根据对象的属性之一对对象的JavaScript数组进行排序[重复]

This question already has an answer here:

这个问题在这里已有答案：

Sort array by firstname (alphabetically) in Javascript 15 answers
在Javascript 15答案中按名字（按字母顺序）对数组进行排序

I've got an array of objects, each of which has a property name, a string. I'd like to sort the array by this property. I'd like them sorted in the following way..

我有一个对象数组，每个对象都有一个属性名称，一个字符串。我想通过这个属性对数组进行排序。我希望他们按以下方式排序..

`ABC`
`abc`
`BAC`
`bac`
etc...

How would I achieve this in JavaScript?

我如何在JavaScript中实现这一目标？

5 个解决方案

#1

There are 2 basic ways:

有两种基本方法：

var arr = [{name:"ABC"},{name:"BAC"},{name:"abc"},{name:"bac"}];

arr.sort(function(a,b){
  var alc = a.name.toLowerCase(), blc = b.name.toLowerCase();
  return alc > blc ? 1 : alc < blc ? -1 : 0;
 });

要么

arr.sort(function(a,b){
  return a.name.toLowerCase().localeCompare(b.name.toLowerCase());
 });

Be aware that the 2nd version ignore diacritics, so a and à will be sorted as the same letter.

请注意，第二个版本会忽略变音符号，因此a和à将被排序为相同的字母。

Now the problem with both these ways is that they will not sort uppercase ABC before lowercase abc, since it will treat them as the same.

现在这两种方式的问题在于它们不会在小写abc之前对大写ABC进行排序，因为它会将它们视为相同。

To fix that, you will have to do it like this:

要解决这个问题，你必须这样做：

arr.sort(function(a,b){
  var alc = a.name.toLowerCase(), blc = b.name.toLowerCase();
  return alc > blc ? 1 : alc < blc ? -1 : a.name > b.name ? 1 : a.name < b.name ? -1 : 0;
});

Again here you could choose to use localeCompare instead if you don't want diacritics to affect the sorting like this:

如果您不希望变音符号影响排序，您可以再次选择使用localeCompare：

arr.sort(function(a,b){
  var lccomp = a.name.toLowerCase().localeCompare(b.name.toLowerCase());
  return lccomp ? lccomp : a.name > b.name ? 1 : a.name < b.name ? -1 : 0;
});

You can read more about sort here: https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Array/sort

您可以在此处阅读有关排序的更多信息：https：//developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Array/sort

#2

You can pass-in a sort function reference to Array.sort.

您可以将一个排序函数引用传递给Array.sort。

#3

objects.sort(function(c, d) {
    return (
        c['name'].toLowerCase() > d['name'].toLowerCase() || 
            c['name'] > d['name']
    ) ? 1 : -1;
});

see there http://jsfiddle.net/p8Gny/1/

看到那里http://jsfiddle.net/p8Gny/1/

#4

You can pass a custom sorting function to the sort() method of an array. The following will do the trick and take your requirements about capitalization into account.

您可以将自定义排序函数传递给数组的sort（）方法。以下将解决这个问题并考虑您对资本化的要求。

objects.sort(function(o1, o2) {
    var n1 = o1.name, n2 = o2.name, ni1 = n1.toLowerCase(), ni2 = n2.toLowerCase();
    return ni1 === ni2 ? (n1 === n2 ? 0 : n1 > n2 ? 1 : -1) : (ni1 > ni2 ? 1 : -1);
});

#5

Slightly modified from Sorting an array of objects,

从排序对象数组略微修改，

yourobject.sort(function(a, b) {
    var nameA = a.name, nameB = b.name
    if (nameA < nameB) //Sort string ascending.
        return -1
    if (nameA > nameB)
        return 1
    return 0 //Default return value (no sorting).
})

#1