逐行读取

我有一个.gz 格式的文件。用于读取这个文件的 java 类是 GZIPInputStream。 但是，这个类不扩展 java 的 BufferedReader 类。结果，我无法逐行读取文件。我需要这样的东西

reader  = new MyGZInputStream( some constructor of GZInputStream)
reader.readLine()...

我想创建我的类，它扩展了 Reader 或 BufferedReader 的 java 类，并使用 GZIPInputStream 作为其变量之一。

import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.Reader;
import java.util.zip.GZIPInputStream;


public class MyGZFilReader extends Reader {


private GZIPInputStream gzipInputStream = null;
char[] buf = new char[1024];


@Override
public void close() throws IOException {
gzipInputStream.close();
}


public MyGZFilReader(String filename)
throws FileNotFoundException, IOException {
gzipInputStream = new GZIPInputStream(new FileInputStream(filename));
}


@Override
public int read(char[] cbuf, int off, int len) throws IOException {
// TODO Auto-generated method stub
return gzipInputStream.read((byte[])buf, off, len);
}


}

但是，这不工作，当我使用

BufferedReader in = new BufferedReader(
new MyGZFilReader("F:/gawiki-20090614-stub-meta-history.xml.gz"));
System.out.println(in.readLine());

有人能告诉我该怎么做吗。

111862 次浏览
小开 
GZIPInputStream gzip = new GZIPInputStream(new FileInputStream("F:/gawiki-20090614-stub-meta-history.xml.gz"));
BufferedReader br = new BufferedReader(new InputStreamReader(gzip));
br.readLine();

小开
最佳答案

The basic setup of decorators is like this:

InputStream fileStream = new FileInputStream(filename);
InputStream gzipStream = new GZIPInputStream(fileStream);
Reader decoder = new InputStreamReader(gzipStream, encoding);
BufferedReader buffered = new BufferedReader(decoder);

The key issue in this snippet is the value of encoding. This is the character encoding of the text in the file. Is it "US-ASCII", "UTF-8", "SHIFT-JIS", "ISO-8859-9", …? there are hundreds of possibilities, and the correct choice usually cannot be determined from the file itself. It must be specified through some out-of-band channel.

For example, maybe it's the platform default. In a networked environment, however, this is extremely fragile. The machine that wrote the file might sit in the neighboring cubicle, but have a different default file encoding.

Most network protocols use a header or other metadata to explicitly note the character encoding.

In this case, it appears from the file extension that the content is XML. XML includes the "encoding" attribute in the XML declaration for this purpose. Furthermore, XML should really be processed with an XML parser, not as text. Reading XML line-by-line seems like a fragile, special case.

Failing to explicitly specify the encoding is against the second commandment. Use the default encoding at your peril!

小开 
BufferedReader in = new BufferedReader(new InputStreamReader(
new GZIPInputStream(new FileInputStream("F:/gawiki-20090614-stub-meta-history.xml.gz"))));


String content;


while ((content = in.readLine()) != null)


System.out.println(content);
小开

You can use the following method in a util class, and use it whenever necessary...

public static List<String> readLinesFromGZ(String filePath) {
List<String> lines = new ArrayList<>();
File file = new File(filePath);


try (GZIPInputStream gzip = new GZIPInputStream(new FileInputStream(file));
BufferedReader br = new BufferedReader(new InputStreamReader(gzip));) {
String line = null;
while ((line = br.readLine()) != null) {
lines.add(line);
}
} catch (FileNotFoundException e) {
e.printStackTrace(System.err);
} catch (IOException e) {
e.printStackTrace(System.err);
}
return lines;
}
小开

here is with one line 

try (BufferedReader br = new BufferedReader(
new InputStreamReader(
new GZIPInputStream(
new FileInputStream(
"F:/gawiki-20090614-stub-meta-history.xml.gz")))))
{br.readLine();}

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