1

我可以使用 HTTP 发布和响应来检索 xml 文件。但是，现在我需要发布一个String参数以及 URL。以下代码告诉我 HTTP 状态代码不正确（即 500），因此它返回null，然后我得到一个NullPointerException.

package com.JobsWebService;
import java.io.IOException;
import java.io.Reader;
import java.io.StringReader;
import java.io.UnsupportedEncodingException;
import java.util.ArrayList;
import java.util.List;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.HttpStatus;
import org.apache.http.NameValuePair;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;
import org.apache.http.protocol.HTTP;
import org.apache.http.util.EntityUtils;
import org.simpleframework.xml.Serializer;
import org.simpleframework.xml.core.Persister;
import android.util.Log;

public class XmlConnection {

private static final String url =         "http://www.accuservlite.com/AccumobileWS/TestService.asmx/RyanMB_GetJobs";
private DefaultHttpClient client = new DefaultHttpClient();
String param= "Johnny";

public List<NewJob> RyanMB_GetJobs() {
    try {
        String xmlData = retrieve(url);
        Serializer serializer = new Persister();
        Reader reader = new StringReader(xmlData);
        ArrayOfNewJob testService = serializer.read(ArrayOfNewJob.class,
                reader, false);
        Log.i("gary", "Worked");
        return testService.NewJob;

    } catch (Exception e) {
        Log.i("gary", e.toString());
    }
    return null;
}

// method retrieve
public String retrieve(String url) throws UnsupportedEncodingException {

    List<NameValuePair> qparams = new ArrayList<NameValuePair>();
    qparams.add(new BasicNameValuePair("Johnny", "Johhny"));

    UrlEncodedFormEntity postEntity = new UrlEncodedFormEntity(qparams,
            HTTP.UTF_8);

    HttpPost getRequest = new HttpPost(url);

    getRequest.setEntity(postEntity);

    try {

        HttpResponse getResponse = client.execute(getRequest);

        final int statusCode = getResponse.getStatusLine().getStatusCode();

        if (statusCode != HttpStatus.SC_OK) {

            return null;
        }

        HttpEntity getResponseEntity = getResponse.getEntity();

        if (getResponseEntity != null) {
            return EntityUtils.toString(getResponseEntity);
        }

    } catch (IOException e) {
        Log.i("gary", "Error for URL " + url, e);
        getRequest.abort();
    }
    return null;
}

}

4

2 回答 2

0

要在您的 HttpPost 对象中添加 POST 参数，您将使用 NameValuPair 列表：

List<NameValuePair> parameters = new ArrayList<NameValuePair>();
parameters.add(new BasicNameValuePair("name_param1", "value_param1"));

并将此参数添加到您的 HttpPost ：

UrlEncodedFormEntity formEntity  = new UrlEncodedFormEntity(parameters);
httpPost.setEntity(formEntity);
于 2012-07-10T17:26:37.633 回答
0

500 是一个服务器错误，我觉得这段代码没有明显错误。您是否针对发布请求测试了服务器？当您给它一个名称值对时，服务器是否需要一个字符串？

您可以尝试使用Wfetch进行测试。

于 2012-07-10T17:23:15.657 回答