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
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;
}
其他回答
此代码段位于\sdk\samples\android-19\connection\NetworkConnect\NetworkConnectSample\src\main\java.com\example\android\NetworkConnect\MainActivity.java中,该代码段根据Apache许可证2.0版授权,由Google编写。
/** Reads an InputStream and converts it to a String.
* @param stream InputStream containing HTML from targeted site.
* @param len Length of string that this method returns.
* @return String concatenated according to len parameter.
* @throws java.io.IOException
* @throws java.io.UnsupportedEncodingException
*/
private String readIt(InputStream stream, int len) throws IOException, UnsupportedEncodingException {
Reader reader = null;
reader = new InputStreamReader(stream, "UTF-8");
char[] buffer = new char[len];
reader.read(buffer);
return new String(buffer);
}
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);
}
}
这是最适合Android和任何其他JVM的纯Java解决方案。
这个解决方案非常好。。。它简单、快速,适用于大小河流!!(见上文第8号基准)
public String readFullyAsString(InputStream inputStream, String encoding)
throws IOException {
return readFully(inputStream).toString(encoding);
}
public byte[] readFullyAsBytes(InputStream inputStream)
throws IOException {
return readFully(inputStream).toByteArray();
}
private ByteArrayOutputStream readFully(InputStream inputStream)
throws IOException {
ByteArrayOutputStream baos = new ByteArrayOutputStream();
byte[] buffer = new byte[1024];
int length = 0;
while ((length = inputStream.read(buffer)) != -1) {
baos.write(buffer, 0, length);
}
return baos;
}
我有log4j可用,所以我可以使用org.apache.log4j.lf5.util.StreamUtils.getBytes来获取字节,我可以使用Stringctor将其转换为字符串
String result = new String(StreamUtils.getBytes(inputStream));
我建议使用StringWriter类来解决这个问题。
StringWriter wt= new StringWriter();
IOUtils.copy(inputStream, wt, encoding);
String st= wt.toString();