假设我们有一组具有混合名称值的对象(nums,cyrillic,english)：
(如果代码不适合你,将undefined更改为’ru’,它也会改变排序结构)
let ascending = true
var items = [
{name: 'c',value: ''},{name: 'b',{name: 'a',{name: 'д',{name: 'в',{name: '41',{name: 'а',{name: 'б',{name: '0',{name: '31',{name: '4',{name: 'г',value: ''}
]
items.sort(function (a,b) {
// ascending order
if (ascending) {
return a.name.localeCompare(b.name,undefined,{ numeric: true });
}
// descending order
else {
return b.name.localeCompare(a.name,{ numeric: true });
}
})
console.log(items);
目前的结果
我按升序获得排序数组,结构如下：
> Nums(升序)
>英语(升序)
>非英语(升序)
升序= true时的结果：
{name: "0",value: ""}
{name: "31",value: ""}
{name: "4",value: ""}
{name: "41",value: ""}
{name: "a",value: ""}
{name: "b",value: ""}
{name: "c",value: ""}
{name: "а",value: ""}
{name: "б",value: ""}
{name: "в",value: ""}
{name: "г",value: ""}
{name: "д",value: ""}
期望的结果
我需要它能够在升序= false时按降序对数组进行排序并保留结构：
> Nums(降序)
>英语(降序)
>非英语(降序)
升序= false时需要的结果：
{name: "41",value: ""}
{name: "0",value: ""}
{name: "c",value: ""}
问题
当我更改ascending = false并使用a.name更改b.name的位置时,它只是将整个数组翻转而不是翻转其“类别”(nums,english,cyrillic)中的值.
我不确定如何正确地做到这一点.我的意思是,反转数组会反转这些值,所以我应该在数组翻转后重组数组“类别”吗？这样的事情可能是：
>使用isNaN()获取带有数字的所有对象,并将它们移到顶部
>然后在[0]处获取仅包含a-z的项目,并将它们移到数字“类别”下方
>其他一切都停留在底部
解决方法
如果您有识别数字,英语和非英语字符串的机制,您可以使用以下想法：
var items = [
{ name: "c",value: "" },{ name: "b",{ name: "a",{ name: "д",{ name: "в",{ name: "41",{ name: "а",{ name: "б",{ name: "0",{ name: "31",{ name: "4",{ name: "г",value: "" }
];
function sortFunctionMaker(ascending) {
function isNumber(str) {
return Number.isNaN(Number(str)) === false;
}
function isEnglish(str) {
return /^[a-zA-Z]+$/.test(str);
}
return function(a,b) {
var aw,bw;
if (isNumber(a.name)) {
aw = 1;
} else if (isEnglish(a.name)) {
aw = 2;
} else {
aw = 3;
}
if (isNumber(b.name)) {
bw = 1;
} else if (isEnglish(b.name)) {
bw = 2;
} else {
bw = 3;
}
if (aw !== bw) {
// a and b belong to different categories
// no further comparison is needed
return aw - bw;
} else if (aw === 1) {
// both are numbers
// sort mathematically
return (ascending ? 1 : -1) * (a.name - b.name);
} else {
// both are english or otherwise
// sort using localeCompare
return (ascending ? 1 : -1) * a.name.localeCompare(b.name);
}
}
}
items.sort(sortFunctionMaker(true));
console.log("Ascending");
items.forEach(function(item) {
console.log(item.name);
});
items.sort(sortFunctionMaker(false));
console.log("Descending");
items.forEach(function(item) {
console.log(item.name);
});
console.groupEnd();
