Java 将文本文件解析为数字输出
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: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.
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");