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
注意:这可能不是个好主意。此方法使用递归,因此将非常快地命中StackOverflowError:
public String read (InputStream is) {
byte next = is.read();
return next == -1 ? "" : next + read(is); // Recursive part: reads next byte recursively
}
其他回答
JDK中最简单的方法是使用以下代码片段。
String convertToString(InputStream in){
String resource = new Scanner(in).useDelimiter("\\Z").next();
return resource;
}
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();
}
}
基于已接受的Apache Commons答案的第二部分,但在始终关闭流的情况下填补了一个小缺口:
String theString;
try {
theString = IOUtils.toString(inputStream, encoding);
} finally {
IOUtils.closeQuietly(inputStream);
}
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;
}