我有一个字符串,比如Hello world我需要替换索引3处的char。如何通过指定索引替换字符?

var str = "hello world";

我需要这样的东西

str.replaceAt(0,"h");

当前回答

你可以试试

var strArr = str.split("");

strArr[0] = 'h';

str = strArr.join("");

其他回答

function dothis() { var x = document.getElementById("x").value; var index = document.getElementById("index").value; var text = document.getElementById("text").value; var length = document.getElementById("length").value; var arr = x.split(""); arr.splice(index, length, text); var result = arr.join(""); document.getElementById('output').innerHTML = result; console.log(result); } dothis(); <input id="x" type="text" value="White Dog" placeholder="Enter Text" /> <input id="index" type="number" min="0"value="6" style="width:50px" placeholder="index" /> <input id="length" type="number" min="0"value="1" style="width:50px" placeholder="length" /> <input id="text" type="text" value="F" placeholder="New character" /> <br> <button id="submit" onclick="dothis()">Run</button> <p id="output"></p>

此方法适用于较小长度的字符串,但对于较大的文本可能很慢。

var x = "White Dog";
var arr = x.split(""); // ["W", "h", "i", "t", "e", " ", "D", "o", "g"]
arr.splice(6, 1, 'F');

/* 
  Here 6 is starting index and 1 is no. of array elements to remove and 
  final argument 'F' is the new character to be inserted. 
*/
var result = arr.join(""); // "White Fog"

这里的方法很复杂。 我会这样做:

var myString = "this is my string";
myString = myString.replace(myString.charAt(number goes here), "insert replacement here");

这很简单。

你可以扩展字符串类型来包含inset方法:

String.prototype.append =函数(索引,值){ 返回this.slice(0,index) + value + this.slice(index); }; var s = "新字符串"; 警报(s。追加(4 "完成"));

然后你可以调用函数:

str = str.split('');
str[3] = 'h';
str = str.join('');

使用字符串的一行程序。替换回调(不支持表情符号):

// 0 - index to replace, 'f' - replacement string
'dog'.replace(/./g, (c, i) => i == 0? 'f': c)
// "fog"

解释道:

//String.replace will call the callback on each pattern match
//in this case - each character
'dog'.replace(/./g, function (character, index) {
   if (index == 0) //we want to replace the first character
     return 'f'
   return character //leaving other characters the same
})