使用nodepad++函数，下面的regex对我来说很好，

发现:{1}\ \ s + K 替换:留空

var string = "The dog      has a long   tail, and it     is RED!";
var replaced = string.replace(/ +/g, " ");

或者如果你也想替换制表符:

var replaced = string.replace(/\s+/g, " ");

更健壮的:

function trim(word)
{
    word = word.replace(/[^\x21-\x7E]+/g, ' '); // change non-printing chars to spaces
    return word.replace(/^\s+|\s+$/g, '');      // remove leading/trailing spaces
}

Def removeblanks(text):返回re.sub(r'\s\s+'，" "，text) 我正在处理一个有很多重复空格的大型文本数据。 上面的RE对我很有用。所有重复的空格都被一个空格取代。

let nameCorrection = function (str) {
  let strPerfect = str.replace(/\s+/g, " ").trim();
  let strSmall = strPerfect.toLowerCase();
  let arrSmall = strSmall.split(" ");
  let arrCapital = [];
  for (let x of arrSmall.values()) {
    arrCapital.push(x[0].toUpperCase() + x.slice(1));
  }

  let result = arrCapital.join(" ");
  console.log(result);
};
nameCorrection("Pradeep kumar dHital");

这个脚本删除了单词和修饰之间的任何空白(多个空格，制表符，返回值等):

// Trims & replaces any wihtespacing to single space between words
String.prototype.clearExtraSpace = function(){
  var _trimLeft  = /^\s+/,
      _trimRight = /\s+$/,
      _multiple  = /\s+/g;

  return this.replace(_trimLeft, '').replace(_trimRight, '').replace(_multiple, ' ');
};