我有一个用例,我必须将任何转义/未转义的字符作为分隔符来分隔句子。到目前为止,我们拥有的未跳过/转义字符是:
" " (space),"\t","|", "\|",";","\;","," etc
到目前为止,它使用的是正则表达式,定义为:
String delimiter = " ";
String regex = "(?:\\.|[^"+ delimiter +"\\]++)*";
输入字符串为:
String input = "234|Tamarind|something interesting ";
现在,下面是分割和打印的代码:
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,还有一些额外的字符串(新行(被意外存储。因此输出看起来像:
Unescaped/escaped test result with size: 5
234|Tamarind|something
interesting
.
有没有更好的方法可以做到这一点,这样就不会有任何额外的字符串?
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+
。请参阅regex演示。
完整的Java演示:
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