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
如果您使用的是AWS SDK v2,请调用IoUtils.toUtf8String():
public String convertStreamToString(InputStream is) {
return IoUtils.toUtf8String(is);
}
其他回答
这个很好,因为:
它可以安全地处理Charset。您可以控制读取缓冲区的大小。您可以设置生成器的长度,而不必是精确的值。不受库依赖关系的影响。适用于Java 7或更高版本。
怎么做?
public static String convertStreamToString(InputStream is) throws IOException {
StringBuilder sb = new StringBuilder(2048); // Define a size if you have an idea of it.
char[] read = new char[128]; // Your buffer size.
try (InputStreamReader ir = new InputStreamReader(is, StandardCharsets.UTF_8)) {
for (int i; -1 != (i = ir.read(read)); sb.append(read, 0, i));
}
return sb.toString();
}
对于JDK 9
public static String inputStreamString(InputStream inputStream) throws IOException {
try (inputStream) {
return new String(inputStream.readAllBytes(), StandardCharsets.UTF_8);
}
}
这是一个改编自org.apache.commons.io.IOUtils源代码的答案,适用于那些想要apache实现但不想要整个库的人。
private static final int BUFFER_SIZE = 4 * 1024;
public static String inputStreamToString(InputStream inputStream, String charsetName)
throws IOException {
StringBuilder builder = new StringBuilder();
InputStreamReader reader = new InputStreamReader(inputStream, charsetName);
char[] buffer = new char[BUFFER_SIZE];
int length;
while ((length = reader.read(buffer)) != -1) {
builder.append(buffer, 0, length);
}
return builder.toString();
}
此代码适用于新的Java学习者:
private String textDataFromFile;
public String getFromFile(InputStream myInputStream) throws FileNotFoundException, IOException {
BufferedReader bufferReader = new BufferedReader(new InputStreamReader(myInputStream));
StringBuilder stringBuilder = new StringBuilder();
String eachStringLine;
while ((eachStringLine = bufferReader.readLine()) != null) {
stringBuilder.append(eachStringLine).append("\n");
}
textDataFromFile = stringBuilder.toString();
return textDataFromFile;
}
InputStream inputStream = null;
BufferedReader bufferedReader = null;
try {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
String stringBuilder = new StringBuilder();
String content;
while((content = bufferedReader.readLine()) != null){
stringBuilder.append(content);
}
System.out.println("content of file::" + stringBuilder.toString());
}
catch (IOException e) {
e.printStackTrace();
}finally{
if(bufferedReader != null){
try{
bufferedReader.close();
}catch(IoException ex){
ex.printStackTrace();
}
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();
}