Scanner scanner = new Scanner(new File(pathf2));
scanner.useDelimiter("########");
String pattern2="^[-a-z0-9R:._-`&=*'`~\"\\+[\\s]]+[\\(]";
Pattern r2 = Pattern.compile(pattern2);
Matcher m2 = r2.matcher(line);
while (m2.find())
{
rel = m2.group();
rel = rel.substring(0, rel.length()-1).trim();
System.out.println("The relation are " + rel);
}
这是我的输入文件....我需要一个正则表达式,在块的每个句子中匹配3个字母..上面的编码只返回“agt”,它匹配块中的第一行。然后它返回“ obj“是第二个块的开始。我需要一个块中每行3个字符....如果我在正则表达式中删除”^“它也会在句子中取第二个单词....请帮助
agt(eat(icl>consume>do,agt>living_thing,obj>concrete_thing,ins>thing).@entry.@present,ram(icl>volatile_storage>thing,equ>random-access_memory))
obj(eat(icl>consume>do,agt>living_thing,obj>concrete_thing,ins>thing).@entry.@present,rice(icl>grain>thing))
########
obj(clear(icl>remove>do,plf>thing,obj>thing,ins>thing).@entry.@past,bloodsworth.@topic)
man(clear(icl>remove>do,plf>thing,obj>thing,ins>thing).@entry.@past,ultimately(icl>how,com>ultimate))
ins(clear(icl>remove>do,plf>thing,obj>thing,ins>thing).@entry.@past,evidence(icl>indication>thing))
obj(gather(icl>do,equ>accumulate,plf>thing,agt>thing,obj>thing,plc>thing).@state,evidence(icl>indication>thing))
plf(gather(icl>do,equ>accumulate,plf>thing,agt>thing,obj>thing,plc>thing).@state,from)
mod(stain(icl>appearance>thing).@indef,semen(icl>liquid_body_substance>thing))
obj(from,stain(icl>appearance>thing).@indef)
plc(gather(icl>do,equ>accumulate,plf>thing,agt>thing,obj>thing,plc>thing).@state,panties.@pl)
pos(panties.@pl,victim(icl>unfortunate>thing).@def)
########
答案 0 :(得分:0)
我认为你需要循环才能获得每一行直到达到EOF
尝试以下代码
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class test {
public static void main(String[] args) throws FileNotFoundException {
Scanner scanner = new Scanner(new File(pathf2));
scanner.useDelimiter("########");
String pattern2 = "^[-a-z0-9R:._-`&=*'`~\"\\+[\\s]]+[\\(]";
while (scanner.hasNextLine()) {
String line = scanner.nextLine();
Pattern r2 = Pattern.compile(pattern2);
Matcher m2 = r2.matcher(line);
while (m2.find()) {
String rel = m2.group();
rel = rel.substring(0, rel.length() - 1).trim();
System.out.println("The relation are " + rel);
}
}
scanner.close();
}
}
答案 1 :(得分:0)
您可以将每个块拆分成行并分别检查每一行:
Scanner scanner = new Scanner(new File("/tmp/x"));
scanner.useDelimiter("########");
String pattern2="(^[a-z]{3})";
Pattern r2 = Pattern.compile(pattern2);
while (scanner.hasNext()) {
System.out.println( "### NEXT BLOCK ###" );
String block = scanner.next();
String [] lines = block.split("\n");
for(String line: lines ) {
Matcher m2 = r2.matcher(line);
if( m2.find() ) {
System.out.println( "\t" + m2.group() );
}
}
}
scanner.close();
答案 2 :(得分:0)
将模式更改为:
String pattern2="(?m)^[-a-z0-9R:._-`&=*'`~\"\\+[\\s]]+[\\(]";