2

我正在尝试触发 POST 请求。我发送的数据是 XML 格式（我通过 JAXB 执行此操作），但我必须发送请求参数。我的问题是关于Content-Type（我在代码中添加了一些注释）。我应该使用哪一个？

请参阅下面我的代码：

       private V callAndGetResponse(K request, Class<K> requestClassType, Class<V> responseClassType) throws Exception {
    JAXBContext jaxbContext = JAXBContext.newInstance(requestClassType, responseClassType);

    Marshaller marshaller = jaxbContext.createMarshaller();
    // set properties on marshaller
    marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
    marshaller.setProperty(Marshaller.JAXB_ENCODING, charset);
    marshaller.setProperty("com.sun.xml.internal.bind.xmlHeaders", String.format(XML_HEADER_FORMAT, dtd));
    marshaller.marshal(request, System.out);

    URL wsUrl = new URL(primaryEndpointUrl);
    HttpURLConnection connection = openAndPrepareConnection(wsUrl);
    tryToMarshallWsRequestToOutputStream(request, marshaller, connection);

    printDebug(connection);

    Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
    Object response = unmarshaller.unmarshal(connection.getInputStream());

    // cleanup
    connection.disconnect();
    return responseClassType.cast(response);
}

private HttpURLConnection openAndPrepareConnection(URL wsUrl) throws IOException {
    HttpURLConnection connection = (HttpURLConnection) wsUrl.openConnection();
    connection.setDoOutput(true);
    connection.setRequestProperty("Accept", "application/xml");
    connection.setRequestProperty("Accept-Charset", charset);
    // Both types ? Doesn't work
    connection.setRequestProperty("Content-Type", "application/xml;application/x-www-form-urlencoded");
    // Only app/xml ? it seems that query param is not added to request
    connection.setRequestProperty("Content-Type", "application/xml");
    // Only app/query param ? it seems that xml is not added to request
    connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");

    return connection;
}

private void tryToMarshallWsRequestToOutputStream(K request, Marshaller jaxbForRequestMarshaller,
        HttpURLConnection connection) throws JAXBException, IOException {
    OutputStream outputStream = null;
    try {
        outputStream = connection.getOutputStream();
        jaxbForRequestMarshaller.marshal(request, outputStream);
        addQueryParameters(outputStream);
    }
    finally {
        tryToClose(outputStream);
    }
}

private void addQueryParameters(OutputStream outputStream) throws IOException {
    String value = "none";
    String query = String.format("xmlmsg=%s", URLEncoder.encode(value, charset));
    outputStream.write(query.getBytes(charset));
}

protected void tryToClose(OutputStream outputStream) throws IOException {
    if (outputStream != null) {
        outputStream.close();
    }
}

非常感谢！

4

1 回答 1

6

好的，您正在处理两个不同的问题：

  1. 正文的内容类型。在这种情况下，如果您的正文是 XML 正文，那么它应该是：

    Content-Type: application/xml; charset="utf-8"

  2. 您的查询参数被剥离（或忽略）的事实。这是一个更复杂的问题。当通过 HTTP Post 提交表单时，它通常作为查询字符串参数发送，这看起来像是可能发生的事情。你能得到正在发送的东西的线路跟踪吗？您可能需要自己解析参数，而不是尝试将它们解析为表单主体。

如果您需要发送两种不同类型的数据，您可以考虑发送多部分表单之类的内容，或者发送一种正文类型，并添加额外的数据作为标题。

于 2013-08-16T18:06:00.410 回答