How to execute url which has xml data as query parameter through Apache HttpClient?

Question

I have a XML like below which is stored as a String in xmlData variable and I need to pass this String to my url in client_data variable -

<?xml version="1.0"?>
<ClientData
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
 xsi:schemaLocation="http://www.google.com model.xsd" 
 xmlns="http://www.google.com"> 
    <client id="100">
        <clock>
                <for>
                    <etc>val(tery) = 1</etc>
                    <while><![CDATA[val(tery)  < 20]]></while>
                </for>
        </clock>
    </model>
</ClientData>

Below is the URL I am hitting on the browser by URL encoding client_data value to get the response back and it works fine.

http://localhost:8080/test_tmp?max_time=30&users=1000&client_data=<?xml version="1.0"?>
<ClientData
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
 xsi:schemaLocation="http://www.google.com model.xsd" 
 xmlns="http://www.google.com"> 
    <client id="100">
        <clock>
                <for>
                    <etc>val(tery) = 1</etc>
                    <while><![CDATA[val(tery)  < 20]]></while>
                </for>
        </clock>
    </model>
</ClientData>

Now I need to do the same thing through Apache HttpClient. What is the best way to do this?

I have got below code but not sure whethere there is any better way to do this? Is there anything I can improve on or some better way of doing it?

public static void main(String[] args) {
    String xmlData = getXMLData();
    String url = generateURL(xmlData);

    HttpClient httpclient = HttpClientBuilder.create().build();
    HttpGet request = new HttpGet(url);

    HttpResponse httpResponse = httpclient.execute(request);

    BufferedReader rd = new BufferedReader(new InputStreamReader(httpResponse.getEntity().getContent()));

    StringBuffer response = new StringBuffer();
    String line = "";
    while ((line = rd.readLine()) != null) {
        response.append(line);
    }

    // prints response
    System.out.println(response);       
}

private static String generateURL(final String clientData) {
    StringBuilder url = new StringBuilder();
    url.append("http://localhost:8080/test_tmp?max_time=30&users=1000&client_data=");
    url.append(URLEncoder.encode(clientData, "UTF-8"));

    return url.toString();
}

TravisWhidden · Accepted Answer · 2015-01-04 21:46:24Z

1

Honestly, I don't think you want to try and pass XML through your URI. There are several things to consider. One, some servers will block long URIs (Google first result: http://www.checkupdown.com/status/E414.html ). Next, you would want to "encode" the URI so it escapes all of your chars that are not ready to be passed raw.

Consider doing a "POST" with the xml the body of the HTTP message.

Hope this helps.

answered Jan 4, 2015 at 21:46

TravisWhidden

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1 Comment

john Over a year ago

Thanks for suggestion. If I am sending POST request, then my web service has to handle that POST request right? Also, can you provide an example how can I send POST request through HttpClient?

Geba · Accepted Answer · 2023-03-01 13:47:20Z

I guess you could make your request without using Apache HttpClient, you could just open stream by url

This is how you could pass XML in the query parameter and read the response as stream:

URI uri = new URIBuilder("http://localhost:8080/test_tmp")
    .addParameter("max_time", "30")
    .addParameter("users", "1000")
    .addParameter("client_data", xmlData)
    .build();

BufferedReader rd = new BufferedReader(
    new InputStreamReader(uri.toURL().openStream())
);

If you insist on using Apache HttpClient, then code could look like:

URI uri = new URIBuilder("http://localhost:8080/test_tmp")
    .addParameter("max_time", "30")
    .addParameter("users", "1000")
    .addParameter("client_data", xmlData)
    .build();

HttpClient httpClient = HttpClients.createDefault();
HttpGet request = new HttpGet(uri);
HttpResponse httpResponse = httpClient.execute(request);

BufferedReader rd = new BufferedReader(
    new InputStreamReader(httpResponse.getEntity().getContent())
);

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