我有一个JavaScript对象数组：
var objs = [
{ first_nom: 'Lazslo', last_nom: 'Jamf' },
{ first_nom: 'Pig', last_nom: 'Bodine' },
{ first_nom: 'Pirate', last_nom: 'Prentice' }
];
如何在JavaScript中按last_nom的值对它们进行排序？
我知道排序（a，b），但这似乎只适用于字符串和数字。是否需要向对象添加toString（）方法？
当前回答
对于工作狂：
const objectSorter = (p) => (a, b) => ((a, b) => a>b ? 1 : a<b ? -1 : 0)(a[p], b[p]);
objs.sort(objectSorter('first_nom'));
其他回答
我遇到了排序对象数组的问题，改变了值的优先级。基本上，我想按年龄，然后按姓氏，或者仅仅按姓氏，名字来对一系列人进行排序。
我认为与其他答案相比，这是最简单的解决方案。
它用于调用sortPeoples（['array'，'of'，'properties']，reverse=false）。
///////////////////////人群示例阵列///////////////////////var人民=[{姓名：“Zach”，姓：“Emergency”，年龄：1}，{姓名：“南希”，姓：“护士”，年龄：1岁}，{姓名：“Ethel”，姓：“Emergency”，年龄：1}，{姓名：“尼娜”，姓：“护士”，年龄：42岁}，{姓名：“Anthony”，姓：“Emergency”，年龄：42岁}，{姓名：“尼娜”，姓：“护士”，年龄：32岁}，{姓名：“Ed”，姓：“Emergency”，年龄：28岁}，{姓名：“彼得”，姓：“医生”，年龄：58岁}，{姓名：“Al”，姓：“Emergency”，年龄：58岁}，{姓名：“Ruth”，姓：“Registration”，年龄：62岁}，{姓名：“Ed”，姓：“Emergency”，年龄：38岁}，{姓名：“Tammy”，姓：“Triage”，年龄：29岁}，{姓名：“Alan”，姓：“Emergency”，年龄：60岁}，{姓名：“尼娜”，姓：“护士”，年龄：58岁}];////////////////////////排序功能/////////////////////函数sortPeoples（propertyArr，reverse）{函数比较（a，b）{变量i=0；while（propertyArr[i]）{如果（a[propertyArr[i]]<b[propertyAr[i]]）返回-1；如果（a[propertyArr[i]]>b[propertyAr[i]]）返回1；i++；}返回0；}人民。排序（比较）；if（反向）{peoples.reverse（）；}};////////////////排序方法结束///////////////函数printPeoples（）{$（“#输出”）.html（“”）；people.forEach（功能（人）{$（'#output'）.append（person.surname+“”+person.name+““”+person.age+“<br>”）；})}<head><script src=“https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js“></script></head><html><body><button onclick=“sortPeoples（['surname']）；printPeoples（）”>仅按姓氏ASC排序会导致同名案例的混乱</button><br><button onclick=“sortPeoples（['surname'，'name]，true）；printPeoples（）”>按姓氏排序，然后按名称DESC</button><br><button onclick=“sortPeoples（['age']）；printPeoples（）”>按age ASC排序。与第一种情况相同的问题</button><br><button onclick=“sortPeoples（['age'，'姓']）；printPeoples（）”>按age和姓氏ASC排序。添加第二个字段修复了它。</button><br><div id=“output”></div></body></html>
使用原型继承简单快速地解决此问题：
Array.prototype.sortBy = function(p) {
return this.slice(0).sort(function(a,b) {
return (a[p] > b[p]) ? 1 : (a[p] < b[p]) ? -1 : 0;
});
}
示例/用法
objs = [{age:44,name:'vinay'},{age:24,name:'deepak'},{age:74,name:'suresh'}];
objs.sortBy('age');
// Returns
// [{"age":24,"name":"deepak"},{"age":44,"name":"vinay"},{"age":74,"name":"suresh"}]
objs.sortBy('name');
// Returns
// [{"age":24,"name":"deepak"},{"age":74,"name":"suresh"},{"age":44,"name":"vinay"}]
更新：不再修改原始数组。
使用xPrototype的sortBy：
var o = [
{ Name: 'Lazslo', LastName: 'Jamf' },
{ Name: 'Pig', LastName: 'Bodine' },
{ Name: 'Pirate', LastName: 'Prentice' },
{ Name: 'Pag', LastName: 'Bodine' }
];
// Original
o.each(function (a, b) { console.log(a, b); });
/*
0 Object {Name: "Lazslo", LastName: "Jamf"}
1 Object {Name: "Pig", LastName: "Bodine"}
2 Object {Name: "Pirate", LastName: "Prentice"}
3 Object {Name: "Pag", LastName: "Bodine"}
*/
// Sort By LastName ASC, Name ASC
o.sortBy('LastName', 'Name').each(function(a, b) { console.log(a, b); });
/*
0 Object {Name: "Pag", LastName: "Bodine"}
1 Object {Name: "Pig", LastName: "Bodine"}
2 Object {Name: "Lazslo", LastName: "Jamf"}
3 Object {Name: "Pirate", LastName: "Prentice"}
*/
// Sort by LastName ASC and Name ASC
o.sortBy('LastName'.asc, 'Name'.asc).each(function(a, b) { console.log(a, b); });
/*
0 Object {Name: "Pag", LastName: "Bodine"}
1 Object {Name: "Pig", LastName: "Bodine"}
2 Object {Name: "Lazslo", LastName: "Jamf"}
3 Object {Name: "Pirate", LastName: "Prentice"}
*/
// Sort by LastName DESC and Name DESC
o.sortBy('LastName'.desc, 'Name'.desc).each(function(a, b) { console.log(a, b); });
/*
0 Object {Name: "Pirate", LastName: "Prentice"}
1 Object {Name: "Lazslo", LastName: "Jamf"}
2 Object {Name: "Pig", LastName: "Bodine"}
3 Object {Name: "Pag", LastName: "Bodine"}
*/
// Sort by LastName DESC and Name ASC
o.sortBy('LastName'.desc, 'Name'.asc).each(function(a, b) { console.log(a, b); });
/*
0 Object {Name: "Pirate", LastName: "Prentice"}
1 Object {Name: "Lazslo", LastName: "Jamf"}
2 Object {Name: "Pag", LastName: "Bodine"}
3 Object {Name: "Pig", LastName: "Bodine"}
*/
我将给您一个实现选择排序算法的解决方案。它简单有效。
var objs = [
{ first_nom: 'Lazslo', last_nom: 'Jamf' },
{ first_nom: 'Pig', last_nom: 'Bodine' },
{ first_nom: 'Pirate', last_nom: 'Prentice' }
];
function selection_Sort(num) {
//console.log(num);
var temp, index;
for (var i = 0; i <= num.length - 1; i++) {
index = i;
for (var j = i + 1; j <= num.length - 1; j++) {
// You can use first_nom/last_nom, any way you choose to sort
if (num[j].last_nom < num[index].last_nom) {
index = j;
}
}
// Below is the swapping part
temp = num[i].last_nom;
num[i].last_nom = num[index].last_nom;
num[index].last_nom = temp;
};
console.log(num);
return num;
}
selection_Sort(objs);
let propName = 'last_nom';
let sorted_obj = objs.sort((a,b) => {
if(a[propName] > b[propName]) {
return 1;
}
if (a[propName] < b[propName]) {
return -1;
}
return 0;
}
//This works because the js built-in sort function allows us to define our
//own way of sorting, this funny looking function is simply telling `sort` how to
//determine what is larger.
//We can use `if(a[propName] > b[propName])` because string comparison is already built into JS
//if you try console.log('a' > 'z' ? 'a' : 'z')
//the output will be 'z' as 'a' is not greater than 'z'
//The return values 0,-1,1 are how we tell JS what to sort on. We're sorting on the last_nom property of the object.
//When sorting a list it comes down to comparing two items and how to determine which one of them is "larger".
//We need a way to tell JS how to determine which one is larger.
//The sort defining function will use the case that returns a 1 to mean that a > b
//and the case that returns -1 to mean that a < b
