Java 主流的 Inputstream 转 String 的方法

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1. Ways to convert an InputStream to a String:

1. Using IOUtils.toString (Apache Utils)

String result = IOUtils.toString(inputStream, StandardCharsets.UTF_8);

2. Using CharStreams (Guava)

String result = CharStreams.toString(new InputStreamReader(inputStream, Charsets.UTF_8));

3. Using Scanner (JDK)

Scanner s = new Scanner(inputStream).useDelimiter("\\A");
String result = s.hasNext() ? s.next() : "";

4. Using Stream API (Java 8). Warning: This solution converts different line breaks (like \r\n) to \n.

String result = new BufferedReader(new InputStreamReader(inputStream))
	.lines().collect(Collectors.joining("\n"));

5. Using parallel Stream API (Java 8). Warning: This solution converts different line breaks (like \r\n) to \n.

String result = new BufferedReader(new InputStreamReader(inputStream))
	.lines().parallel().collect(Collectors.joining("\n"));

6. Using InputStreamReader and StringBuilder (JDK)

int bufferSize = 1024;
char[] buffer = new char[bufferSize];
StringBuilder out = new StringBuilder();
Reader in = new InputStreamReader(stream, StandardCharsets.UTF_8);
for (int numRead; (numRead = in.read(buffer, 0, buffer.length)) > 0; ) {
    out.append(buffer, 0, numRead);
}
return out.toString();

7. Using StringWriter and IOUtils.copy (Apache Commons)

StringWriter writer = new StringWriter();
IOUtils.copy(inputStream, writer, "UTF-8");
return writer.toString();

8. Using ByteArrayOutputStream and inputStream.read (JDK)

ByteArrayOutputStream result = new ByteArrayOutputStream();
byte[] buffer = new byte[1024];
for (int length; (length = inputStream.read(buffer)) != -1; ) {
    result.write(buffer, 0, length);
}
// StandardCharsets.UTF_8.name() > JDK 7
return result.toString("UTF-8");

9. Using BufferedReader (JDK). Warning: This solution converts different line breaks (like \n\r) to line.separator system property (for example, in Windows to “\r\n”).

String newLine = System.getProperty("line.separator");
BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream));
StringBuilder result = new StringBuilder();
for (String line; (line = reader.readLine()) != null; ) {
    if (result.length() > 0) {
        result.append(newLine);
    }
    result.append(line);
}
return result.toString();

10. Using BufferedInputStream and ByteArrayOutputStream (JDK)

BufferedInputStream bis = new BufferedInputStream(inputStream);
ByteArrayOutputStream buf = new ByteArrayOutputStream();
for (int result = bis.read(); result != -1; result = bis.read()) {
    buf.write((byte) result);
}
// StandardCharsets.UTF_8.name() > JDK 7
return buf.toString("UTF-8");

11. Using inputStream.read() and StringBuilder (JDK). Warning: This solution has problems with Unicode, for example with Russian text (works correctly only with non-Unicode text)

StringBuilder sb = new StringBuilder();
for (int ch; (ch = inputStream.read()) != -1; ) {
    sb.append((char) ch);
}
return sb.toString();

Warning:

1. Solutions 4, 5 and 9 convert different line breaks to one.

2. Solution 11 can’t work correctly with Unicode text

Performance tests

Performance tests for small String (length = 175), url in github (mode = Average Time, system = Linux, score 1,343 is the best):

              Benchmark                         Mode  Cnt   Score   Error  Units
 8. ByteArrayOutputStream and read (JDK)        avgt   10   1,343 ± 0,028  us/op
 6. InputStreamReader and StringBuilder (JDK)   avgt   10   6,980 ± 0,404  us/op
10. BufferedInputStream, ByteArrayOutputStream  avgt   10   7,437 ± 0,735  us/op
11. InputStream.read() and StringBuilder (JDK)  avgt   10   8,977 ± 0,328  us/op
 7. StringWriter and IOUtils.copy (Apache)      avgt   10  10,613 ± 0,599  us/op
 1. IOUtils.toString (Apache Utils)             avgt   10  10,605 ± 0,527  us/op
 3. Scanner (JDK)                               avgt   10  12,083 ± 0,293  us/op
 2. CharStreams (guava)                         avgt   10  12,999 ± 0,514  us/op
 4. Stream Api (Java 8)                         avgt   10  15,811 ± 0,605  us/op
 9. BufferedReader (JDK)                        avgt   10  16,038 ± 0,711  us/op
 5. parallel Stream Api (Java 8)                avgt   10  21,544 ± 0,583  us/op

Performance tests for big String (length = 50100), url in github (mode = Average Time, system = Linux, score 200,715 is the best):

               Benchmark                        Mode  Cnt   Score        Error  Units
 8. ByteArrayOutputStream and read (JDK)        avgt   10   200,715 ±   18,103  us/op
 1. IOUtils.toString (Apache Utils)             avgt   10   300,019 ±    8,751  us/op
 6. InputStreamReader and StringBuilder (JDK)   avgt   10   347,616 ±  130,348  us/op
 7. StringWriter and IOUtils.copy (Apache)      avgt   10   352,791 ±  105,337  us/op
 2. CharStreams (guava)                         avgt   10   420,137 ±   59,877  us/op
 9. BufferedReader (JDK)                        avgt   10   632,028 ±   17,002  us/op
 5. parallel Stream Api (Java 8)                avgt   10   662,999 ±   46,199  us/op
 4. Stream Api (Java 8)                         avgt   10   701,269 ±   82,296  us/op
12. BufferedInputStream, ByteArrayOutputStream  avgt   10   740,837 ±    5,613  us/op
 3. Scanner (JDK)                               avgt   10   751,417 ±   62,026  us/op
11. InputStream.read() and StringBuilder (JDK)  avgt   10  2919,350 ± 1101,942  us/op

Graphs (performance tests depending on Input Stream length in Windows 7 system)

Performance test (Average Time) depending on Input Stream length in Windows 7 system:

 length  182    546     1092    3276    9828    29484   58968

 test8  0.38    0.938   1.868   4.448   13.412  36.459  72.708
 test4  2.362   3.609   5.573   12.769  40.74   81.415  159.864
 test5  3.881   5.075   6.904   14.123  50.258  129.937 166.162
 test9  2.237   3.493   5.422   11.977  45.98   89.336  177.39
 test6  1.261   2.12    4.38    10.698  31.821  86.106  186.636
 test7  1.601   2.391   3.646   8.367   38.196  110.221 211.016
 test1  1.529   2.381   3.527   8.411   40.551  105.16  212.573
 test3  3.035   3.934   8.606   20.858  61.571  118.744 235.428
 test2  3.136   6.238   10.508  33.48   43.532  118.044 239.481
 test10 1.593   4.736   7.527   20.557  59.856  162.907 323.147
 test11 3.913   11.506  23.26   68.644  207.591 600.444 1211.545

2. nio 使用 Paths.get 的 java.nio.file.FileSystemNotFoundException 分析

Question:

I have a Maven project and inside a method I want to create a path for a directory in my resources folder. This is done like this:

try {
    final URI uri = getClass().getResource("/my-folder").toURI();
    Path myFolderPath = Paths.get(uri);
} catch (final URISyntaxException e) {
    ...
}

The generated URI looks like jar:file:/C:/path/to/my/project.jar!/my-folder.

The stacktrace is as following:

Exception in thread "pool-4-thread-1" java.nio.file.FileSystemNotFoundException
    at com.sun.nio.zipfs.ZipFileSystemProvider.getFileSystem(ZipFileSystemProvider.java:171)
    at com.sun.nio.zipfs.ZipFileSystemProvider.getPath(ZipFileSystemProvider.java:157)
    at java.nio.file.Paths.get(Paths.java:143)

The URI seems to be valid. The part before ! points to the generated jar-file and the part after it to my-folder in the root of the archive. I have used this instructions before to create paths to my resources. Why am I getting an exception now?

Answer:

You need to create the file system before you can access the path within the zip like

final URI uri = getClass().getResource("/my-folder").toURI();
Map<String, String> env = new HashMap<>(); 
env.put("create", "true");
FileSystem zipfs = FileSystems.newFileSystem(uri, env);
Path myFolderPath = Paths.get(uri);

This is not done automatically.

See http://docs.oracle.com/javase/7/docs/technotes/guides/io/fsp/zipfilesystemprovider.html

or

private FileSystem initFileSystem(URI uri) throws IOException {
    try {
        return FileSystems.newFileSystem(uri, Collections.emptyMap());
    }catch(IllegalArgumentException e) {
        return FileSystems.getDefault();
    }
}

or

private FileSystem initFileSystem(URI uri) throws IOException {
    try {
        return FileSystems.getFileSystem(uri);
    } catch( FileSystemNotFoundException e ) {
        Map<String, String> env = new HashMap<>();
        env.put("create", "true");
        return FileSystems.newFileSystem(uri, env);
    }
}

Calling this with the URI you are about to load will ensure the filesystem is in working condition. I always call FileSystem.close() after using it:

FileSystem zipfs = initFileSystem(fileURI);
filePath = Paths.get(fileURI);
// Do whatever you need and then close the filesystem
zipfs.close();

Careful, a ZipFileSystem can be closed, but a WindowsFileSystem will complain.

3. 在使用nio加载文件时，在idea中运行没有问题，但打成jar包后在windows和linux下都有问题

public void test() throws Exception {
    URI        uri      = getClass().getClassLoader().getResource("conf/sh.txt").toURI();
    FileSystem aDefault = FileSystems.getDefault();
    System.out.println(aDefault.getClass());
    FileSystemProvider provider = FileSystems.getDefault().provider();
    System.out.println(provider.getClass());
    System.out.println("====================" + uri.getScheme());
    List<FileSystemProvider> fileSystemProviders = FileSystemProvider.installedProviders();
    fileSystemProviders.forEach(p -> System.out.println(p.getClass()));
    Path path = Paths.get(uri);
}

这种情况下在idea中没有问题:

class sun.nio.fs.WindowsFileSystemclass sun.nio.fs.WindowsFileSystemProvider
====================fileclass sun.nio.fs.WindowsFileSystemProviderclass
com.sun.nio.zipfs.ZipFileSystemProvider

但是在打成jar包运行时Path path = Paths.get(uri)这一行会抛出异常：

Exception in thread "pool-4-thread-1" java.nio.file.FileSystemNotFoundException
    at com.sun.nio.zipfs.ZipFileSystemProvider.getFileSystem(ZipFileSystemProvider.java:171)
    at com.sun.nio.zipfs.ZipFileSystemProvider.getPath(ZipFileSystemProvider.java:157)
    at java.nio.file.Paths.get(Paths.java:143)

究其原因，是FileSystemProvider的使用问题，先看java.nio.file.Paths#get(java.net.URI)：

public static Path get(URI uri) {
    String scheme = uri.getScheme();
    if (scheme == null) throw new IllegalArgumentException("Missing scheme");
    // check for default provider to avoid loading of installed providers
    if (scheme.equalsIgnoreCase("file"))
        return FileSystems.getDefault().provider().getPath(uri);
    // try to find provider
    for (FileSystemProvider provider : FileSystemProvider.installedProviders()) {
        if (provider.getScheme().equalsIgnoreCase(scheme)) {
            return provider.getPath(uri);
        }
    }
    throw new FileSystemNotFoundException("Provider " + scheme + " not installed");
}

• uri.getScheme()在idea中是file，在打成jar包后变成了jar。
• 当前缀以file开头时，会使用FileSystems.getDefault().provider()来处理，这个provider在windows环境下是WindowsFileSystemProvider， 在linux环境下是LinuxFileSystemProvider。
• FileSystemProvider.installedProviders()对应windows中的WindowsFileSystemProvider和ZipFileSystemProvider，对应linux中的LinuxFileSystemProvider和ZipFileSystemProvider。
• 当前缀不以file开头时，会使用FileSystemProvider.installedProviders()中与uri.getScheme()匹配的provider来处理，对应的就是ZipFileSystemProvider。
• ZipFileSystemProvider对应的FileSystem需要自己创建，使用和创建方式参考：https://docs.oracle.com/javase/8/docs/technotes/guides/io/fsp/zipfilesystemprovider.html

解决办法：

在 Path path = Paths.get(uri) 中进行处理

Path path = null;
try {
    path = Paths.get(uri);
} catch (Exception e) {
    // @see https://stackoverflow.com/questions/25032716/getting-filesystemnotfoundexception-from-zipfilesystemprovider-when-creating-a-p
    // @see http://docs.oracle.com/javase/7/docs/technotes/guides/io/fsp/zipfilesystemprovider.html
    Map<String, String> env = new HashMap<>();
    env.put("create", "true");
    FileSystem zipfs = FileSystems.newFileSystem(uri, env);
    path = Paths.get(uri);
}

或者使用其他办法加载资源文件:

byte[] data;
try (InputStream in = getClass().getResourceAsStream("/elasticsearch/segmentsIndex.json")) {
    data = IOUtils.toByteArray(in);
}

4. 获取 InputStream 的方式

/**
 * 直接通过文件名+getPath()来获取路径
 *
 * @param fileName
 * @throws IOException
 */
public void func00(String fileName) throws IOException {
    String path = this.getClass().getClassLoader().getResource(fileName).getPath(); // 注意getResource("")里面是空字符串
    System.out.println(path);
    String filePath = URLDecoder.decode(path, "UTF-8"); // 如果路径中带有中文会被URLEncoder,因此这里需要解码
    System.out.println(filePath);
    getFileContent(filePath);
}

/**
 * 直接通过文件名+getFile()来获取
 * 
 * url.getFile()=/pub/files/foobar.txt?id=123456
 * url.getPath()=/pub/files/foobar.txt
 *
 * @param fileName
 * @throws IOException
 */
public void func01(String fileName) throws IOException {
    String path = this.getClass().getClassLoader().getResource(fileName).getFile(); // 注意getResource("")里面是空字符串
    System.out.println(path);
    String filePath = URLDecoder.decode(path, "UTF-8"); // 如果路径中带有中文会被URLEncoder,因此这里需要解码
    System.out.println(filePath);
    getFileContent(filePath);
}

/**
 * 直接使用getResourceAsStream方法获取流
 * springboot项目中需要使用此种方法，因为jar包中没有一个实际的路径存放文件
 *
 * @param fileName
 * @throws IOException
 */
public void func02(String fileName) throws IOException {
	String filePath = URLDecoder.decode(fileName, "UTF-8"); // 如果路径中带有中文会被URLEncoder,因此这里需要解码
    InputStream in = this.getClass().getClassLoader().getResourceAsStream(filePath);
    getFileContent(in);
}

/**
 * 通过ClassPathResource类获取，建议SpringBoot中使用
 * springboot项目中需要使用此种方法，因为jar包中没有一个实际的路径存放文件
 *
 * @param fileName
 * @throws IOException
 */
public void func03(String fileName) throws IOException {
    ClassPathResource classPathResource = new ClassPathResource(fileName);
    InputStream inputStream = classPathResource.getInputStream();
    getFileContent(inputStream);
}

/**
 * 通过绝对路径获取项目中文件的位置（通过new File("")获取当前的绝对路径，只是本地绝对路径，不能用于服务器）
 * 
 * @param fileName
 * @throws IOException
 */
public void func04(String fileName) throws IOException {
    // 参数为空
    File directory = new File("");
    // 规范路径：getCanonicalPath() 方法返回绝对路径，会把 ..\ 、.\ 这样的符号解析掉
    String rootCanonicalPath = directory.getCanonicalPath();
    // 绝对路径：getAbsolutePath() 方法返回文件的绝对路径，如果构造的时候是全路径就直接返回全路径，如果构造时是相对路径，就返回当前目录的路径 + 构造 File 对象时的路径
    String rootAbsolutePath =directory.getAbsolutePath();
    System.out.println(rootCanonicalPath);
    System.out.println(rootAbsolutePath);
    String filePath = rootCanonicalPath + "\\chapter-2-springmvc-quickstart\\src\\main\\resources\\"+fileName;
    getFileContent(filePath);
}

5. 读取 InputStream 的方式

// BufferedReader
String inContent = new BufferedReader(new InputStreamReader(inputStream, StandardCharsets.UTF_8))
						.lines()
						// .parallel() // 可选
						.collect(Collectors.joining("\n"));

// StringBuilder
final int bufferSize = 4 * 0x400; // 4KB
char[] buffer = new char[bufferSize];
StringBuilder sb = new StringBuilder();
Reader in = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
for (int numRead; (numRead = in.read(buffer, 0, bufferSize) > 0; ) {
    sb.append(buffer, 0, numRead);
}
String inContent = sb.toString();

6. ClassLoader#getResourceAsStream 不能使用前导斜杠 /

No leading “/” (all names are absolute)

https://stackoverflow.com/questions/47900677/where-does-leading-slash-in-java-class-loader-getresource-leads-to
Leading slash works only for class.getResource() to override its default behavior. There is no leading slash concept for class.getClassLoader().getResource(), so it always returns null.

https://stackoverflow.com/questions/3803326/this-getclass-getclassloader-getresource-and-nullpointerexception
The reason you can’t use a leading / in the ClassLoader path is because all ClassLoader paths are absolute and so / is not a valid first character in the path.

Reference

• stackoverflow
• Paths.get(uri)
• nio使用中的java.nio.file.FileSystemNotFoundException分析

文章来源:https://blog.csdn.net/howeres/article/details/137222042
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