如果您有java.io.InputStream对象,应该如何处理该对象并生成String?
假设我有一个包含文本数据的InputStream,我想将其转换为String,例如,我可以将其写入日志文件。
获取InputStream并将其转换为字符串的最简单方法是什么?
public String convertStreamToString(InputStream is) {
// ???
}
如果您有java.io.InputStream对象,应该如何处理该对象并生成String?
假设我有一个包含文本数据的InputStream,我想将其转换为String,例如,我可以将其写入日志文件。
获取InputStream并将其转换为字符串的最简单方法是什么?
public String convertStreamToString(InputStream is) {
// ???
}
当前回答
JDK 7/8应答,关闭流并仍然抛出IOException:
StringBuilder build = new StringBuilder();
byte[] buf = new byte[1024];
int length;
try (InputStream is = getInputStream()) {
while ((length = is.read(buf)) != -1) {
build.append(new String(buf, 0, length));
}
}
其他回答
以下是我经过一些实验后提出的最优雅、纯Java(无库)解决方案:
public static String fromStream(InputStream in) throws IOException
{
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
StringBuilder out = new StringBuilder();
String newLine = System.getProperty("line.separator");
String line;
while ((line = reader.readLine()) != null) {
out.append(line);
out.append(newLine);
}
return out.toString();
}
如果您正在使用Google Collections/Guava,您可以执行以下操作:
InputStream stream = ...
String content = CharStreams.toString(new InputStreamReader(stream, Charsets.UTF_8));
Closeables.closeQuietly(stream);
请注意,InputStreamReader的第二个参数(即Charsets.UTF_8)不是必需的,但如果您知道编码(您应该这样做!)
如果不能使用Commons IO(FileUtils/IOUtils/CopyUtils),下面是一个使用BufferedReader逐行读取文件的示例:
public class StringFromFile {
public static void main(String[] args) /*throws UnsupportedEncodingException*/ {
InputStream is = StringFromFile.class.getResourceAsStream("file.txt");
BufferedReader br = new BufferedReader(new InputStreamReader(is/*, "UTF-8"*/));
final int CHARS_PER_PAGE = 5000; //counting spaces
StringBuilder builder = new StringBuilder(CHARS_PER_PAGE);
try {
for(String line=br.readLine(); line!=null; line=br.readLine()) {
builder.append(line);
builder.append('\n');
}
}
catch (IOException ignore) { }
String text = builder.toString();
System.out.println(text);
}
}
或者,如果你想要原始速度,我会根据Paul de Vrieze的建议(避免使用StringWriter(内部使用StringBuffer))提出一个变体:
public class StringFromFileFast {
public static void main(String[] args) /*throws UnsupportedEncodingException*/ {
InputStream is = StringFromFileFast.class.getResourceAsStream("file.txt");
InputStreamReader input = new InputStreamReader(is/*, "UTF-8"*/);
final int CHARS_PER_PAGE = 5000; //counting spaces
final char[] buffer = new char[CHARS_PER_PAGE];
StringBuilder output = new StringBuilder(CHARS_PER_PAGE);
try {
for(int read = input.read(buffer, 0, buffer.length);
read != -1;
read = input.read(buffer, 0, buffer.length)) {
output.append(buffer, 0, read);
}
} catch (IOException ignore) { }
String text = output.toString();
System.out.println(text);
}
}
我会使用一些Java8技巧。
public static String streamToString(final InputStream inputStream) throws Exception {
// buffering optional
try
(
final BufferedReader br
= new BufferedReader(new InputStreamReader(inputStream))
) {
// parallel optional
return br.lines().parallel().collect(Collectors.joining("\n"));
} catch (final IOException e) {
throw new RuntimeException(e);
// whatever.
}
}
基本上与其他一些答案相同,只是更简洁。
下面的代码对我有用。
URL url = MyClass.class.getResource("/" + configFileName);
BufferedInputStream bi = (BufferedInputStream) url.getContent();
byte[] buffer = new byte[bi.available() ];
int bytesRead = bi.read(buffer);
String out = new String(buffer);
请注意,根据Java文档,available()方法可能不适用于InputStream,但始终适用于BufferedInputStream。如果您不想使用available()方法,我们可以始终使用以下代码
URL url = MyClass.class.getResource("/" + configFileName);
BufferedInputStream bi = (BufferedInputStream) url.getContent();
File f = new File(url.getPath());
byte[] buffer = new byte[ (int) f.length()];
int bytesRead = bi.read(buffer);
String out = new String(buffer);
我不确定是否会有任何编码问题。如果代码有任何问题,请发表评论。