有没有一种方法可以检查 BufferedReader
对象中是否有可读内容?类似于 C++ 中的 cin.peek()
。谢谢。
有没有一种方法可以检查 BufferedReader
对象中是否有可读内容?类似于 C++ 中的 cin.peek()
。谢谢。
PushbackReader pr = new PushbackReader(reader);
char c = (char)pr.read();
// do something to look at c
pr.unread((int)c); //pushes the character back into the buffer
您可以尝试使用"boolean ready()"方法。 来自Java 6 API文档的描述:如果缓冲区不为空或底层字符流已准备好,则缓冲字符流是就绪的。
BufferedReader r = new BufferedReader(reader);
if(r.ready())
{
r.read();
}
BufferedReader bReader = new BufferedReader(inputStream);
bReader.mark(1);
int byte1 = bReader.read();
bReader.reset();
BufferedReader#readLine()
是否返回null
。如果到达流的末尾(例如文件结尾,套接字关闭等),则它会返回null
。BufferedReader reader = new BufferedReader(someReaderSource);
String line = null;
while ((line = reader.readLine()) != null) {
// ...
}
如果你不想逐行读取(这也是选择BufferedReader
的主要原因),那么可以使用BufferedReader#ready()
代替:
BufferedReader reader = new BufferedReader(someReaderSource);
while (reader.ready()) {
int data = reader.read();
// ...
}
BufferedReader br = new BufferedReader(reader);
br.mark(1);
int firstByte = br.read();
br.reset();
public class PeekBufferedReader extends BufferedReader{
private Queue<String> buf;
private int bufSize;
public PeekBufferedReader(Reader reader, int bufSize) throws IOException {
super(reader);
this.bufSize = bufSize;
buf = Queues.newArrayBlockingQueue(bufSize);
}
/**
* readAheadLimit is set to 1048576. Line which has length over readAheadLimit
* will cause IOException.
* @throws IOException
**/
//public String peekLine() throws IOException {
// super.mark(1048576);
// String peekedLine = super.readLine();
// super.reset();
// return peekedLine;
//}
/**
* This method can be implemented by mark and reset methods. But performance of
* this implementation is better ( about 2times) than using mark and reset
**/
public String peekLine() throws IOException {
if (buf.isEmpty()) {
while (buf.size() < bufSize) {
String readLine = super.readLine();
if (readLine == null) {
break;
} else {
buf.add(readLine);
}
}
} else {
return buf.peek();
}
if (buf.isEmpty()) {
return null;
} else {
return buf.peek();
}
}
public String readLine() throws IOException {
if (buf.isEmpty()) {
while (buf.size() < bufSize) {
String readLine = super.readLine();
if (readLine == null) {
break;
} else {
buf.add(readLine);
}
}
} else {
return buf.poll();
}
if (buf.isEmpty()) {
return null;
} else {
return buf.poll();
}
}
public boolean isEmpty() throws IOException {
if (buf.isEmpty()) {
while (buf.size() < bufSize) {
String readLine = super.readLine();
if (readLine == null) {
break;
} else {
buf.add(readLine);
}
}
} else {
return false;
}
if (buf.isEmpty()) {
return true;
} else {
return false;
}
}
}
PushBackReader
来读取一个字符,然后“推回”它。这样你就可以确定有东西存在,而不会影响其整体状态——一种“窥视”。pgmura的答案(依赖于ready()方法)简单且有效。但请记住,这是因为Sun公司对该方法的实现;它并不完全符合文档。如果此行为很关键,我不会依赖它。 在这里查看http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4090471 我更倾向于选择PushbackReader选项。