无法从文件读取数据

Question

我正在尝试从com.example软件包中的CSV文件读取值。 但是当我使用以下语法运行代码时：

DataModel model = new FileDataModel(new File("Dataset.csv"));

它说：

java.io.FileNotFoundException：Dataset.csv

我也尝试过使用：

DataModel model = new FileDataModel(new File("/com/example/Dataset.csv"));

还是行不通。 任何帮助都会有所帮助。 谢谢。

Answer 1

public class ReadCVS {

  public static void main(String[] args) {

    ReadCVS obj = new ReadCVS();
    obj.run();

  }

  public void run() {

    String csvFile = "file path of csv";
    BufferedReader br = null;
    String line = "";
    String cvsSplitBy = ",";

    try {

        br = new BufferedReader(new FileReader(csvFile));
        while ((line = br.readLine()) != null) {

                // Do stuff here

        }

    } catch (FileNotFoundException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        if (br != null) {
            try {
                br.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }

    System.out.println("Done");
  }

}

Answer 2

包com.example中存在的CSV文件

您可以使用getResource()或getResourceAsStream()从包中访问资源。 例如

InputStream is = getClass().getResourceAsStream("/com/example/Dataset.csv");//uses absolute (package root) path
BufferedReader br = new BufferedReader(new InputStreamReader(is));
//read from BufferedReader

（请注意，为简洁起见，上面省略了异常处理和文件关闭）

Answer 3

如果这是来自org.apache.mahout.cf.taste.impl.model.file的FileDataModel ，则它不能接受输入流，而只需要一个文件。 问题是您不能认为该文件对您来说很容易使用（请参阅此问题的答案 ）。

最好读取文件的内容并将其保存到临时文件，然后将该临时文件传递给FileDataModel 。

InputStream initStream = getClass().getClasLoader().getResourceAsStream("Dataset.csv");
//simplistic approach is to put all the contents of the file stream into memory at once
//  but it would be smarter to buffer and do it in chunks
byte[] buffer = new byte[initStream.available()];
initStream.read(buffer);

//now save the file contents in memory to a temporary file on the disk
//choose your own temporary location - this one is typical for linux
String tempFilePath = "/tmp/Dataset.csv";  
File tempFile = new File(tempFilePath);
OutputStream outStream = new FileOutputStream(tempFile);
outStream.write(buffer);

DataModel model = new FileDataModel(new File(tempFilePath));
...