Android请求中的Post参数
我正在做一个Android应用程序，但在针对自己的服务器执行请求时遇到了问题。我使用Play Framework制作了服务器，并从Json获取参数：Android请求中的Post参数,android,post,parameters,request,Android,Post,Parameters,Request,我正在做一个Android应用程序，但在针对自己的服务器执行请求时遇到了问题。我使用Play Framework制作了服务器，并从Json获取参数： response.setContentTypeIfNotSet("application/json; charset=utf-8"); JsonParser jsonParser = new JsonParser(); JsonElement jsonElement = jsonParser.parse(getBody(re
response.setContentTypeIfNotSet("application/json; charset=utf-8");
JsonParser jsonParser = new JsonParser();
JsonElement jsonElement = jsonParser.parse(getBody(request.body));
Long id =jsonElement.getAsJsonObject().get("id").getAsLong();
HttpPost postRequest = new HttpPost(host);
// Add headers
for(NameValuePair h : headers)
{
postRequest.addHeader(h.getName(), h.getValue());
}
if(!params.isEmpty())
{
postRequest.setEntity(new UrlEncodedFormEntity(params, HTTP.UTF_8));
}
executeRequest(postRequest, host);
break;
HttpParams HttpParams=新的BasicHttpParams（） 还有不同的错误，这取决于我的请求。它们都是“play.exceptions.JavaExecutionException”：
- 'com.google.gson.stream.MalformedJsonException'
- '这不是JSON对象'
- '找到预期对象：“id”'
我希望有人能帮助我。这里有一个发送HTTP帖子的简单方法
HttpPost httppost = new HttpPost("Your URL here");
httppost.setEntity(new StringEntity(paramsJson));
httppost.addHeader("content-type", "application/json");
HttpResponse response = httpclient.execute(httppost);
您最好直接使用JSON字符串，而不是在这里解析它。希望它有帮助。这里是发送HTTP帖子的简单方法
HttpPost httppost = new HttpPost("Your URL here");
httppost.setEntity(new StringEntity(paramsJson));
httppost.addHeader("content-type", "application/json");
HttpResponse response = httpclient.execute(httppost);
Try this,It may help u
public void executeHttpPost(String string) throws Exception
{
//This method for HttpConnection
try
{
HttpClient client = new DefaultHttpClient();
HttpPost request = new HttpPost("URL");
List<NameValuePair> value=new ArrayList<NameValuePair>();
value.add(new BasicNameValuePair("Name",string));
UrlEncodedFormEntity entity=new UrlEncodedFormEntity(value);
request.setEntity(entity);
client.execute(request);
System.out.println("after sending :"+request.toString());
}
catch(Exception e) {System.out.println("Exp="+e);
}
}
Try this,It may help u
public void executeHttpPost(String string) throws Exception
{
//This method for HttpConnection
try
{
HttpClient client = new DefaultHttpClient();
HttpPost request = new HttpPost("URL");
List<NameValuePair> value=new ArrayList<NameValuePair>();
value.add(new BasicNameValuePair("Name",string));
UrlEncodedFormEntity entity=new UrlEncodedFormEntity(value);
request.setEntity(entity);
client.execute(request);
System.out.println("after sending :"+request.toString());
}
catch(Exception e) {System.out.println("Exp="+e);
}
}
试试这个，它可能会对你有所帮助
public void executehttpost（字符串）引发异常
{
//此方法适用于HttpConnection
尝试
{
HttpClient=new DefaultHttpClient（）；
HttpPost请求=新的HttpPost（“URL”）；
列表值=新的ArrayList（）；
add（新的BasicNameValuePair（“名称”，字符串））；
UrlEncodedFormEntity实体=新的UrlEncodedFormEntity（值）；
请求。设置实体（实体）；
执行（请求）；
System.out.println（“发送后：+request.toString（））；
}
catch（异常e）{System.out.println（“Exp=”+e）；
}
}