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Java 将文本文件解析为数字输出_Java_Regex_Parsing_Lex - Fatal编程技术网

Java 将文本文件解析为数字输出

Java 将文本文件解析为数字输出,java,regex,parsing,lex,Java,Regex,Parsing,Lex,我想逐行读取文件并对其进行解析 我的目标是找到这样一条线: Vx=51:52.77牛顿001:13.33瓦 数字可能不同的地方 这是我目前的代码: package kml_gen; import java.io.IOException; import java.nio.file.Path; import java.nio.file.Paths; import java.util.Scanner; import java.util.regex.Matcher; import java.util.

我想逐行读取文件并对其进行解析

我的目标是找到这样一条线:

Vx=51:52.77牛顿001:13.33瓦

数字可能不同的地方

这是我目前的代码:

package kml_gen;

import java.io.IOException;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Kml_gen {

private final Path fFilePath;

public static void main(String... aArgs) throws IOException {
    Kml_gen parser = new Kml_gen("C:\\Users\\DSR\\Documents\\master\\kml_gen\\uk_air_2002.txt");
    parser.processLineByLine();
    log("Done.");
}

/**
 * Constructor.
 *
 * @param aFileName full name of an existing, readable file.
 */
public Kml_gen(String aFileName) {
    fFilePath = Paths.get(aFileName);
}

/**
 * Template method that calls {@link #processLine(String)}.
 *
 * @throws java.io.IOException
 */
public final void processLineByLine() throws IOException {
    try (Scanner scanner = new Scanner(fFilePath)) {
        while (scanner.hasNextLine()) {
            processLine(scanner.nextLine());
        }
    }
}

/*
 * @param aLine
 */
protected void processLine(String aLine) {
    //use a second Scanner to parse the content of each line
    //regex [V]\\s*[X][=]\\d+\\:\\d+\\.\\d+\\s*[N]\\s*\\d+\\:\\d+\\.\\d+\\s*[W]   
    String patternstr = "[V]\\s[X][=]\\d*\\:\\d*\\.\\d*\\s[N]\\s\\d*\\:\\d*\\.\\d*\\s[W]";
    Pattern pattern = Pattern.compile(patternstr);
    Matcher matcher = pattern.matcher(aLine);
    boolean matches = matcher.matches();
    if (matcher.find()) {
        /* example of text to be parsed: 
         V X=51:52.77 N 001:13.33 W */
        if (matches) {
            System.out.println("V X=" + matcher.group() + ":" + matcher.group() + "." + matcher.group() + " N "
                    + matcher.group() + ":" + matcher.group() + "." + matcher.group() + " W");
        }
    } else {
        log("Empty or invalid line. Unable to process.");
    }
}

private static void log(Object aObject) {
    System.out.println(String.valueOf(aObject));
}
因为我的输出总是“空或无效行。无法处理。”。我真的很感激任何帮助,谢谢你

编辑:这里有一些要处理的文本,我只考虑以“vx=…”开头的行

交流Q

帕拉韦斯顿酒店

美国证监会

AH FL120

Vx=51:52.77牛顿001:13.33瓦


DC 1.5

尝试执行
aLine=aLine.trim()
'\n'
字符可能会弄乱您的正则表达式。您是否有一些示例输出—不仅是您要查找的内容,而且还有来自实际文件的行—两行都与正则表达式匹配,还有一些行与正则表达式不匹配?@budi在
processLine
的开头,我应该在哪里插入命令。
//.matches() and .find() can't be used twice,should remove .matches()
String patternstr = "V\\sX=(\\d+):(\\d+)\\.(\\d+)\\sN\\s(\\d+):(\\d+)\\.(\\d+)\\sW";
//matcher.group();  match all the line
System.out.println("V X=" + matcher.group(1) + ":" + matcher.group(2) 
                    + "." + matcher.group(3) + " N " + matcher.group(4) + ":" 
                          + matcher.group(5) + "." + matcher.group(6) + " W");