你可以使用这样的函数:

function fixStr(str) {
    var out = str.replace(/^\s*/, "");  // strip leading spaces
    out = out.replace(/^[a-z]|[^\s][A-Z]/g, function(str, offset) {
        if (offset == 0) {
            return(str.toUpperCase());
        } else {
            return(str.substr(0,1) + " " + str.substr(1).toUpperCase());
        }
    });
    return(out);
}

"hello World" ==> "Hello World"
"HelloWorld" ==> "Hello World"
"FunInTheSun" ==? "Fun In The Sun"

带有一堆测试字符串的代码:http://jsfiddle.net/jfriend00/FWLuV/。

保留前导空格的替代版本:http://jsfiddle.net/jfriend00/Uy2ac/。

我的分裂案例解决方案的行为方式，我想:

const splitCase = s => !s || s.indexOf(' ') >= 0 ? s :
    (s.charAt(0).toUpperCase() + s.substring(1))
        .split(/(?<=[a-z])(?=[A-Z])|(?<=[A-Z])(?=[A-Z][a-z])/g)
        .map(x => x.replace(/([0-9]+)/g,'$1 '))
        .join(' ')

输入

'a,abc,TheId,TheID,TheIDWord,TheID2Word,Leave me Alone!'
.split(',').map(splitCase)
.forEach(x => console.log(x))

输出

A
Abc
The Id
The ID
The ID Word
The ID2 Word
Leave me Alone!

由于上述函数需要在JS中使用Lookbehind，而目前Safari中还没有实现，因此我重写了实现，以不使用下面的RegEx:

const isUpper = c => c >= 'A' && c <= 'Z'
const isDigit = c => c >= '0' && c <= '9'
const upperOrDigit = c => isUpper(c) || isDigit(c)

function splitCase(s) {
    let to = []
    if (typeof s != 'string') return to
    let lastSplit = 0
    for (let i=0; i<s.length; i++) {
        let c = s[i]
        let prev = i>0 ? s[i-1] : null
        let next = i+1 < s.length ? s[i+1] : null
        if (upperOrDigit(c) && (!upperOrDigit(prev) || !upperOrDigit(next))) {
            to.push(s.substring(lastSplit, i))
            lastSplit = i
        }
    }
    to.push(s.substring(lastSplit, s.length))
    return to.filter(x => !!x)
}

试试这个库

http://sugarjs.com/api/String/titleize

'man from the boondocks'.titleize()>"Man from the Boondocks"
'x-men: the last stand'.titleize()>"X Men: The Last Stand"
'TheManWithoutAPast'.titleize()>"The Man Without a Past"
'raiders_of_the_lost_ark'.titleize()>"Raiders of the Lost Ark"

使用JS的String.prototype.replace()和String.prototype.toUpperCase()

const str = "thisIsATestString"; const res = str.replace (/ ^ [a - z] | [a - z] / g (c i) = >(我吗?": "") + c.toUpperCase()); console.log (res);// "This Is A Test String"

添加了另一个ES6解决方案，我更喜欢上面的一些想法。

https://codepen.io/902Labs/pen/mxdxRv?editors=0010#0

const camelize = (str) => str
    .split(' ')
    .map(([first, ...theRest]) => (
        `${first.toUpperCase()}${theRest.join('').toLowerCase()}`)
    )
    .join(' ');

连续大写单词的最兼容的答案是:

常量文本 = 'theKD'; const result = text.replace（/（[A-Z]{1，}）/g， “$1”）; const finalResult = result.charAt（0）.toUpperCase（） + result.slice（1）; 控制台.log（最终结果）;

它也与KD兼容，不会将其转换为KD。