返回该排序结果并转到下一个记录。

我在5000条记录上测试了它(实际上使用了更复杂的排序逻辑),它在眨眼之间就做到了。如果实际向客户端加载超过1000条记录,则可能应该使用服务器端排序和过滤。

这段代码不区分大小写,我把它留给读者来处理这个微不足道的修改。

其他回答

按多个字段排序对象数组的最简单方法:

 let 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) => (a.city > b.city) ? 1 : -1);

输出: “Bevery山” “达拉斯” “达拉斯” “达拉斯” “纽约”

我认为这可能是最简单的方法。

https://coderwall.com/p/ebqhca/javascript-sort-by-two-fields

这真的很简单,我尝试了3个不同的键值对,它工作得很好。

这是一个简单的例子,查看链接了解更多细节

testSort(data) {
    return data.sort(
        a['nameOne'] > b['nameOne'] ? 1
        : b['nameOne'] > a['nameOne'] ? -1 : 0 ||
        a['date'] > b['date'] ||
        a['number'] - b['number']
    );
}

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);
  };
}

我喜欢snowburn的方法,但它需要调整来测试城市的等效性,而不是差异。

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

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;
   });