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Home/ Questions/Q 6693779
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T06:01:31+00:00 2026-05-26T06:01:31+00:00

I have the following code: HttpGet httpGet = new HttpGet(serverAdress + /rootservices); httpGet.setHeader(Accept, text/xml);

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I have the following code:

HttpGet httpGet = new HttpGet(serverAdress + "/rootservices");
        httpGet.setHeader("Accept", "text/xml");
        HttpResponse response = client.execute(httpGet, localContext);

        String projectURL = XMLDocumentParser.parseDocument(response.getEntity().getContent(), "oslc_scm:scmServiceProviders", "rdf:resource");
        String workItemURL = XMLDocumentParser.parseDocument(response.getEntity().getContent(), "oslc_cm:cmServiceProviders", "rdf:resource");

The problem here is that I read two times the HttpResponse object. So the second time I get the exception. But although I know the problem, I can´t find an easy solution. So what is a good way to solve that problem?

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  1. Editorial Team
    Editorial Team
    2026-05-26T06:01:32+00:00Added an answer on May 26, 2026 at 6:01 am

    Read the input stream returned by response.getEntity().getContent() into a byte[], stored in a local variable. See Convert InputStream to byte array in Java.

    byte[] content = IOUtils.toByteArray(response.getEntity().getContent());
    
    String projectURL = XMLDocumentParser.parseDocument(new ByteArrayInputStream(content), "oslc_scm:scmServiceProviders", "rdf:resource");
    String workItemURL = XMLDocumentParser.parseDocument(new ByteArrayInputStream(content), "oslc_cm:cmServiceProviders", "rdf:resource");
    
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