我是初学Android程序员,我正在开发一个从本机设备电话簿中读取联系人并将其写入数据库()的程序。
我可以通过手机阅读和阅读联系人,但我仍然坚持“将联系人写入数据库”。 这是我必须遵循的准则:
write function:
write:
[{'activity':'writeData',
'firstname':'Janis',
'lastname':'Berzins',
'telnr': '12312312'}]
request should be sent as: [{'activity':'readData'}]
example for read: [{"firstname":"Vards","lastname":"Uzvards","telnr":"12345678"},{"firstname":"Viesturs","lastname":"Lapsa","telnr":"11223344"}]
我研究了无数教程,文档等等四天,这是我认为应该用于发送到数据库部分的工作:
public static HttpResponse doPost(String url, JSONObject c) throws ClientProtocolException, IOException
{
HttpParams httpParams = new BasicHttpParams();
HttpClient httpclient = new DefaultHttpClient(httpParams);
HttpPost request = new HttpPost(url);
StringEntity s = new StringEntity(c.toString());
s.setContentEncoding("UTF-8");
s.setContentType(new BasicHeader(HTTP.CONTENT_TYPE,"application/json"));
request.setEntity(s);
request.addHeader("accept", "application/json");
return httpclient.execute(request);
}
为了创建一个简单的测试Json,我使用了这个类:
public class RestPost extends AsyncTask<String, Void, String>
{
@Override
protected String doInBackground(String... params) {
JSONObject json = new JSONObject();
try
{
json.put("activity", "writeData");
json.put("firstname", "Janis");
json.put("lastname", "Berzins");
json.put("telnr", "123123123");
RestMethods.doPost(Main.server, json);
}
catch (JSONException e)
{
// TODO Auto-generated catch block
e.printStackTrace();
Log.i("Error: ", e.toString());
}
catch (ClientProtocolException e)
{
// TODO Auto-generated catch block
e.printStackTrace();
Log.i("Error: ", e.toString());
}
catch (IOException e)
{
// TODO Auto-generated catch block
e.printStackTrace();
Log.i("Error: ", e.toString());
}
Log.i("DONE ", " I GUES");
return null;
}
}
但是在我的Android应用程序执行此功能后 - 数据库()中没有任何变化。
那么,拜托,任何人都可以帮我弄清楚,我做错了什么?
谢谢!
答案 0 :(得分:0)
你需要一个json数组(参见规范中json周围的“[”和“]”字符。
JSONArray jsonArray = new JSONArray();
// your code to create the JSONObject
// ...
jsonArray.put(json);
// use this within your RestMethods call
RestMethods.doPost(Main.server, jsonArray);
你可能可以同时向服务器发送多个对象 - 即使“规范”没有提到它,使用数组也是一种暗示。
答案 1 :(得分:0)
好的,我让它发送到数据库! :) 现在我正在使用Get方法..这就是我得到的 - 现在似乎没有工作,可能是因为数据库中的数据是作为JSONArray发送的?
public static JSONObject doGet(String url)
{
JSONObject json = null;
HttpClient httpclient = new DefaultHttpClient();
HttpGet httpget = new HttpGet(url);
httpget.addHeader("accept", "application/json");
HttpResponse response;
try {
response = httpclient.execute(httpget);
HttpEntity entity = response.getEntity();
if (entity != null)
{
InputStream instream = entity.getContent();
String result= convertStreamToString(instream);
json=new JSONObject(result);
instream.close();
}
}
catch (ClientProtocolException e)
{
// TODO Auto-generated catch block
e.printStackTrace();
Log.i("Error: ", e.toString());
}
catch (IOException e)
{
// TODO Auto-generated catch block
e.printStackTrace();
Log.i("Error: ", e.toString());
}
catch (JSONException e)
{
// TODO Auto-generated catch block
e.printStackTrace();
Log.i("Error: ", e.toString());
}
return json;
}
public static String convertStreamToString(InputStream is) {
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();
String line = null;
try
{
while ((line = reader.readLine()) != null)
{
sb.append(line + "\n");
}
}
catch (IOException e)
{
e.printStackTrace();
Log.i("Error: ", e.toString());
}
finally
{
try
{
is.close();
}
catch (IOException e)
{
e.printStackTrace();
}
}
return sb.toString();
}