GZIPInputStream to String

Question

First of all, I'm sorry if my terminology is a bit amateur, try to bear with me ;)

I am attempting to convert the gzipped body of a HTTP response to plaintext. I've taken the byte array of this response and converted it to a ByteArrayInputStream. I've then converted this to a GZIPInputStream. I now want to read the GZIPInputStream and store the final decompressed HTTP response body as a plaintext String.

This code will store the final decompressed contents in an OutputStream, but I want to store the contents as a String:

public static int sChunk = 8192;
ByteArrayInputStream bais = new ByteArrayInputStream(responseBytes);
GZIPInputStream gzis = new GZIPInputStream(bais);
byte[] buffer = new byte[sChunk];
int length;
while ((length = gzis.read(buffer, 0, sChunk)) != -1) {
        out.write(buffer, 0, length);
}

Answer 1

To decode bytes from an InputStream, you can use an InputStreamReader. Then, a BufferedReader will allow you to read your stream line by line.

Your code will look like:

ByteArrayInputStream bais = new ByteArrayInputStream(responseBytes);
GZIPInputStream gzis = new GZIPInputStream(bais);
InputStreamReader reader = new InputStreamReader(gzis);
BufferedReader in = new BufferedReader(reader);

String readed;
while ((readed = in.readLine()) != null) {
    System.out.println(readed);
}

Answer 2

You can use the StringWriter to write to String

Answer 3

you can also do

try (GZIPInputStream gzipIn = new GZIPInputStream(new ByteArrayInputStream(pByteArray)))
{
....
}

AutoClosable is a good thing https://docs.oracle.com/javase/tutorial/essential/exceptions/tryResourceClose.html

Answer 4

Use the try-with-resources idiom (which automatically closes any resources opened in try(...) on exit from the block) to make code cleaner.

Use Apache IOUtils to convert inputStream to String using default CharSet.

import org.apache.commons.io.IOUtils;
public static String gzipFileToString(File file) throws IOException {
    try(GZIPInputStream gzipIn = new GZIPInputStream(new FileInputStream(file))) {
        return IOUtils.toString(gzipIn);
    }
}

Answer 5

You should rather have obtained the response as an InputStream instead of as byte[]. Then you can ungzip it using GZIPInputStream and read it as character data using InputStreamReader and finally write it as character data into a String using StringWriter.

String body = null;
String charset = "UTF-8"; // You should determine it based on response header.

try (
    InputStream gzippedResponse = response.getInputStream();
    InputStream ungzippedResponse = new GZIPInputStream(gzippedResponse);
    Reader reader = new InputStreamReader(ungzippedResponse, charset);
    Writer writer = new StringWriter();
) {
    char[] buffer = new char[10240];
    for (int length = 0; (length = reader.read(buffer)) > 0;) {
        writer.write(buffer, 0, length);
    }
    body = writer.toString();
}

// ...

See also:

• Java IO tutorial
• How to use URLConnecion to fire/handle HTTP requests

---

If your final intent is to parse the response as HTML, then I strongly recommend to just use a HTML parser for this like Jsoup. It's then as easy as:

String html = Jsoup.connect("http://google.com").get().html();

Answer 6

Use Apache Commons to convert GzipInputStream to byteArray.

import java.io.InputStream;
import java.util.zip.GZIPInputStream;
import org.apache.commons.io.IOUtils;

public static byte[] decompressContent(byte[] pByteArray) throws IOException {
        GZIPInputStream gzipIn = null;
        try {
            gzipIn = new GZIPInputStream(new ByteArrayInputStream(pByteArray));
            return IOUtils.toByteArray(gzipIn);
        } finally {
            if (gzipIn != null) {
                gzipIn.close();
            }
        }

To convert byte array uncompressed content to String, do something like this :

String uncompressedContent = new String(decompressContent(inputStream));