JavaScript 字符串全部替换

Question

我有以下字符串，我想替换字符串的所有特殊字符。

let data = "`test@#$demo%^here[sf.s]\|demo{delata}*testing?+$^()`"

特殊字符：

let spcial=[".", "+", "*", "?", "^", "$", "(", ")", "[", "]", "{", "}", "|", "\"];

最终输出：

test@#\\$demo%\\^here\\[sf\\.s\\]\\\\\|demo\\{delata\\}\\*testing\\?\\+\\$\\^\\(\\)

简而言之，我想在特殊字符前添加\，我该怎么做？

Answer 1

let data = "`test@#$demo%^here[sf.s]\|demo{delata}*testing?+$^()`"
let spcial=[".", "+", "*", "?", "^", "$", "(", ")", "[", "]", "{", "}", "|", "\\"]    
for (let c of data) {
  if(spcial.includes(c) ) {
    data = data.replace(c, '\\'+c)
  }
}
console.log(data)

Answer 2

let data = "`test@#$demo%^here[sf.s]\\|demo{delata}*testing?+$^()`"
let spcial=[".", "+", "*", "?", "^", "$", "(", ")", "[", "]", "{", "}", "|", "\\"]

const res = data.split("").map(ch => spcial.includes(ch) ? "\\\\" + ch : ch).join("")

console.log(res)

Answer 3

let data = "`test@#$demo%^here[sf.s]\|demo{delata}*testing?+$^()`";
let special=[".", "+", "*", "?", "^", "$", "(", ")", "[", "]", "{", "}", "|", "\\"];
let [...chars] = data;
chars = chars.map(char => {
    if (special.includes(char)) return `\\${char}`;
    return char;
}
data = chars.join("");

Answer 4

str = "test@#$demo%^here[sf.s]\|demo{delata}*testing?+$^()";
const spcial=[".", "+", "*", "?", "^", "$", "(", ")", "[", "]", "{", "}", "|", "\\"];
spcial.forEach(myFunction);

function myFunction(value) {
  str = str.replaceAll(value, "\\" + value);
}

console.log(str)

Answer 5

我认为从长远来看，学习一些正则表达式 (regex/regexp) 对你有好处。我用来查找特殊字符的正则表达式是/[^a-zA-Z0-9]/g。如果你使用带有回调函数的replace方法，你可以把匹配的特殊字符m替换成\\${m}。

在此处阅读有关 RegEx 的更多信息：https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp

在此处阅读有关replace 的更多信息：https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace

要完全按照上面的要求做，这将是一个正则表达式实现：

const data = "`test@#$demo%^here[sf.s]|demo{delata}*testing?+$^()`";
const specialCharacters = /\.|\+|\*|\?|\^|\$|\(|\)|\[|\]|\{|\}|\||\\/g;
const output = data.replace(specialCharacters, (m) => `\\${m}`);

console.log(output);

我建议检查更多特殊字符。

const data = "`test@#$demo%^here[sf.s]|demo{delata}*testing?+$^()`";
const output = data.replace(/[^a-zA-Z0-9]/g, (m) => `\\${m}`);
console.log(output);

Answer 6

方法 1：从固定的特殊字符列表中

你可以一次全部替换，用这样的正则表达式：

let data = "test@#$demo%^here[sf.s]\|demo{delata}*testing?+$^()";

let replaced = data.replace(/[\.\+\*\?\^\$\(\)\[\]\{\}\|\\]/g, '\\$&');

console.log(replaced);

请注意，在正则表达式中列出了特殊字符，并使用反斜杠进行转义。 如果您想详细了解其工作原理，请查看here, in the "Explain" section。

方法 2：来自特殊字符的动态列表

如果你想从一个数组动态构建这个正则表达式，你可以这样做：

let data = "test@#$demo%^here[sf.s]\|demo{delata}*testing?+$^()";
let specialChars = [".", "+", "*", "?", "^", "$", "(", ")", "[", "]", "{", "}", "|", "\\"];

let regexp = new RegExp(`[\\${specialChars.join('\\')}]`, 'g');
let replaced = data.replace(regexp, '\\$&');

console.log(replaced);

结果

在这两个例子中，结果都是：

test@#\$demo%\^here\[sf\.s\]\|demo\{delata\}\*testing\?\+\$\^\(\)

使用全局替换（最后用g 表示），您可以一次替换所有字符，无需循环。