Java : Pattern matcher returns new lines unexpectedly

Question

我有一个用例，我必须处理任何转义/非转义字符作为分隔符来拆分句子。到目前为止，我们拥有的未转义/转义字符是:

" " (space),"\\t","|", "\\|",";","\\;","," etc

到目前为止，它使用正则表达式，定义为:

String delimiter = " ";
String regex = "(?:\\\\.|[^"+ delimiter +"\\\\]++)*";

输入的字符串是:

String input = "234|Tamarind|something interesting ";

现在，下面是拆分和打印的代码:

 List<String> matchList = new ArrayList<>(  );
 Matcher regexMatcher = pattern.matcher( input );
 while ( regexMatcher.find() )
 {
     matchList.add( regexMatcher.group() );
 }

 System.out.println( "Unescaped/escaped test result with size: " + matchList.size() );
 matchList.stream().forEach( System.out::println );

但是，意外地存储了额外的字符串(新行)。所以输出看起来像:

Unescaped/escaped test result with size: 5
234|Tamarind|something

interesting

.

有没有更好的方法来做到这一点，这样就不会有任何额外的字符串？

Answer 1

这很简单:确保至少匹配一个字符。这意味着您可以删除 ++ 量词并将 * 替换为 +。查看regex demo .

完整 Java demo :

String delimiter = " ";
String regex = "(?:\\\\.|[^"+ delimiter +"\\\\])+";
// System.out.println(regex); // => (?:\\.|[^ \\])+
Pattern pattern = Pattern.compile(regex, Pattern.DOTALL);
String input = "234|Tamarind|something interesting ";
List<String> matchList = new ArrayList<>(  );
Matcher regexMatcher = pattern.matcher( input );
while ( regexMatcher.find() )
{
    // System.out.println("'"+regexMatcher.group()+"'");
    matchList.add( regexMatcher.group() );
}

System.out.println( "Unescaped/escaped test result with size: " + matchList.size() );
matchList.stream().forEach( System.out::println );

输出:

Unescaped/escaped test result with size: 2
234|Tamarind|something
interesting