Java 解析.txt文件

Question

我试图在 Netbeans IDE 8.0.2 中运行以下文件 TemplateMaker.java，但遇到了以下错误消息。 Netbeans 没有显示任何红色指示器供我修复。请帮忙。

Exception in thread "main" java.util.NoSuchElementException
    at java.util.Scanner.throwFor(Scanner.java:907)
    at java.util.Scanner.next(Scanner.java:1416)
    at templatemaker.TemplateMaker.processLine(TemplateMaker.java:48)
    at templatemaker.TemplateMaker.processLineByLine(TemplateMaker.java:35)
    at templatemaker.TemplateMaker.main(TemplateMaker.java:17)
Java Result: 1

这是我的源代码：

package templatemaker;


import java.io.IOException;
import java.nio.charset.Charset;
import java.nio.charset.StandardCharsets;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.util.Scanner;


public class TemplateMaker {

        public static void main(String [] args)
 throws IOException {
    TemplateMaker parser = new TemplateMaker("Book1.txt");
    parser.processLineByLine();
    log("Done.");
  }

  /**
   Constructor.
   @param aFileName full name of an existing, readable file.
  */
  public TemplateMaker(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, ENCODING.name())){
      while (scanner.hasNextLine()){
        processLine(scanner.nextLine());
      }      
    }
  }


  protected void processLine(String aLine){
    //use a second Scanner to parse the content of each line 
    Scanner scanner = new Scanner(aLine);
    scanner.useDelimiter("=");
    if (scanner.hasNext()){
      //assumes the line has a certain structure
      String name = scanner.next();
      String value = scanner.next();
      log("Name is : " + quote(name.trim()) + ", and Value is : " + quote(value.trim()));
    }
    else {
      log("Empty or invalid line. Unable to process.");
    }
  }

  // PRIVATE 
  private final Path fFilePath;
  private final static Charset ENCODING = StandardCharsets.UTF_8;  

  private static void log(Object aObject){
    System.out.println(String.valueOf(aObject));
  }

  private String quote(String aText){
    String QUOTE = "'";
    return QUOTE + aText + QUOTE;
  }
}

Answer 1

如堆栈跟踪所示，调用 scanner.next() 时抛出异常

at java.util.Scanner.next(Scanner.java:1416)

if (scanner.hasNext()){
      //assumes the line has a certain structure
      String name = scanner.next(); // checked by hasNext()
      String value = scanner.next(); // not checked by hasNext()
      log("Name is : " + quote(name.trim()) + ", and Value is : " + quote(value.trim()));
}

错误必须在第二个.next() 中。您检查扫描器是否有下一个标记，但您调用了 .next() 两次。所以我假设还剩下 1 个令牌，你读了两次。同样来自 API next() 方法：

Throws:
    NoSuchElementException - if no more tokens are available
    IllegalStateException - if this scanner is closed

您可以通过添加 System.out.println 语句轻松检查并检查异常前的最后一个语句是什么（在第一次或第二次调用 next() 之后）

Answer 2

您的 processLine() 期待一对 "name=value"。正如 MightyPork 所说，您正在检查 hasNext() 一次，然后读取两次。因此，如果该行没有 = 符号，这将中断，因为扫描器不会获得 next() 标记。您应该添加两个 hasNext() 检查。理想情况下，您在这里不需要扫描仪。由于您总是期望由 = 分隔的两个标记，因此您可以简单地依赖 java.util.StringTokenizer 作为

protected void processLine(String aLine){
    StringTokenizer st = new StringTokenizer(aLine, "=");
    if(st.countTokens() == 2) {
        log("Name is : " + quote(st.nextToken().trim()) + ", and Value is : " + quote(st.nextToken().trim()));
    } else {
        log("Empty or invalid line. Unable to process.");
    }
}

Java 解析.txt文件

Java parse .txt file

java

java.util.scanner