我有一个多维数组。主要数组是一个数组
[publicationID][publication_name][ownderID][owner_name]
我想要做的是对数组进行排序owner_name，然后再排序publication_name。我知道在 JavaScript 中你有Array.sort()，你可以在其中放置一个自定义函数，在我的情况下，我有：
function mysortfunction(a, b) {
var x = a[3].toLowerCase();
var y = b[3].toLowerCase();
return ((x < y) ? -1 : ((x > y) ? 1 : 0));
}
这是刚刚排序的一列，即OWNER_NAME很好，但我怎样修改它进行排序owner_name，然后publication_name？
如果所有者名称不同，请按它们排序。否则，对决胜局使用出版物名称。
function mysortfunction(a, b) {
var o1 = a[3].toLowerCase();
var o2 = b[3].toLowerCase();
var p1 = a[1].toLowerCase();
var p2 = b[1].toLowerCase();
if (o1 < o2) return -1;
if (o1 > o2) return 1;
if (p1 < p2) return -1;
if (p1 > p2) return 1;
return 0;
}
我想你要找的是 thenBy.js：https : //github.com/Teun/thenBy.js
它允许您使用标准的 Array.sort，但具有firstBy().thenBy().thenBy()样式。
对许多字符串字段进行排序的一种好方法是使用toLocaleCompare和 布尔运算符||。
就像是：
// Sorting record releases by name and then by title.
releases.sort((oldRelease, newRelease) => {
const compareName = oldRelease.name.localeCompare(newRelease.name);
const compareTitle = oldRelease.title.localeCompare(newRelease.title);
return compareName || compareTitle;
})
如果你想对更多的字段进行排序，你可以简单地用更多的布尔运算符将它们从 return 语句链接起来。
遇到需要按键进行 SQL 风格的混合 asc 和 desc 对象数组排序的需求。
上面 kennebec 的解决方案帮助我解决了这个问题：
Array.prototype.keySort = function(keys) {
keys = keys || {};
// via
// https://stackoverflow.com/questions/5223/length-of-javascript-object-ie-associative-array
var obLen = function(obj) {
var size = 0, key;
for (key in obj) {
if (obj.hasOwnProperty(key))
size++;
}
return size;
};
// avoiding using Object.keys because I guess did it have IE8 issues?
// else var obIx = function(obj, ix){ return Object.keys(obj)[ix]; } or
// whatever
var obIx = function(obj, ix) {
var size = 0, key;
for (key in obj) {
if (obj.hasOwnProperty(key)) {
if (size == ix)
return key;
size++;
}
}
return false;
};
var keySort = function(a, b, d) {
d = d !== null ? d : 1;
// a = a.toLowerCase(); // this breaks numbers
// b = b.toLowerCase();
if (a == b)
return 0;
return a > b ? 1 * d : -1 * d;
};
var KL = obLen(keys);
if (!KL)
return this.sort(keySort);
for ( var k in keys) {
// asc unless desc or skip
keys[k] =
keys[k] == 'desc' || keys[k] == -1 ? -1
: (keys[k] == 'skip' || keys[k] === 0 ? 0
: 1);
}
this.sort(function(a, b) {
var sorted = 0, ix = 0;
while (sorted === 0 && ix < KL) {
var k = obIx(keys, ix);
if (k) {
var dir = keys[k];
sorted = keySort(a[k], b[k], dir);
ix++;
}
}
return sorted;
});
return this;
};
示例用法：
var obja = [
{USER:"bob", SCORE:2000, TIME:32, AGE:16, COUNTRY:"US"},
{USER:"jane", SCORE:4000, TIME:35, AGE:16, COUNTRY:"DE"},
{USER:"tim", SCORE:1000, TIME:30, AGE:17, COUNTRY:"UK"},
{USER:"mary", SCORE:1500, TIME:31, AGE:19, COUNTRY:"PL"},
{USER:"joe", SCORE:2500, TIME:33, AGE:18, COUNTRY:"US"},
{USER:"sally", SCORE:2000, TIME:30, AGE:16, COUNTRY:"CA"},
{USER:"yuri", SCORE:3000, TIME:34, AGE:19, COUNTRY:"RU"},
{USER:"anita", SCORE:2500, TIME:32, AGE:17, COUNTRY:"LV"},
{USER:"mark", SCORE:2000, TIME:30, AGE:18, COUNTRY:"DE"},
{USER:"amy", SCORE:1500, TIME:29, AGE:19, COUNTRY:"UK"}
];
var sorto = {
SCORE:"desc",TIME:"asc", AGE:"asc"
};
obja.keySort(sorto);
产生以下结果：
0: { USER: jane; SCORE: 4000; TIME: 35; AGE: 16; COUNTRY: DE; }
1: { USER: yuri; SCORE: 3000; TIME: 34; AGE: 19; COUNTRY: RU; }
2: { USER: anita; SCORE: 2500; TIME: 32; AGE: 17; COUNTRY: LV; }
3: { USER: joe; SCORE: 2500; TIME: 33; AGE: 18; COUNTRY: US; }
4: { USER: sally; SCORE: 2000; TIME: 30; AGE: 16; COUNTRY: CA; }
5: { USER: mark; SCORE: 2000; TIME: 30; AGE: 18; COUNTRY: DE; }
6: { USER: bob; SCORE: 2000; TIME: 32; AGE: 16; COUNTRY: US; }
7: { USER: amy; SCORE: 1500; TIME: 29; AGE: 19; COUNTRY: UK; }
8: { USER: mary; SCORE: 1500; TIME: 31; AGE: 19; COUNTRY: PL; }
9: { USER: tim; SCORE: 1000; TIME: 30; AGE: 17; COUNTRY: UK; }
keySort: { }
（使用这里的打印功能）
这对于所有大小的 alpha 类型都很方便。将要排序的索引按顺序传递给它作为参数。
Array.prototype.deepSortAlpha= function(){
var itm, L=arguments.length, order=arguments;
var alphaSort= function(a, b){
a= a.toLowerCase();
b= b.toLowerCase();
if(a== b) return 0;
return a> b? 1:-1;
}
if(!L) return this.sort(alphaSort);
this.sort(function(a, b){
var tem= 0, indx=0;
while(tem==0 && indx<L){
itm=order[indx];
tem= alphaSort(a[itm], b[itm]);
indx+=1;
}
return tem;
});
return this;
}
var arr= [[ "Nilesh","Karmshil"], ["Pranjal","Deka"], ["Susants","Ghosh"],
["Shiv","Shankar"], ["Javid","Ghosh"], ["Shaher","Banu"], ["Javid","Rashid"]];
arr.deepSortAlpha(1,0);