我使用JSON从android向php发送数据。为此,我使用了以下代码
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://futuretime.in/reg.php");
httppost.setHeader("content-type", "application/json");
JSONObject dataJson = new JSONObject();
dataJson.put("password", password);
dataJson.put("number", Integer.parseInt("5556"));
StringEntity entity = new StringEntity(dataJson.toString());
httppost.setEntity(entity);
HttpResponse response = httpclient.execute(httppost);
text=response.toString();
当我执行此操作时,我收到的消息如下:org.apache.http.message.basichttpresponse 4fb06e5e
答案 0 :(得分:0)
答案 1 :(得分:0)
你应该致电 - response.getEntity().getContent()
答案 2 :(得分:0)
试试这个只是在这个方法中传递你的网址
public void getUrlData(String url)
{
String result="";
//Making HTTP request
try
{
HttpParams httpParameters = new BasicHttpParams();
HttpConnectionParams.setConnectionTimeout(httpParameters, 30000);
HttpConnectionParams.setSoTimeout(httpParameters, 30000);
//defaultHttpClient
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
//httpPost.setEntity(new UrlEncodedFormEntity(params));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
inputStream = httpEntity.getContent();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try
{
BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line);
}
inputStream.close();
result = sb.toString();
}
catch (Exception e)
{
url_error = 1;
System.out.println(e.toString());
}
}
然后从String结果中检索您的数据,您可以将结果放入日志中以知道数据