CODEWITHSUNDEEP
javagziphttpclient
Alex Bliskovsky
Alex Bliskovsky

Reputation: 6323

GZip POST request with HTTPClient in Java

I need to send a POST request to a web server which includes a gzipped request parameter. I'm using Apache HttpClient and I've read that it supports Gzip out of the box, but I can't find any examples of how to do what I need. I'd appreciate it if anyone could post some examples of this.

Upvotes: 16

Views: 33294

Answers (2)

e-venter
e-venter

Reputation: 11

Try the GzipCompressingEntity class. If I'm zipping the body of a post e.g. for a JSON object I would go:

    // json payload
if (jsonBody != null) {
    
    post.setHeader("Content-Type", "application/json");
    StringEntity requestEntity = new StringEntity( jsonBody, ContentType.APPLICATION_JSON);
    
    if (gzipBody) {
        GzipCompressingEntity gzippedEntity = new GzipCompressingEntity(requestEntity);
        post.setEntity(gzippedEntity);
    }else {
        post.setEntity(requestEntity);
    }
    
}

Haven't tested but I assume for adding parameters you'd do:

    // add parameters
if (parameters != null && parameters.length > 0){

    List<NameValuePair> urlParameters = new ArrayList<NameValuePair>();
    for (int i = 0; i < parameters.length; i++){
        urlParameters.add(new BasicNameValuePair(parameters[i][0], parameters[i][1]));
    }
    post.setEntity(new GzipCompressingEntity(new UrlEncodedFormEntity(urlParameters)));
}

Upvotes: 1

BalusC
BalusC

Reputation: 1109422

You need to turn that String into a gzipped byte[] or (temp) File first. Let's assume that it's not an extraordinary large String value so that a byte[] is safe enough for the available JVM memory:

String foo = "value";
ByteArrayOutputStream baos = new ByteArrayOutputStream();

try (GZIPOutputStream gzos = new GZIPOutputStream(baos)) {
    gzos.write(foo.getBytes("UTF-8"));
}

byte[] fooGzippedBytes = baos.toByteArray();

Then, you can send it as a multipart body using HttpClient as follows:

MultipartEntity entity = new MultipartEntity();
entity.addPart("foo", new InputStreamBody(new ByteArrayInputStream(fooGzippedBytes), "foo.txt"));

HttpPost post = new HttpPost("http://example.com/some");
post.setEntity(entity);

HttpClient client = new DefaultHttpClient();
HttpResponse response = client.execute(post);
// ...

Note that HttpClient 4.1 supports the new ByteArrayBody which can be used as follows:

entity.addPart("foo", new ByteArrayBody(fooGzippedBytes, "foo.txt"));

Upvotes: 19

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