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Home/ Questions/Q 8384459
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T17:24:40+00:00 2026-06-09T17:24:40+00:00

I created a webservice using servlet and tomcat 6.0. I created sample java application

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I created a webservice using servlet and tomcat 6.0. I created sample java application to call the web service and its working fine. i need to send some data while calling web service. I created in java application as follows

StringEntity zStringEntityL = new StringEntity(zAPIInputStringP.toString());
zStringEntityL.setContentType(new BasicHeader(HTTP.CONTENT_TYPE,
        "application/json"));
HttpParams aHttpParamsL = new BasicHttpParams();
HttpProtocolParams.setVersion(aHttpParamsL, HttpVersion.HTTP_1_1);
HttpProtocolParams.setContentCharset(aHttpParamsL, HTTP.UTF_8);

SchemeRegistry aSchemeRegistryL = new SchemeRegistry();
aSchemeRegistryL.register(new Scheme("http", PlainSocketFactory.getSocketFactory(), 80));

ClientConnectionManager ccm = new ThreadSafeClientConnManager(aHttpParamsL, aSchemeRegistryL);
DefaultHttpClient client = new DefaultHttpClient(ccm, aHttpParamsL);
HttpPost aHttpPostL = new HttpPost(URL + zAPIName);
aHttpPostL.setHeader("Authorization", "Basic");



aHttpPostL.setEntity(zStringEntityL);
HttpResponse aHttpResponseL;
aHttpResponseL = client.execute(aHttpPostL); 

Here “zAPIInputStringP” is my data in JSON format.
In webservice how i need to get those data? I checked in debug mode in eclispe, i cant able to find it.

    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException 
{
    //How to get data?
}

Please help me out.

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  1. Editorial Team
    Editorial Team
    2026-06-09T17:24:41+00:00Added an answer on June 9, 2026 at 5:24 pm

    When you send the data to a servlet via post method, the data is available via input stream. Following is how your post method should look like.

    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    
    String zAPIInputStringP = "";
    
    BufferedReader in = new BufferedReader(new InputStreamReader(
                    request.getInputStream()));
    String line = in.readLine();
    while (line != null) {
        zAPIInputStringP += line;
        line = in.readLine();
    }
    
    
    }
    

    You JSON string is contained in zAPIInputStringP.

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