Question

我有一个这样的字符串：

import multiprocessing as mp

def func(val):
    print "\nInside Function.....\n"
    return val*val

if __name__ == '__main__':
    cpu_count = mp.cpu_count()
    pool = mp.Pool(processes = cpu_count)

    results = []
    num = 1
    while cpu_count >= 1:
        results.append(pool.apply_async(func, (num,)))
        cpu_count = cpu_count - 1
        num = num + 1

    output = [p.get() for p in results]
    print output

    pool.close()
    pool.join()

现在我想要这个：

var str = "this is [link1][1]
           this is [link2][2]

           this is [link3][3]
           this is [link4][4]

           [1]: http://example1.com

           [2]: http://example2.com
           [3]: http://example1.com


           [4]: http://example4.com";

正如您在上面的示例中所看到的，有两件事：

在所有这些网址（仅网址）

var str = "this is [link1][1]
           this is [link2][2]

           this is [link3][1]
           this is [link4][4]

           [1]: http://example1.com
           [2]: http://example2.com
           [4]: http://example4.com";

删除重复的网址并替换引用号码。（换句话说，在上面的示例中，因为第一个和第三个网址相同，然后移除\n并将[3]: http://example1.com替换为{{1} }）

有一个正则表达式将[3]及其[1]分为两组：

[any digit]:

在url中，正则表达式匹配：

/(\[[0-9]*]:)\s*(.*)\r?\n?/gm

另外还有另一个正则表达式可以删除仅链接之间的所有浪费str：

//group1
$1: [1]:
    [2]:
    [3]:
    [4]:

//group2
$2: http://example1.com
    http://example2.com
    http://example1.com
    http://example4.com

嗯，我怎么能这样做？

Answer 1

这可能是解决问题的方法：

function removeMultipleMatkdownLinks(markownString) {
  var seperateLinks = /(\[[0-9]*]):\s*(.*)\r?\n?/gm;
  var removeNewLines = /(^\[[0-9]*]:.*)\n+/gm; 

  var result;
  var urls = [], ids = [];
  var formattedString = str;
  while ((result = seperateLinks.exec(str)) !== null) {
      if (result.index === seperateLinks.lastIndex) {
          result.lastIndex++;
      }
      //check if link already exists
      var index = urls.indexOf(result[2]);
      if(index < 0) {
        urls.push(result[2]);
        ids.push(result[1]);

      } else { //remove links and replace ids
        var removeLink = new RegExp("(\\" + result[1] + ":.*\\r?\\n?)", "gm"); ///(\[1\]:.*\n)/gm
        var changeNumber = new RegExp("(\\" + result[1] + ")", "gm");
        formattedString = formattedString
          .replace(removeLink, "")
          .replace(changeNumber, ids[index]);
      }
  }

  return formattedString.replace(removeNewLines, "$1\n");
}

JSFiddle

如何为字符串中的所有相同URL定义相同的数字？

1 个答案: