2

这是我的 android 应用程序中的 httppost 方法。它不接受冗长的网址。冗长的网址没有响应/例外。当我在浏览器中手动输入相同的网址时，它工作正常。谁能在这里指出这个问题？

    try {
        // defaultHttpClient
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(url);

        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();           

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

更新：添加了一个示例网址。当在浏览器中手动输入相同的 url 并给出响应时，它可以正常工作。

 url.com/data?format=json&pro={%22merchanturl%22:%22http://url.com/logo.pn‌​g%22,%22price%22:599,%22productDesc%22:%22Apple%2032GBBlack%22,%22prodID%22:%2291‌​3393%22,%22merchant%22:%224536%22,%22prourl%22:%22http://url.com/data%22,%22name%‌​22:%22Apple%2032GB%20%2D%20Black%22,%22productUrl%22:%22http://www.url.com/image.‌​jpg%22,%22myprice%22:550,%22mercname%22:%22hello%22,%22mybool%22:false}
4

3 回答 3

1

我想您的网址包含类似index.php?call=getUsers&something=bla

要解决此问题，您可以使用NameValuePair：

String url = "http://example.com/index.php";

ArrayList<NameValuePair> nvp = new ArrayList<NameValuePair>();
nvp.add(new BasicNameValuePair("call", "getUsers"));
nvp.add(new BasicNameValuePair("something", "bla"));

try {
    HttpClient client = new DefaultHttpClient();
    HttpPost post = new HttpPost(url);
    post.setEntity(new UrlEncodedFormEntity(nvp));
    HttpResponse response = client.execute(post);
    HttpEntity entity = response.getEntity();
    [...]
} catch (Exception e) {
    [...]
}
于 2012-12-09T15:39:55.837 回答
1 
public void postData() {
// Create a new HttpClient and Post Header
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://www.yoursite.com/script.php");

try {
    // Add your data
    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
    nameValuePairs.add(new BasicNameValuePair("id", "12345"));
    nameValuePairs.add(new BasicNameValuePair("stringdata", "AndDev is Cool!"));
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

    // Execute HTTP Post Request
    HttpResponse response = httpclient.execute(httppost);

} catch (ClientProtocolException e) {
    // TODO Auto-generated catch block
} catch (IOException e) {
    // TODO Auto-generated catch block
}
}
于 2012-12-09T15:45:30.267 回答
0

您可以尝试使用以下代码。你应该有Json API。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.Reader;
import java.net.URL;
import java.nio.charset.Charset;

import org.json.JSONException;
import org.json.JSONObject;

public class JsonReader {

  private static String readAll(Reader rd) throws IOException {
    StringBuilder sb = new StringBuilder();
    int cp;
    while ((cp = rd.read()) != -1) {
  sb.append((char) cp);
}
return sb.toString();
  }

  public static JSONObject readJsonFromUrl(String url) throws IOException, JSONException {
    InputStream is = new URL(url).openStream();
    try {
      BufferedReader rd = new BufferedReader(new InputStreamReader(is, Charset.forName("UTF-8")));
  String jsonText = readAll(rd);
  JSONObject json = new JSONObject(jsonText);
  return json;
} finally {
  is.close();
    }
  }

  public static void main(String[] args) throws IOException, JSONException {
    JSONObject json = readJsonFromUrl("https://graph.facebook.com/19292868552");
    System.out.println(json.toString());
    System.out.println(json.get("id"));
  }
于 2012-12-09T16:26:40.750 回答