5 回答

• 2

  我用这个正则表达式：

  ^(.*?)(?: (AND) (.*))?$
  

  explanation:

  The regular expression:
  
  (?-imsx:^(.*?)(?: (AND) (.*))?$)
  
  matches as follows:
  
  NODE                     EXPLANATION
  ----------------------------------------------------------------------
  (?-imsx:                 group, but do not capture (case-sensitive)
                           (with ^ and $ matching normally) (with . not
                           matching \n) (matching whitespace and #
                           normally):
  ----------------------------------------------------------------------
    ^                        the beginning of the string
  ----------------------------------------------------------------------
    (                        group and capture to \1:
  ----------------------------------------------------------------------
      .*?                      any character except \n (0 or more times
                               (matching the least amount possible))
  ----------------------------------------------------------------------
    )                        end of \1
  ----------------------------------------------------------------------
    (?:                      group, but do not capture (optional
                             (matching the most amount possible)):
  ----------------------------------------------------------------------
                               ' '
  ----------------------------------------------------------------------
      (                        group and capture to \2:
  ----------------------------------------------------------------------
        AND                      'AND'
  ----------------------------------------------------------------------
      )                        end of \2
  ----------------------------------------------------------------------
                               ' '
  ----------------------------------------------------------------------
      (                        group and capture to \3:
  ----------------------------------------------------------------------
        .*                       any character except \n (0 or more
                                 times (matching the most amount
                                 possible))
  ----------------------------------------------------------------------
      )                        end of \3
  ----------------------------------------------------------------------
    )?                       end of grouping
  ----------------------------------------------------------------------
    $                        before an optional \n, and the end of the
                             string
  ----------------------------------------------------------------------
  )                        end of grouping
  ----------------------------------------------------------------------
  

  回复于 2024-04-28T11:34:13+08:00
• 0

  如何使用

  String split[] = sentence.split("AND");
  

  这将用你的词分开句子，并给你一个子部分列表 .

  回复于 2024-04-28T11:34:13+08:00
• 2

  说明

  此正则表达式将请求的字符串部分返回到请求的组 . and 是可选的，如果在字符串中找不到，那么整个字符串将被放入组1.所有 \s*? 强制捕获的组自动修剪其空白区域 .

  ^\s*?\b(.*?)\b\s*?(?:\b(and)\b\s*?\b(.*?)\b\s*?)?$

  

  团体

  0获取整个匹配字符串

  • 在分隔单词 and 之前获取字符串，如果没有 and 则整个字符串出现在此处

  • 获取分隔词，在这种情况下它是 and

  • 获取字符串的第二部分

  Java代码示例：

  情况1

  import java.util.regex.Pattern;
  import java.util.regex.Matcher;
  class Module1{
    public static void main(String[] asd){
    String sourcestring = "foo sentence A AND foo sentence B";
    Pattern re = Pattern.compile("^\\s*?\\b(.*?)\\b\\s*?(?:\\b(and)\\b\\s*?\\b(.*?)\\b\\s*?)?$",Pattern.CASE_INSENSITIVE);
    Matcher m = re.matcher(sourcestring);
      if(m.find()){
        for( int groupIdx = 0; groupIdx < m.groupCount()+1; groupIdx++ ){
          System.out.println( "[" + groupIdx + "] = " + m.group(groupIdx));
        }
      }
    }
  }
  
  $matches Array:
  (
      [0] => foo sentence A AND foo sentence B
      [1] => foo sentence A
      [2] => AND
      [3] =>  foo sentence B
  )
  

  案例2，使用相同的正则表达式

  import java.util.regex.Pattern;
  import java.util.regex.Matcher;
  class Module1{
    public static void main(String[] asd){
    String sourcestring = "foo sentence A";
    Pattern re = Pattern.compile("^\\s*?\\b(.*?)\\b\\s*?(?:\\b(and)\\b\\s*?\\b(.*?)\\b\\s*?)?$",Pattern.CASE_INSENSITIVE);
    Matcher m = re.matcher(sourcestring);
      if(m.find()){
        for( int groupIdx = 0; groupIdx < m.groupCount()+1; groupIdx++ ){
          System.out.println( "[" + groupIdx + "] = " + m.group(groupIdx));
        }
      }
    }
  }
  
  $matches Array:
  (
      [0] => foo sentence A
      [1] => foo sentence A
  )
  

  回复于 2024-04-28T11:34:13+08:00
• 0

  你的情况2有点奇怪......

  但我会这样做

  String[] parts = sentence.split("(?<=AND)|(?=AND)"));
  

  你检查 parts.length . 如果length == 1，则为case2 . 你只是在数组中有句子，你可以添加空字符串作为"group2/3"

  如果在case1中你直接 parts ：

  [foo sentence A , AND,  foo sentence B]
  

  回复于 2024-04-28T11:34:13+08:00
• 2

  更改你的正则表达式，以便在他的第一句可选后创建空格：

  (.*\\S) ?(AND (.*))?$
  

  或者您可以使用 split() 来使用 AND 以及任何周围空格：

  String sentences = sentence.spli("\\s*AND\\s*");
  

  回复于 2024-04-28T11:34:13+08:00