java将大文件拆分为较小的文件，同时拆分多行记录，而不会中断处于不完整状态的记录

Question

我在一个文件中将一条记录拆分为多行。识别记录结尾的唯一方法是当新记录以ABC开头时。下面是示例。文件大小可以是5-10 GB，我正在寻找一个有效的java逻辑只拆分文件(不需要读取每一行)，但拆分逻辑应该检查开始一个新的文件与新的记录，这应该以"ABC“在这种情况下。

添加了更多的细节，我只是在寻找拆分文件，同时拆分最后一条记录应该在一个文件中正确结束。

有没有人能提个建议？

HDR
ABCline1goesonforrecord1   //first record 
line2goesonForRecord1      
line3goesonForRecord1          
line4goesonForRecord1
ABCline2goesOnForRecord2  //second record
line2goesonForRecord2
line3goesonForRecord2
line4goesonForRecord2
line5goesonForRecord2
ABCline2goesOnForRecord3     //third record
line2goesonForRecord3
line3goesonForRecord3
line4goesonForRecord3
TRL

Answer 1

所以，这就是你需要的代码。我在一个10 to的文件上进行了测试，拆分文件需要64秒

import java.io.BufferedWriter;
import java.io.IOException;
import java.io.UncheckedIOException;
import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.StandardOpenOption;
import java.util.concurrent.TimeUnit;

public class FileSplitter {

    private final Path filePath;
    private BufferedWriter writer;
    private int fileCounter = 1;

    public static void main(String[] args) throws Exception {
        long startTime = System.nanoTime();
        new FileSplitter(Path.of("/tmp/bigfile.txt")).split();
        System.out.println("Time to split " + TimeUnit.NANOSECONDS.toSeconds(System.nanoTime() - startTime));
    }

    private static void generateBigFile() throws Exception {
        var writer = Files.newBufferedWriter(Path.of("/tmp/bigfile.txt"), StandardOpenOption.CREATE, StandardOpenOption.TRUNCATE_EXISTING);
        for (int i = 0; i < 100_000; i++) {
            writer.write(String.format("ABCline1goesonforrecord%d\n", i + 1));
            for (int j = 0; j < 10_000; j++) {
                writer.write(String.format("line%dgoesonForRecord%d\n", j + 2, i + 1));
            }
        }

        writer.flush();
        writer.close();
    }

    public FileSplitter(Path filePath) {
        this.filePath = filePath;
    }

    void split() throws IOException {
        try (var stream = Files.lines(filePath, StandardCharsets.UTF_8)) {
            stream.forEach(line -> {
                if (line.startsWith("ABC")) {
                    closeWriter();
                    openWriter();
                }
                writeLine(line);
            });
        }
        closeWriter();
    }

    private void writeLine(String line) {
        if (writer != null) {
            try {
                writer.write(line);
                writer.write("\n");
            } catch (IOException e) {
                throw new UncheckedIOException("Failed to write line to file part", e);
            }
        }
    }

    private void openWriter() {
        if (this.writer == null) {
            var filePartName = filePath.getFileName().toString().replace(".", "_part" + fileCounter + ".");
            try {
                writer = Files.newBufferedWriter(Path.of("/tmp/split", filePartName), StandardOpenOption.CREATE, StandardOpenOption.TRUNCATE_EXISTING);
            } catch (IOException e) {
                throw new UncheckedIOException("Failed to write line to file", e);
            }
            fileCounter++;
        }
    }

    private void closeWriter() {
        if (writer != null) {
            try {
                writer.flush();
                writer.close();
                writer = null;
            } catch (IOException e) {
                throw new UncheckedIOException("Failed to close writer", e);
            }
        }
    }
}

顺便说一句，扫描仪的解决方案也是可行的。

关于没有读完所有的行，我不明白你为什么不想要这个。如果您选择不读取所有行(这是可能的)，那么，第一，您将使解决方案过于复杂，第二，我非常确定您将失去性能，因为您必须将逻辑合并到拆分中。

Answer 2

我没有测试它，但是像这样的东西应该可以工作，你不是一次只读一行内存中的整个文件，所以它应该不是坏事。

public void spiltRecords(String filename) {
        /*
            HDR
            ABCline1goesonforrecord1   //first record
            line2goesonForRecord1
            line3goesonForRecord1
            line4goesonForRecord1
            ABCline2goesOnForRecord2  //second record
            line2goesonForRecord2
            line3goesonForRecord2
            line4goesonForRecord2
            line5goesonForRecord2
            ABCline2goesOnForRecord3     //third record
            line2goesonForRecord3
            line3goesonForRecord3
            line4goesonForRecord3
            TRL
         */
        try {
            Scanner scanFile = new Scanner(new File(filename));
            // now you do not want to edit the existing file in case things go wrong. one way is to get list of index
            // where a new record starts.
            LinkedList<Long> startOfRecordIndexes = new LinkedList<>();
            long index = 0;
            while (scanFile.hasNext()) {
                if (scanFile.nextLine().startsWith("ABC")) {
                    startOfRecordIndexes.add(index);
                }
                index++;
            }

            // Once you have the starting index for all records you can iterate through the list and create new records
            scanFile = scanFile.reset();
            index = 0;

            BufferedWriter writer = null;
            
            while (scanFile.hasNext()) {
                if (!startOfRecordIndexes.isEmpty() && index == startOfRecordIndexes.peek()) {
                    if(writer != null) {
                        writer.write("TRL");
                        writer.close();
                    }
                    writer = new BufferedWriter(new OutputStreamWriter(
                        new FileOutputStream("Give unique filename"), StandardCharsets.UTF_8));
                    writer.write("HDR");
                    writer.write(scanFile.nextLine());

                    startOfRecordIndexes.remove();
                } else {
                    writer.write(scanFile.nextLine());
                }
            }
            // Close the last record
            if(writer != null) {
                writer.write("TRL");
                writer.close();
            }
        } catch (IOException e) {
            // deal with exception
        }
    }