Question

我有以下数据。我想做的是将每个读数分成不同的输出，但是不起作用。它仅显示“ null”。我希望工作的是：

输出： C.txt
1 1000 1000
2 2000 2000

输出：B.txt
1 2 90.000 2
2 3 180.000 2

输出：D.txt
1 2 100.1 0.038
2 3 200.1 0.038

Input.txt中的数据：
C; 1; 1000; 1000
C; 2; 2000; 2000

B; 1; 2; 90.00; 2
B; 2; 3; 180.00; 2

D; 1; 2; 100.1; 0.038
D; 2; 3; 200.1; 0.038

import java.io.*;
import java.util.StringTokenizer;

public class ReadFile {

  public static void main(String[] args) {
    BufferedReader input = null; //read
    PrintWriter outC = null; //write output
    PrintWriter outB = null;
    PrintWriter outD = null;
    try {
      input = new BufferedReader(new FileReader("C:\\Users\\PC\\Desktop\\FYP\\Input.txt"));
      outC = new PrintWriter(new BufferedWriter(new FileWriter("C:\\Users\\PC\\Desktop\\FYP_Test\\C.txt")));
      outB = new PrintWriter(new BufferedWriter(new FileWriter("C:\\Users\\PC\\Desktop\\FYP_Test\\B.txt")));
      outD = new PrintWriter(new BufferedWriter(new FileWriter("C:\\Users\\PC\\Desktop\\FYP_Test\\D.txt")));

      String inputData = null;
      int C = 0;
      int B = 0;
      int D = 0;
      while ((inputData = input.readLine()) != null) {
        StringTokenizer tokenizer = new StringTokenizer(inputData, ";");
        String id = tokenizer.nextToken();
        String StnFrom = tokenizer.nextToken();
        String NorthingTo = tokenizer.nextToken();
        String EastingDistBrg = tokenizer.nextToken();
        String StdError = tokenizer.nextToken();

        if (id.equalsIgnoreCase("C")) {
          C++;
          outC.println(StnFrom + " " + NorthingTo + " " + EastingDistBrg);
        } else if (id.equalsIgnoreCase("B")) {
          B++;
          outB.println(StnFrom + " " + NorthingTo + " " + EastingDistBrg + " " + StdError);
        } else if (id.equalsIgnoreCase("D")) {
          D++;
          outB.println(StnFrom + " " + NorthingTo + " " + EastingDistBrg + " " + StdError);
        }
      }
      input.close();
      outC.close();
      outB.close();
      outD.close();
    } catch (FileNotFoundException fe) {
      System.out.println(fe.getMessage());
    } catch (IOException iox) {
      System.out.println(iox.getMessage());
    } catch (Exception e) {
      System.out.println(e.getMessage());
    }
  }
}

Answer 1

tokenizer.nextToken()将在令牌生成器的字符串中没有令牌时抛出NoSuchElementException。

您的示例输入（如果提供）将抛出“ NoSuchElementException”，因为“ C”的“ Input.txt”中的数据错误。在您的程序中，您要调用“ nextToken”五次，而“ C”的数据仅包含4个值（C; 1; 1000; 1000）。

下面是改进的“输入”数据。

C;1;1000;1000;1
C;2;2000;2000;1

B;1;2;90.00;2
B;2;3;180.00;2

D;1;2;100.1;0.038
D;2;3;200.1;0.038

此外，您需要改进while循环以读取空行。当前，它将引发错误。考虑下面的while循环：

while ((inputData = input.readLine()) != null) {
        if(inputData.length() != 0) {
          StringTokenizer tokenizer = new StringTokenizer(inputData, ";");
          String id = tokenizer.nextToken();
          String StnFrom = tokenizer.nextToken();
          String NorthingTo = tokenizer.nextToken();
          String EastingDistBrg = tokenizer.nextToken();
          String StdError = tokenizer.nextToken();

          if (id.equalsIgnoreCase("C")) {
            C++;
            outC.println(StnFrom + " " + NorthingTo + " " + EastingDistBrg);
          } else if (id.equalsIgnoreCase("B")) {
            B++;
            outB.println(StnFrom + " " + NorthingTo + " " + EastingDistBrg + " " + StdError);
          } else if (id.equalsIgnoreCase("D")) {
            D++;
            outD.println(StnFrom + " " + NorthingTo + " " + EastingDistBrg + " " + StdError);
          }
        }
      }

读取文件并提取到不同的输出

1 个答案: