Question

我需要从Spintax中提取最长的结果。但是我找不到添加非旋转文本的方法。当有选择可用时，我的代码将提取最长的选项，但它忽略了spintax大括号之外的内容。

var totalcount = 0;
var text = "{This is|Here} a {sample|demo} of a sentence {made with|created using} spintax";

var longtext = "";
var matches, options, random;
var regEx = new RegExp(/{([^{}]+?)}/);

while((matches = regEx.exec(text)) !== null) {  

  options = matches[1].split("|");
  var long1 = matches[1].split("|");
  var longest = long1[0];
  var strcount = 0;

  for (i = 0; i < long1.length; i++) {
    if (long1[i].length > longest.length) {
        longest = long1[i];
        }
       strcount = longest.length;
     }
      totalcount = totalcount + strcount;
      longtext = longtext + ' ' + longest;


  random = Math.floor(Math.random() * options.length);
  text = text.replace(matches[0], options[random]);
}

document.write('Random spin:<br>'+text+'<br><br>');
document.write('Longest spin ('+strcount+' chars):<br>'+longtext);

我用来计算最长选项的循环仅计算了spintax的变化。因此它跳过了“ a”，“ of a句子”和“ spintax”

我该如何添加它们，使它们也算在内？

预先感谢...，对于代码混乱，我们深表歉意！

Answer 1

Facebook上的一位朋友为我回答了这个问题：)因此，对于任何有相同问题的人，他的解决方案都是：

<script>
var matches, options, random;

var spintaxString = "{London|Birmingham|Ely} is a {nice|wonderful|fabulous} place with some great {cafes | bars | restaurants}";
var longestString = spintaxString;
var regEx = new RegExp(/{([^{}]+?)}/);


while((matches = regEx.exec(spintaxString)) !== null) {
    options = matches[1].split("|");
    random = Math.floor(Math.random() * options.length);

    var longest = options[0];
    for (i = 0; i < options.length; i++) {
    if (options[i].length > longest.length) {
        longest = options[i];
        }
     }

    longestString = longestString.replace(matches[0], longest);
    spintaxString = spintaxString.replace(matches[0], options[random]);
}

document.write('Random: '+spintaxString+' ('+spintaxString.length+' chars)<br>');
document.write('Longest: '+longestString+'  ('+longestString.length+' chars)<br>');

</script>

如何在字符串长度计数中包括非固定码内容？

1 个答案: