Question

您好，我正在申请中使用肥皂服务。 soap服务使用mozilla firefox REST客户端和SOA客户端。

请求正文：

<s:Envelope xmlns:s="http://www.w3.org/2003/05/soap-envelope" xmlns:a="http://www.w3.org/2005/08/addressing">
<s:Header>
<a:Action s:mustUnderstand="1">http://tempuri.org/IService/GetData</a:Action>
</s:Header>
<s:Body>
<GetData xmlns="http://tempuri.org/"><value>Hello</value></GetData>
</s:Body>
</s:Envelope>

代码：

private final String raw1 = "<s:Envelope xmlns:s=\"http://www.w3.org/2003/05/soap-envelope\" xmlns:a=\"http://www.w3.org/2005/08/addressing\">" +
            "<s:Header>" +
            "<a:Action s:mustUnderstand=\"1\">http://tempuri.org/IService/GetData</a:Action>" +
            "</s:Header>" +
            "<s:Body>" +
            "<GetData xmlns=\"http://tempuri.org/\">" +
            "<value>Hello</value>" +
            "</GetData>" +
            "</s:Body>" +
            "</s:Envelope>";
HttpClient client = new DefaultHttpClient();
HttpPost post = new HttpPost(URL);
post.setHeader("Content-type", "application/soap+xml");
post.setHeader("charset", "utf-8");
try {
    post.setEntity(new StringEntity(raw1, "UTF8"));
} catch (UnsupportedEncodingException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}
try {
    HttpResponse response = client.execute(post);
    Log.d("test", ""+response.getStatusLine());
    InputStream is = response.getEntity().getContent();
    BufferedReader br = new BufferedReader(new InputStreamReader(is));
    String str, res = ""; 
    while((str = br.readLine()) != null) {
        res += str;
    }
    br.close();
    is.close();
    Log.d("test", ""+res);
} catch(Exception e) {
    e.printStackTrace();
}

如果我使用这个原始请求正文并手动创建一个http帖子我能够在我的Android应用程序中点击服务，但如果我使用ksoap2

private String NAME_SPACE = "http://tempuri.org";
private String METHOD_NAME = "GetData";
private final String URL =  "http://example.amazonaws.com/PresenceServices/Service.svc";
private final String SOAP_ACTION = "http://tempuri.org/IService/GetData";

SoapObject request = new SoapObject(NAME_SPACE, METHOD_NAME);
request.addProperty("value", "hello");
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER12);
envelope.dotNet = true;
envelope.setOutputSoapObject(request);
HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
try {
    androidHttpTransport.call(SOAP_ACTION, envelope);
    SoapObject response = (SoapObject)envelope.getResponse();
    Log.e("Object response", response.toString());
    return response;
} catch (Exception e) {
    e.printStackTrace();
}

抛出 IOException Http请求失败Http状态500

请帮我在ksoap2中实现这一目标。

Answer 1

请在Working with SOAP in Android using KSOAP2

上找到我的教程

希望它会帮助你。

Answer 2

我认为问题是你没有将标题添加到信封中。（行动标题）

尝试

private Element[] buildHeader(String URL, String SOAP_ACTION) {
    List<Element> headers = new ArrayList<Element>();
    Element action = new Element().createElement(
            "http://www.w3.org/2005/08/addressing", "Action");
    action.addChild(Node.TEXT, SOAP_ACTION);
    action.setAttribute("http://www.w3.org/2005/08/addressing",
            "mustUnderstand", "1");
    headers.add(action);

    int size = headers.size();
    Element[] array = new Element[size];
    for (int i = 0; i < size; i++)
        array[i] = headers.get(i);
    return array;
}

并且

Element[] e = buildHeader(url, SOAP_ACTION);
soapEnvelope.headerOut = e;

ksoap2 android无法访问soap服务？

2 个答案: