您可以简单地执行以下操作：
- 如果
is_folder属性相等，则它们都是文件，或者都是文件夹。比较他们的名字 - 否则排序
is_folder === "Y"更高
上升：
const arr = [
{ "name": "Folder 3", "is_folder": "Y" },
{ "name": "2.jpg", "is_folder": "N" },
{ "name": "Folder 2", "is_folder": "Y" },
{ "name": "Folder 10", "is_folder": "Y" },
{ "name": "Folder 1", "is_folder": "Y" },
{ "name": "15.jpg", "is_folder": "N" },
{ "name": "3.jpg", "is_folder": "N" },
{ "name": "abc.txt", "is_folder": "N" },
{ "name": "1.jpg", "is_folder": "N" },
];
arr.sort((a, b) => {
if (a.is_folder === b.is_folder)
return a.name.localeCompare(b.name, undefined, {numeric: true});
if (a.is_folder === "Y")
return -1;
if (b.is_folder === "Y")
return 1;
return 0;
});
console.log(arr)降序：
const arr = [
{ "name": "Folder 3", "is_folder": "Y" },
{ "name": "2.jpg", "is_folder": "N" },
{ "name": "Folder 2", "is_folder": "Y" },
{ "name": "Folder 10", "is_folder": "Y" },
{ "name": "Folder 1", "is_folder": "Y" },
{ "name": "15.jpg", "is_folder": "N" },
{ "name": "3.jpg", "is_folder": "N" },
{ "name": "abc.txt", "is_folder": "N" },
{ "name": "1.jpg", "is_folder": "N" },
];
arr.sort((a, b) => {
if (a.is_folder === b.is_folder)
return b.name.localeCompare(a.name, undefined, {numeric: true}); //<-- flip `a` and `b`
if (a.is_folder === "Y")
return -1;
if (b.is_folder === "Y")
return 1;
return 0;
});
console.log(arr)在TypeScript Playground上查看（包括类型）
使用localeCompare与数字整理选项可以确保数字将被正确排序，其中10为后 2，例如。如果不使用它，将会发生以下情况：
const arr = [
{ "name": "Folder 3", "is_folder": "Y" },
{ "name": "2.jpg", "is_folder": "N" },
{ "name": "Folder 2", "is_folder": "Y" },
{ "name": "Folder 10", "is_folder": "Y" },
{ "name": "Folder 1", "is_folder": "Y" },
{ "name": "15.jpg", "is_folder": "N" },
{ "name": "3.jpg", "is_folder": "N" },
{ "name": "abc.txt", "is_folder": "N" },
{ "name": "1.jpg", "is_folder": "N" },
];
arr.sort((a, b) => {
if (a.is_folder === b.is_folder)
return a.name.localeCompare(b.name); //<-- no numeric collation
if (a.is_folder === "Y")
return -1;
if (b.is_folder === "Y")
return 1;
});
console.log(arr)如您所见，我们基本上重复了所有代码以进行升序/降序排序。这使维护变得更加困难，但是我们可以改进它。我们可以将每个部分提取到一个单独的函数中：
- 排序相似项目的名称-升序
- 排序相似项目的名称-降序
- 在文件之前对文件夹进行排序
幸运的是，对于的升/降，翻转和的顺序a与b乘以相同，-1因为它localeCompare返回一个数字-正，负或零。因此，我们只能使用一次逻辑，而不能重复两次：
const compareFoldersFirst = (a, b) => {
if (a.is_folder === "Y")
return -1;
if (b.is_folder === "Y")
return 1;
return 0;
}
const compareNameAsc = (a, b) => {
if (a.is_folder === b.is_folder)
return a.name.localeCompare(b.name, undefined, {numeric: true});
return 0;
}
const compareNameDesc = (a, b) => compareNameAsc(a, b) * -1;
实际上，我们可以泛化其中使用的逻辑compareNameDesc-它只运行带有两个参数的函数并将其乘以-1，因此我们可以创建一个reverse可以颠倒任何排序顺序的泛型函数：
const reverse = compareFn => (a, b) => compareFn(a, b) * -1;
const compareNameDesc = reverse(compareNameAsc);
另外，我们可以略微更改每个比较的逻辑以使其完全自给自足，因为现在名称的排序取决于某个东西是否是文件夹。
const compareFoldersFirst = (a, b) => {
if (a.is_folder === b.is_folder)
return 0;
if (a.is_folder === "Y")
return -1;
if (b.is_folder === "Y")
return 1;
};
用布尔值和数字转换规则的“创造性用法”来表达的时间更短：
const compareFoldersFirst = (a, b) => Number(b.is_folder === "Y") - Number(a.is_folder === "Y");
无论如何，这使我们能够is_folder从名称比较中删除检查，而我们仅剩下以下比较器：
const compareFoldersFirst = (a, b) => Number(b.is_folder === "Y") - Number(a.is_folder === "Y");
const compareNameAsc = (a, b) => a.name.localeCompare(b.name, undefined, {numeric: true});
const compareNameDesc = reverse(compareNameAsc);
我们几乎拥有所有需要的工具，只需一次添加更多的逻辑，然后再进行逆向转换，就可以生成所需的任何排序顺序。我们只需要能够轻松组成不同的比较器即可。为此，我们可以将排序归纳为以下内容：获得任意数量的排序函数。我们产生了一个新函数，它将逐个运行它们，直到一个返回非零结果为止。可以这样做
const comparer = (...comparers) =>
(a, b) => {
for(let compareFn of comparers){
const result = compareFn(a, b);
if (result !== 0)
return result;
}
}
但是可以使用使其更紧凑Array#reduce。最后，使用helper compare函数可以使代码更易于维护和组合：
const arr = [
{ "name": "Folder 3", "is_folder": "Y" },
{ "name": "2.jpg", "is_folder": "N" },
{ "name": "Folder 2", "is_folder": "Y" },
{ "name": "Folder 10", "is_folder": "Y" },
{ "name": "Folder 1", "is_folder": "Y" },
{ "name": "15.jpg", "is_folder": "N" },
{ "name": "3.jpg", "is_folder": "N" },
{ "name": "abc.txt", "is_folder": "N" },
{ "name": "1.jpg", "is_folder": "N" },
];
//helper function that takes any amount of compare functions
//produces a function that runs each until a non-zero result
const compare = (...comparers) =>
(a, b) => comparers.reduce(
(result, compareFn) => result || compareFn(a, b),
0
);
//reverse the result of any compare function after it runs:
const reverse = compareFn => (a, b) => compareFn (a, b) * -1;
//the basic comparer functions:
const compareFoldersFirst = (a, b) => Number(b.is_folder === "Y") - Number(a.is_folder === "Y");
const compareNameAsc = (a, b) => a.name.localeCompare(b.name, undefined, {numeric: true});
const compareNameDesc = reverse(compareNameAsc);
//final comparison function derived form the basic ones
const asc = compare(
compareFoldersFirst,
compareNameAsc
);
const desc = compare(
compareFoldersFirst,
compareNameDesc
);
console.log("--- Ascending sort ---\n", arr.sort(asc));
console.log("--- Descending sort ---\n", arr.sort(desc));
我已经应用了您的原始答案（请参阅更新的问题），但这似乎并未对文件进行排序。它只对文件夹排序
@J。你有错字吗
a.is_foler->a.is_folder确实是一个错字。我的错。面容
@ J.Do是否已添加指向TS游乐场的链接，该链接使用类型进行比较。这很有用，因为
(a: any, b: any)允许像您必须滑过的小错字，如果您限制类型，则会发现错误，因为该属性不存在。