javascript中的“翻译”功能就像在PHP中一样?

时间:2018-05-07 21:48:13

标签: javascript

是的,我希望javascript中的字符“翻译”功能就像在php中一样。 我做了以下,但它是可怕的。当然必须有更好的方法 - 使用正则表达式?

<html>
<head>
<script>
window.onload = function() { 
  "use strict";
  console.log(translate("abcdefg", "bdf", "XYZ")); // gives aXcYeZg -=-=-
}
function translate(v1, xlatfrom, xlatto) {
  var ch, ipos, retstr = "";
  if (xlatfrom.length != xlatto.length) return ""; // lengths must be =
  for (var i1=0; i1<v1.length; i1+=1) { // go through string
    ch = v1.substring(i1, i1+1); // character by character
    ipos = xlatfrom.indexOf(ch); // ck if in xlatfrom
    if (ipos >= 0) ch = xlatto.substring(ipos, ipos+1); // if yes, replace
    retstr += ch; } // build up return string
  return retstr;
}
</script>
</head>
<body>
</body>
</html>
编辑:我接受了@ dani-sc的回答。我不打算追求表现。但它真是太过分了!并感谢“传播运营商”信息。以下是我如何使用他的答案:

function translate(v1, xlatfrom, xlatto) { // like the PHP translate
 var mapobj = strsToObject(xlatfrom, xlatto); // make str1.ch's:str2ch's object
 return [...v1].map(ch => mapobj[ch] || ch).join(''); // ... is js "spread operator"
}
function strsToObject(str1, str2) { // make object from strings
  if (str1.length != str2.length) return {}; // lengths must be =
  var retobj = {};
  for (var i1=0; i1<str1.length; i1+=1) { // just str[i1]: str2[i1]
    retobj[str1.substring(i1, i1+1)] = str2.substring(i1, i1+1); }
  return retobj;
}

或(这太棒了!谢谢!)

function translate(v1, xlatfrom, xlatto) { // like the PHP translate
  if (xlatfrom.length != xlatto.length) return ""; // lengths must be =
  var mapobj = {}; // make object for mapping
  for (var i1=0; i1<xlatfrom.length; i1+=1) { // just str[i1]: str2[i1]
    mapobj[xlatfrom.substring(i1, i1+1)] = xlatto.substring(i1, i1+1); }
  return [...v1].map(ch => mapobj[ch] || ch).join(''); // ... is js "spread operator"
}

2 个答案:

答案 0 :(得分:2)

好吧,如果你愿意,你可以使用这样的正则表达式:

&#13;
&#13;
function translate(input, oldCharacters, newCharacters) {
  let output = input;
  const oldChArr = [...oldCharacters];
  const newChArr = [...newCharacters];
  for (let i = 0; i < oldChArr.length; i += 1) {
    output = output.replace(new RegExp(oldChArr[i], 'g'), newChArr[i]);
  }
  return output;
}

function translateFixed(input, replacements) {
  return input.replace(/./g, ch => replacements[ch] || ch);
}

function translateFixedNoRegEx(input, replacements) {
  return [...input].map(ch => replacements[ch] || ch).join('');
}

console.log(translate("abcdefgbdb", "bdf", "XYZ"));
console.log(translate("abcdefg", "cde", "dec"));
console.log(translateFixed("abcdefg", {c: 'd', d: 'e', e: 'c'}));
console.log(translateFixedNoRegEx("abcdefg", {c: 'd', d: 'e', e: 'c'}));
&#13;
&#13;
&#13;

如果您可以更改方法的签名,那么当然可以更加简洁。

编辑:我已经添加了两种实际实现您正在寻找的方法。仅供参考,我也将原始方法translate留在那里。

translateFixed使用正则表达式匹配每个字符,如果在replacements参数中指定,则替换它。 translateFixedNoRegex只是从input字符串中创建一个字符数组并迭代它们。如果字符chreplacements参数中的字符匹配,则将其替换,否则保持不变。之后,我们会通过连接字符将其转换回字符串。

你问过[...数组]:它是ES6引入的spread operator。在字符串上使用时,它只占用每个字符并将其作为单个条目放入数组中。这意味着,这两行都是等价的:

&#13;
&#13;
console.log([..."mystring"]);
console.log("mystring".split(''));
&#13;
&#13;
&#13;

答案 1 :(得分:-1)

function translate(val, xlatfrom, xlatto) { // 
  if (xlatfrom.length !== xlatto.length) return "";
  
  Array.from(xlatfrom).forEach((key, index) =>  {
    val = val.replace(key, xlatto[index]);
  })
  return val;
}

console.log(translate("abcdefg", "bdf", "XYZ"));