使用正则表达式提取ISBN [英] Extract an ISBN with regex

问题描述

我有一个非常长的字符串，我想将其解析为出现在子字符串"ISBN"之后的数值.但是，这组13位数字可以通过-"字符进行不同的排列.示例:(这些都是有效的ISBN) 123-456-789-123-4 ，或 1-2-3-4-5-67891234 ，或12-34-56-78-91-23-4 .本质上，我想在潜在的ISBN上使用正则表达式模式匹配器，以查看是否存在有效的13位ISBN.如何忽略"-"字符，以便仅对 \ d {13} 模式进行正则表达式?我的功能:

I have an extremely long string that I want to parse for a numeric value that occurs after the substring "ISBN". However, this grouping of 13 digits can be arranged differently via the "-" character. Examples: (these are all valid ISBNs) 123-456-789-123-4, OR 1-2-3-4-5-67891234, OR 12-34-56-78-91-23-4. Essentially, I want to use a regex pattern matcher on the potential ISBN to see if there is a valid 13 digit ISBN. How do I 'ignore' the "-" character so I can just regex for a \d{13} pattern? My function:

public String parseISBN (String sourceCode) {
  int location = sourceCode.indexOf("ISBN") + 5;
  String ISBN = sourceCode.substring(location); //substring after "ISBN" occurs
  int i = 0;
  while ( ISBN.charAt(i) != ' ' )
    i++;
  ISBN = ISBN.substring(0, i); //should contain potential ISBN value
  Pattern pattern = Pattern.compile("\\d{13}"); //this clearly will find 13 consecutive numbers, but I need it to ignore the "-" character
  Matcher matcher = pattern.matcher(ISBN); 
  if (matcher.find()) return ISBN;
  else return null;
}

推荐答案

• 替代方法1:

  • Alternative 1:

    pattern.matcher(ISBN.replace("-", ""))
    

  • 替代方法2:类似的东西

  • Alternative 2: Something like

    Pattern.compile("(\\d-?){13}")
    

  第二个替代项的演示

  String ISBN = "ISBN: 123-456-789-112-3, ISBN: 1234567891123";
  
  Pattern pattern = Pattern.compile("(\\d-?){13}");
  Matcher matcher = pattern.matcher(ISBN);
  
  while (matcher.find())
      System.out.println(matcher.group());
  

  输出:

  123-456-789-112-3
  1234567891123
  

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