在基于unix的系统上，在命令行上使用wc命令。

要知道文件里有多少行，唯一的方法就是数数。当然，你可以从你的数据中创建一个指标，给你一个平均长度的一行，然后得到文件大小，并除以平均长度，但这不会是准确的。

这是我迄今为止发现的最快的版本，大约比readLines快6倍。对于150MB的日志文件，这需要0.35秒，而在使用readLines()时需要2.40秒。只是为了好玩，linux的wc -l命令需要0.15秒。

public static int countLinesOld(String filename) throws IOException {
    InputStream is = new BufferedInputStream(new FileInputStream(filename));
    try {
        byte[] c = new byte[1024];
        int count = 0;
        int readChars = 0;
        boolean empty = true;
        while ((readChars = is.read(c)) != -1) {
            empty = false;
            for (int i = 0; i < readChars; ++i) {
                if (c[i] == '\n') {
                    ++count;
                }
            }
        }
        return (count == 0 && !empty) ? 1 : count;
    } finally {
        is.close();
    }
}

编辑，9年半后:我几乎没有java经验，但无论如何，我试图将这段代码与下面的LineNumberReader解决方案进行基准测试，因为没有人这样做让我感到困扰。似乎对于大文件，我的解决方案更快。虽然它似乎需要几次运行，直到优化器做一个像样的工作。我已经玩了一些代码，并产生了一个新版本，始终是最快的:

public static int countLinesNew(String filename) throws IOException {
    InputStream is = new BufferedInputStream(new FileInputStream(filename));
    try {
        byte[] c = new byte[1024];
        
        int readChars = is.read(c);
        if (readChars == -1) {
            // bail out if nothing to read
            return 0;
        }
        
        // make it easy for the optimizer to tune this loop
        int count = 0;
        while (readChars == 1024) {
            for (int i=0; i<1024;) {
                if (c[i++] == '\n') {
                    ++count;
                }
            }
            readChars = is.read(c);
        }
        
        // count remaining characters
        while (readChars != -1) {
            for (int i=0; i<readChars; ++i) {
                if (c[i] == '\n') {
                    ++count;
                }
            }
            readChars = is.read(c);
        }
        
        return count == 0 ? 1 : count;
    } finally {
        is.close();
    }
}

1.3GB文本文件的基准测试结果，y轴以秒为单位。我已经对同一个文件执行了100次运行，并使用System.nanoTime()对每次运行进行了测量。您可以看到countLinesOld有一些异常值，而countLinesNew没有异常值，虽然它只是稍微快一点，但差异在统计上是显著的。LineNumberReader显然更慢。

如果没有任何索引结构，就无法读取完整的文件。但是您可以通过避免逐行读取并使用正则表达式来匹配所有行结束符来优化它。

上面的count()方法给出的答案是，如果文件末尾没有换行符，就会出现行数错误——它无法计算文件中的最后一行。

这个方法更适合我:

public int countLines(String filename) throws IOException {
    LineNumberReader reader  = new LineNumberReader(new FileReader(filename));
int cnt = 0;
String lineRead = "";
while ((lineRead = reader.readLine()) != null) {}

cnt = reader.getLineNumber(); 
reader.close();
return cnt;
}

如果你用这个

public int countLines(String filename) throws IOException {
    LineNumberReader reader  = new LineNumberReader(new FileReader(filename));
    int cnt = 0;
    String lineRead = "";
    while ((lineRead = reader.readLine()) != null) {}

    cnt = reader.getLineNumber(); 
    reader.close();
    return cnt;
}

你不能运行到大num行，比如100K行，因为从读取器返回。getLineNumber是int。你需要长类型的数据来处理最多的行。

我已经实现了这个问题的另一个解决方案，我发现它在计算行数时更有效:

try
(
   FileReader       input = new FileReader("input.txt");
   LineNumberReader count = new LineNumberReader(input);
)
{
   while (count.skip(Long.MAX_VALUE) > 0)
   {
      // Loop just in case the file is > Long.MAX_VALUE or skip() decides to not read the entire file
   }

   result = count.getLineNumber() + 1;                                    // +1 because line index starts at 0
}

我知道这是一个老问题，但公认的解决方案并不完全符合我所需要的。因此，我将其改进为接受各种行结束符(而不仅仅是换行)并使用指定的字符编码(而不是ISO-8859-n)。所有在一个方法(适当重构):

public static long getLinesCount(String fileName, String encodingName) throws IOException {
    long linesCount = 0;
    File file = new File(fileName);
    FileInputStream fileIn = new FileInputStream(file);
    try {
        Charset encoding = Charset.forName(encodingName);
        Reader fileReader = new InputStreamReader(fileIn, encoding);
        int bufferSize = 4096;
        Reader reader = new BufferedReader(fileReader, bufferSize);
        char[] buffer = new char[bufferSize];
        int prevChar = -1;
        int readCount = reader.read(buffer);
        while (readCount != -1) {
            for (int i = 0; i < readCount; i++) {
                int nextChar = buffer[i];
                switch (nextChar) {
                    case '\r': {
                        // The current line is terminated by a carriage return or by a carriage return immediately followed by a line feed.
                        linesCount++;
                        break;
                    }
                    case '\n': {
                        if (prevChar == '\r') {
                            // The current line is terminated by a carriage return immediately followed by a line feed.
                            // The line has already been counted.
                        } else {
                            // The current line is terminated by a line feed.
                            linesCount++;
                        }
                        break;
                    }
                }
                prevChar = nextChar;
            }
            readCount = reader.read(buffer);
        }
        if (prevCh != -1) {
            switch (prevCh) {
                case '\r':
                case '\n': {
                    // The last line is terminated by a line terminator.
                    // The last line has already been counted.
                    break;
                }
                default: {
                    // The last line is terminated by end-of-file.
                    linesCount++;
                }
            }
        }
    } finally {
        fileIn.close();
    }
    return linesCount;
}

这个解决方案在速度上与公认的解决方案相当，在我的测试中大约慢了4%(尽管Java中的计时测试是出了名的不可靠)。

接受的答案有一个错误关闭多行文件，不以换行符结束。一个没有换行符的单行文件将返回1，但是一个没有换行符的两行文件也将返回1。下面是解决这个问题的公认解决方案的实现。endsWithoutNewLine检查对于除最终读取外的所有内容都是浪费的，但与整个函数相比，应该是微不足道的时间。

public int count(String filename) throws IOException {
    InputStream is = new BufferedInputStream(new FileInputStream(filename));
    try {
        byte[] c = new byte[1024];
        int count = 0;
        int readChars = 0;
        boolean endsWithoutNewLine = false;
        while ((readChars = is.read(c)) != -1) {
            for (int i = 0; i < readChars; ++i) {
                if (c[i] == '\n')
                    ++count;
            }
            endsWithoutNewLine = (c[readChars - 1] != '\n');
        }
        if(endsWithoutNewLine) {
            ++count;
        } 
        return count;
    } finally {
        is.close();
    }
}

从Java代码中使用Process类如何?然后读取命令的输出。

Process p = Runtime.getRuntime().exec("wc -l " + yourfilename);
p.waitFor();

BufferedReader b = new BufferedReader(new InputStreamReader(p.getInputStream()));
String line = "";
int lineCount = 0;
while ((line = b.readLine()) != null) {
    System.out.println(line);
    lineCount = Integer.parseInt(line);
}

不过需要尝试一下。将公布结果。

在java-8中，你可以使用流:

try (Stream<String> lines = Files.lines(path, Charset.defaultCharset())) {
  long numOfLines = lines.count();
  ...
}

一个直接的方式使用扫描器

static void lineCounter (String path) throws IOException {

        int lineCount = 0, commentsCount = 0;

        Scanner input = new Scanner(new File(path));
        while (input.hasNextLine()) {
            String data = input.nextLine();

            if (data.startsWith("//")) commentsCount++;

            lineCount++;
        }

        System.out.println("Line Count: " + lineCount + "\t Comments Count: " + commentsCount);
    }

/**
 * Count file rows.
 *
 * @param file file
 * @return file row count
 * @throws IOException
 */
public static long getLineCount(File file) throws IOException {

    try (Stream<String> lines = Files.lines(file.toPath())) {
        return lines.count();
    }
}

在JDK8_u31上测试。但与此方法相比，性能确实较慢:

/**
 * Count file rows.
 *
 * @param file file
 * @return file row count
 * @throws IOException
 */
public static long getLineCount(File file) throws IOException {

    try (BufferedInputStream is = new BufferedInputStream(new FileInputStream(file), 1024)) {

        byte[] c = new byte[1024];
        boolean empty = true,
                lastEmpty = false;
        long count = 0;
        int read;
        while ((read = is.read(c)) != -1) {
            for (int i = 0; i < read; i++) {
                if (c[i] == '\n') {
                    count++;
                    lastEmpty = true;
                } else if (lastEmpty) {
                    lastEmpty = false;
                }
            }
            empty = false;
        }

        if (!empty) {
            if (count == 0) {
                count = 1;
            } else if (!lastEmpty) {
                count++;
            }
        }

        return count;
    }
}

经过测试，非常快。

我的结论是wc -l:s计算换行的方法是好的，但是在最后一行不以换行符结束的文件上返回非直观的结果。

和@。基于LineNumberReader的vikas解决方案，但在行数中添加一个，在最后一行以换行符结束的文件上返回非直观的结果。

因此我做了一个算法，处理如下:

@Test
public void empty() throws IOException {
    assertEquals(0, count(""));
}

@Test
public void singleNewline() throws IOException {
    assertEquals(1, count("\n"));
}

@Test
public void dataWithoutNewline() throws IOException {
    assertEquals(1, count("one"));
}

@Test
public void oneCompleteLine() throws IOException {
    assertEquals(1, count("one\n"));
}

@Test
public void twoCompleteLines() throws IOException {
    assertEquals(2, count("one\ntwo\n"));
}

@Test
public void twoLinesWithoutNewlineAtEnd() throws IOException {
    assertEquals(2, count("one\ntwo"));
}

@Test
public void aFewLines() throws IOException {
    assertEquals(5, count("one\ntwo\nthree\nfour\nfive\n"));
}

它是这样的:

static long countLines(InputStream is) throws IOException {
    try(LineNumberReader lnr = new LineNumberReader(new InputStreamReader(is))) {
        char[] buf = new char[8192];
        int n, previousN = -1;
        //Read will return at least one byte, no need to buffer more
        while((n = lnr.read(buf)) != -1) {
            previousN = n;
        }
        int ln = lnr.getLineNumber();
        if (previousN == -1) {
            //No data read at all, i.e file was empty
            return 0;
        } else {
            char lastChar = buf[previousN - 1];
            if (lastChar == '\n' || lastChar == '\r') {
                //Ending with newline, deduct one
                return ln;
            }
        }
        //normal case, return line number + 1
        return ln + 1;
    }
}

如果你想要直观的结果，你可以用这个。如果您只想要wc -l兼容性，只需使用@er即可。Vikas解决方案，但不添加一个到结果，并重试跳过:

try(LineNumberReader lnr = new LineNumberReader(new FileReader(new File("File1")))) {
    while(lnr.skip(Long.MAX_VALUE) > 0){};
    return lnr.getLineNumber();
}

这个有趣的解决方案真的很好!

public static int countLines(File input) throws IOException {
    try (InputStream is = new FileInputStream(input)) {
        int count = 1;
        for (int aChar = 0; aChar != -1;aChar = is.read())
            count += aChar == '\n' ? 1 : 0;
        return count;
    }
}

EOF处没有换行符('\n')的多行文件的最佳优化代码。

/**
 * 
 * @param filename
 * @return
 * @throws IOException
 */
public static int countLines(String filename) throws IOException {
    int count = 0;
    boolean empty = true;
    FileInputStream fis = null;
    InputStream is = null;
    try {
        fis = new FileInputStream(filename);
        is = new BufferedInputStream(fis);
        byte[] c = new byte[1024];
        int readChars = 0;
        boolean isLine = false;
        while ((readChars = is.read(c)) != -1) {
            empty = false;
            for (int i = 0; i < readChars; ++i) {
                if ( c[i] == '\n' ) {
                    isLine = false;
                    ++count;
                }else if(!isLine && c[i] != '\n' && c[i] != '\r'){   //Case to handle line count where no New Line character present at EOF
                    isLine = true;
                }
            }
        }
        if(isLine){
            ++count;
        }
    }catch(IOException e){
        e.printStackTrace();
    }finally {
        if(is != null){
            is.close();    
        }
        if(fis != null){
            fis.close();    
        }
    }
    LOG.info("count: "+count);
    return (count == 0 && !empty) ? 1 : count;
}

我测试了上面的方法来计数行，这里是我对不同方法的观察，在我的系统上进行了测试

文件大小:1.6 Gb 方法:

使用扫描仪:大约35秒 使用BufferedReader:大约5s 使用Java 8: 5s左右 使用LineNumberReader:大约5s

此外，Java8方法似乎非常方便:

Files.lines(Paths.get(filePath), Charset.defaultCharset()).count()
[Return type : long]

扫描与regex:

public int getLineCount() {
    Scanner fileScanner = null;
    int lineCount = 0;
    Pattern lineEndPattern = Pattern.compile("(?m)$");  
    try {
        fileScanner = new Scanner(new File(filename)).useDelimiter(lineEndPattern);
        while (fileScanner.hasNext()) {
            fileScanner.next();
            ++lineCount;
        }   
    }catch(FileNotFoundException e) {
        e.printStackTrace();
        return lineCount;
    }
    fileScanner.close();
    return lineCount;
}

还没记。

似乎有几种不同的方法可以使用LineNumberReader。

我是这样做的:

int lines = 0;

FileReader input = new FileReader(fileLocation);
LineNumberReader count = new LineNumberReader(input);

String line = count.readLine();

if(count.ready())
{
    while(line != null) {
        lines = count.getLineNumber();
        line = count.readLine();
    }
    
    lines+=1;
}
    
count.close();

System.out.println(lines);

更简单的是，可以使用Java BufferedReader lines()方法返回元素流，然后使用stream count()方法对所有元素进行计数。然后只需在输出中添加1，就可以得到文本文件中的行数。

为例:

FileReader input = new FileReader(fileLocation);
LineNumberReader count = new LineNumberReader(input);

int lines = (int)count.lines().count() + 1;
    
count.close();

System.out.println(lines);