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
这个问题的解决方案不是最简单的,但由于没有提到NIO流和通道,这里有一个使用NIO通道和ByteBuffer将流转换为字符串的版本。
public static String streamToStringChannel(InputStream in, String encoding, int bufSize) throws IOException {
ReadableByteChannel channel = Channels.newChannel(in);
ByteBuffer byteBuffer = ByteBuffer.allocate(bufSize);
ByteArrayOutputStream bout = new ByteArrayOutputStream();
WritableByteChannel outChannel = Channels.newChannel(bout);
while (channel.read(byteBuffer) > 0 || byteBuffer.position() > 0) {
byteBuffer.flip(); //make buffer ready for write
outChannel.write(byteBuffer);
byteBuffer.compact(); //make buffer ready for reading
}
channel.close();
outChannel.close();
return bout.toString(encoding);
}
下面是如何使用它的示例:
try (InputStream in = new FileInputStream("/tmp/large_file.xml")) {
String x = streamToStringChannel(in, "UTF-8", 1);
System.out.println(x);
}
对于大型文件,此方法的性能应该很好。
其他回答
下面是如何仅使用JDK使用字节数组缓冲区来实现这一点。这实际上就是commons-ioIOUItils.copy()方法的工作原理。如果要从Reader而不是InputStream复制,可以用char[]替换byte[]。
import java.io.ByteArrayOutputStream;
import java.io.InputStream;
...
InputStream is = ....
ByteArrayOutputStream baos = new ByteArrayOutputStream(8192);
byte[] buffer = new byte[8192];
int count = 0;
try {
while ((count = is.read(buffer)) != -1) {
baos.write(buffer, 0, count);
}
}
finally {
try {
is.close();
}
catch (Exception ignore) {
}
}
String charset = "UTF-8";
String inputStreamAsString = baos.toString(charset);
一个很好的方法是使用Apache Commons IOUItils将InputStream复制到StringWriter中。。。类似于
StringWriter writer = new StringWriter();
IOUtils.copy(inputStream, writer, encoding);
String theString = writer.toString();
甚至
// NB: does not close inputStream, you'll have to use try-with-resources for that
String theString = IOUtils.toString(inputStream, encoding);
或者,如果不想混合流和写入器,可以使用ByteArrayOutputStream。
如果使用流读取器,请确保在结束时关闭流
private String readStream(InputStream iStream) throws IOException {
//build a Stream Reader, it can read char by char
InputStreamReader iStreamReader = new InputStreamReader(iStream);
//build a buffered Reader, so that i can read whole line at once
BufferedReader bReader = new BufferedReader(iStreamReader);
String line = null;
StringBuilder builder = new StringBuilder();
while((line = bReader.readLine()) != null) { //Read till end
builder.append(line);
builder.append("\n"); // append new line to preserve lines
}
bReader.close(); //close all opened stuff
iStreamReader.close();
//iStream.close(); //EDIT: Let the creator of the stream close it!
// some readers may auto close the inner stream
return builder.toString();
}
编辑:在JDK7+上,可以使用trywithresources构造。
/**
* Reads the stream into a string
* @param iStream the input stream
* @return the string read from the stream
* @throws IOException when an IO error occurs
*/
private String readStream(InputStream iStream) throws IOException {
//Buffered reader allows us to read line by line
try (BufferedReader bReader =
new BufferedReader(new InputStreamReader(iStream))){
StringBuilder builder = new StringBuilder();
String line;
while((line = bReader.readLine()) != null) { //Read till end
builder.append(line);
builder.append("\n"); // append new line to preserve lines
}
return builder.toString();
}
}
String inputStreamToString(InputStream inputStream, Charset charset) throws IOException {
try (
final StringWriter writer = new StringWriter();
final InputStreamReader reader = new InputStreamReader(inputStream, charset)
) {
reader.transferTo(writer);
return writer.toString();
}
}
纯Java标准库解决方案-无库自Java 10以来-Reader#transferTo(Java.io.Writer)无环溶液无新行字符处理
此外,您还可以从指定的资源路径获取InputStream:
public static InputStream getResourceAsStream(String path)
{
InputStream myiInputStream = ClassName.class.getResourceAsStream(path);
if (null == myiInputStream)
{
mylogger.info("Can't find path = ", path);
}
return myiInputStream;
}
要从特定路径获取InputStream,请执行以下操作:
public static URL getResource(String path)
{
URL myURL = ClassName.class.getResource(path);
if (null == myURL)
{
mylogger.info("Can't find resource path = ", path);
}
return myURL;
}