如何解析字符串以进行换行

Question

我想有一个函数，它有3个参数:句子(字符串)，数字( maxCharLen=20 )，分隔符(字符串)

并根据所述参数对句子进行变换。

示例

var sentence = "JavaScript is a prototype-based scripting language that is dynamic, weakly typed and has first-class functions."

var newSentence = breakSentence(sentence, maxCharLen=20, separator="<br>");

newSentence // JavaScript is a prototype-based  <br> scripting language that is dynamic, <br> weakly typed and has first-class functions.

附言：

这是我尝试过的：

 var breakSentence = function (sentence, maxCharLen, separator)
 {
   sentence = sentence || "javascript is a language" ;
   maxCharLen = 10 || maxCharLen; // max numb of chars for line
   separator = "<br>" || separator;

   var offset;
   var nbBreak = sentence.length // maxCharLen;
   var newSentence = "";

   for (var c = 0; c < nbBreak; c += 1)
   {
    offset = c * maxCharLen;
    newSentence += sentence.substring(offset, offset + maxCharLen) + separator;
   }

   return newSentence;
}

它是这样工作的：

 breakSentence()  // "javascript<br> is a lang<br>uage<br>"

 it should be:

 breakSentence()  // "javascript<br>is a <br>language"

Answer 1

这里有一个解决方案：http://snippets.dzone.com/posts/show/869

//+ Jonas Raoni Soares Silva
//@ http://jsfromhell.com/string/wordwrap [v1.0]

String.prototype.wordWrap = function(m, b, c){
    var i, j, l, s, r;
    if(m < 1)
        return this;
    for(i = -1, l = (r = this.split("\n")).length; ++i < l; r[i] += s)
        for(s = r[i], r[i] = ""; s.length > m; r[i] += s.slice(0, j) + ((s = s.slice(j)).length ? b : ""))
            j = c == 2 || (j = s.slice(0, m + 1).match(/\S*(\s)?$/))[1] ? m : j.input.length - j[0].length
            || c == 1 && m || j.input.length + (j = s.slice(m).match(/^\S*/)).input.length;
    return r.join("\n");
};

用法：

var sentence = "JavaScript is a prototype-based scripting language that is dynamic, weakly typed and has first-class functions."
sentence.wordWrap(20, "<br>",true)
// Output "JavaScript is a <br>prototype-based <br>scripting language <br>that is dynamic, <br>weakly typed and has<br> first-class <br>functions."

Answer 2

我会像这样试一试(未测试)：

var breakSentence = function (sentence, maxCharLen, separator)
{
    var result = "";
    var index = 0;
    while (sentence.length - index > maxCharLen)
    {
        var spaceIndex = sentence.substring(index, index + maxCharLen).lastIndexOf(' ');
        if (spaceIndex < 0)    //no spaces
        {
            alert('Don\'t know what do do with substring with one long word');
            spaceIndex = maxCharLen;    //assume you want to break anyway to avoid infinite loop
        }
        result += sentence.substring(index, index + spaceIndex + 1)  + separator;
        index += spaceIndex + 1;
    }
    return result;
}

现在应该只在空格后中断...

Answer 3

这是我尝试获得它的方法。它有两件事你应该注意：

• 它首先删除所有分隔符实例(因此重新排序是完全新的)
• 它不会将单词拆分超过maxCharLen字符。

它在节点0.6.10中工作

var breakSentence = function (sentence, maxCharLen, separator) {
  var words = []   // array of words
    , i            // iterator
    , len          // loop
    , current = '' // current line
    , lines = []    // lines split
    ;
  sentence = sentence || "javascript is a language";
  maxCharLen = 10 || maxCharLen;
  separator = separator || "<br>";
  sentence = sentence.replace(separator, '');
  if (sentence.length < maxCharLen) return [sentence]; // no need to work if we're already within limits
  words = sentence.split(' ');
  for (i = 0, len = words.length; i < len; i += 1) {
    // lets see how we add the next word. if the current line is blank, just add it and move on.
    if (current == '') {
      current += words[i];
    // if it's not like that, we need to see if the next word fits the current line
    } else if (current.length + words[i].length <= maxCharLen) { // if the equal part takes the space into consideration
      current += ' ' + words[i];
    // if the next word doesn't fit, start on the next line.
    } else {
      lines.push(current);
      current = words[i];
      // have to check if this is the last word
      if (i === len -1) {
        lines.push(current);
      }
    }
  }
  // now assemble the result and return it.
  sentence = '';
  for (i = 0, len = lines.length; i < len; i += 1) {
    sentence += lines[i];
    // add separator if not the last line
    if (i < len -1) {
      sentence += separator;
    }
  }
  return sentence;
}