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。
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);
这是一个完全的欺骗,但我认为它为这个问题增加了价值,因为它基本上是一个罐装的库函数,你可以开箱即用。
如果你的代码可以访问lodash或者一个与lodash兼容的库,比如下划线,那么你可以使用_。sortBy方法。下面的代码片段直接复制自lodash文档。
示例中的注释结果看起来像是返回数组的数组,但这只是显示了顺序,而不是实际的结果,它是一个对象数组。
var users = [
{ 'user': 'fred', 'age': 48 },
{ 'user': 'barney', 'age': 36 },
{ 'user': 'fred', 'age': 40 },
{ 'user': 'barney', 'age': 34 }
];
_.sortBy(users, [function(o) { return o.user; }]);
// => objects for [['barney', 36], ['barney', 34], ['fred', 48], ['fred', 40]]
_.sortBy(users, ['user', 'age']);
// => objects for [['barney', 34], ['barney', 36], ['fred', 40], ['fred', 48]]
这是一个递归算法,按多个字段排序,同时有机会在比较之前格式化值。
var data = [
{
"id": 1,
"ship": null,
"product": "Orange",
"quantity": 7,
"price": 92.08,
"discount": 0
},
{
"id": 2,
"ship": "2017-06-14T23:00:00.000Z".toDate(),
"product": "Apple",
"quantity": 22,
"price": 184.16,
"discount": 0
},
...
]
var sorts = ["product", "quantity", "ship"]
// comp_val formats values and protects against comparing nulls/undefines
// type() just returns the variable constructor
// String.lower just converts the string to lowercase.
// String.toDate custom fn to convert strings to Date
function comp_val(value){
if (value==null || value==undefined) return null
var cls = type(value)
switch (cls){
case String:
return value.lower()
}
return value
}
function compare(a, b, i){
i = i || 0
var prop = sorts[i]
var va = comp_val(a[prop])
var vb = comp_val(b[prop])
// handle what to do when both or any values are null
if (va == null || vb == null) return true
if ((i < sorts.length-1) && (va == vb)) {
return compare(a, b, i+1)
}
return va > vb
}
var d = data.sort(compare);
console.log(d);
如果a和b相等,它将尝试下一个字段,直到没有可用字段。
另一种方式
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"}
];
function sortBy(ar) {
return ar.sort((a, b) => a.city === b.city ?
b.price.toString().localeCompare(a.price) :
a.city.toString().localeCompare(b.city));
}
console.log(sortBy(homes));
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”
哇,这里有一些复杂的解。如此复杂,我决定想出一些更简单但也相当强大的东西。在这里;
function sortByPriority(data, priorities) {
if (priorities.length == 0) {
return data;
}
const nextPriority = priorities[0];
const remainingPriorities = priorities.slice(1);
const matched = data.filter(item => item.hasOwnProperty(nextPriority));
const remainingData = data.filter(item => !item.hasOwnProperty(nextPriority));
return sortByPriority(matched, remainingPriorities)
.sort((a, b) => (a[nextPriority] > b[nextPriority]) ? 1 : -1)
.concat(sortByPriority(remainingData, remainingPriorities));
}
这里有一个如何使用它的例子。
const data = [
{ id: 1, mediumPriority: 'bbb', lowestPriority: 'ggg' },
{ id: 2, highestPriority: 'bbb', mediumPriority: 'ccc', lowestPriority: 'ggg' },
{ id: 3, mediumPriority: 'aaa', lowestPriority: 'ggg' },
];
const priorities = [
'highestPriority',
'mediumPriority',
'lowestPriority'
];
const sorted = sortByPriority(data, priorities);
这将首先根据属性的优先级排序,然后根据属性的值进行排序。