Bri*_*ian 5 java regex string parsing java.util.scanner
想要解析以下文本文件:
示例文本文件:
<2008-10-07>text entered by user<Ted Parlor><2008-11-26>additional text entered by user<Ted Parlor>
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我想解析上面的文本,以便我可以有三个变量:
v1 = 2008-10-07
v2 = text entered by user
v3 = Ted Parlor
v1 = 2008-11-26
v2 = additional text entered by user
v3 = Ted Parlor
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我试图使用扫描仪和useDelimiter,但是,我有问题如何设置它以获得如上所述的结果.这是我的第一次尝试:
import java.io.*;
import java.util.Scanner;
public class ScanNotes {
public static void main(String[] args) throws IOException {
Scanner s = null;
try {
//String regex = "(?<=\\<)([^\\>>*)(?=\\>)";
s = new Scanner(new BufferedReader(new FileReader("cur_notes.txt")));
s.useDelimiter("[<]+");
while (s.hasNext()) {
String v1 = s.next();
String v2= s.next();
System.out.println("v1= " + v1 + " v2=" + v2);
}
} finally {
if (s != null) {
s.close();
}
}
}
}
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结果如下:
v1= 2008-10-07>text entered by user v2=Ted Parlor>
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我想要的是:
v1= 2008-10-07 v2=text entered by user v3=Ted Parlor
v1= 2008-11-26 v2=additional text entered by user v3=Ted Parlor
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任何可以让我分别提取所有三个字符串的帮助将不胜感激.
您可以使用\s*[<>]\s*分隔符.也就是说,任何<或>任何先前和后续的空格.
为此,除了用于标记输入中的日期和用户字段的输入(即消息中没有)之外,不得有任何输入<或>输入I <3 U!!.
此分隔符允许条目中的空字符串部分,但它也在任意两个条目之间留下空字符串标记,因此必须手动丢弃它们.
import java.util.Scanner;
public class UseDelim {
public static void main(String[] args) {
String content = " <2008-10-07>text entered by user <Ted Parlor>"
+ " <2008-11-26> additional text entered by user <Ted Parlor>"
+ " <2008-11-28><Parlor Ted> ";
Scanner sc = new Scanner(content).useDelimiter("\\s*[<>]\\s*");
while (sc.hasNext()) {
System.out.printf("[%s|%s|%s]%n",
sc.next(), sc.next(), sc.next());
// if there's a next entry, discard the empty string token
if (sc.hasNext()) sc.next();
}
}
}
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这打印:
[2008-10-07|text entered by user|Ted Parlor]
[2008-11-26|additional text entered by user|Ted Parlor]
[2008-11-28||Parlor Ted]
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