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
您可以使用Apache Commons。
在IOUItils中,您可以找到toString方法以及三个有用的实现。
public static String toString(InputStream input) throws IOException {
return toString(input, Charset.defaultCharset());
}
public static String toString(InputStream input) throws IOException {
return toString(input, Charset.defaultCharset());
}
public static String toString(InputStream input, String encoding)
throws IOException {
return toString(input, Charsets.toCharset(encoding));
}
其他回答
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;
}
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);
}
}
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!)
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));
}
}
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();
}