我想在数据库中插入一些项目。在主要活动中,我从用户检索信息并将其传递给另一个类进行一些解析。我的JSONObject一直显示为NULL。
如果我不清楚这个问题,我很抱歉。我试图尽可能地解释它。
以下是您欢迎输入的代码
public class MainActivity extends Activity {
/** THE FOLLOWING STRINGS WILL BE DISPLAYED IN LOGCAT */
final String TAG = "########-------MAIN ACTIVITY: LOGIN--------######";
final String URL = "http://46.51.193.32/timereport/ses/sessions";
UserHelper userAdapter;
UserHelper db;
EditText edit_password,edit_username,edit_company;
String regName;
int duration = Toast.LENGTH_LONG;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
db = new UserHelper(this);
userAdapter = new UserHelper(this);
edit_password = (EditText)findViewById(R.id.password);
edit_username = (EditText)findViewById(R.id.user_name);
edit_company = (EditText)findViewById(R.id.company_string);
Button login = (Button)findViewById(R.id.login_button);
login.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
JSONObject jsonobj = new JSONObject();
try{
JSONObject subJson = new JSONObject();
subJson.put("username", edit_username.getText().toString());
subJson.put("password", edit_password.getText().toString());
subJson.put("company", edit_company.getText().toString());
jsonobj.put("user", subJson);
}
catch(JSONException e) {
Log.i("","#####-----error at catch jsonexception-----#####");
}
HandleJSON.SendHttpPost(URL, jsonobj);
String regNameSplit[] = regName.split("-");
try{
userAdapter.openDatabase();
long id = db.insertIntoDatabase(edit_username.getText().toString(),edit_company.getText().toString(), edit_password.getText().toString(),regNameSplit[0], regNameSplit[2]);
Toast.makeText(getApplicationContext(), "You have successfully logged in as: " +"\n" +regNameSplit[0], duration).show();
Log.i(TAG, "Printing value of id which will be inserted only to remove warnings "+id);
userAdapter.closeDatabase();
}
catch(SQLiteException e){
e.printStackTrace();
}
}
});
}
}
这是我要发送要解析的JSON对象的类
public class HandleJSON{
UserHelper userAdapter;
private static final String TAG = "&&----HTTPClient-----**";
public static String SendHttpPost (String URL, JSONObject jsonobj) {
String regName = "";
try{
Log.v("Json object request is ",jsonobj.toString());
DefaultHttpClient httpClientInstance = GetHttpClient.getHttpClientInstance();
HttpPost httpPostRequest = new HttpPost(URL);
Log.v(TAG,"The url is "+URL);
StringEntity se;
se = new StringEntity(jsonobj.toString());
httpPostRequest.setEntity(se);
httpPostRequest.setHeader("Accept", "application/json");
httpPostRequest.setHeader("Content-type", "application/json");
long t = System.currentTimeMillis();
HttpResponse response = (HttpResponse) httpClientInstance.execute(httpPostRequest);
Log.i(TAG, "HTTPRESPONSE RECIEVED" +(System.currentTimeMillis()-t) + "ms");
String resultString = convertStreamToString(response.getEntity().getContent());
Log.v(TAG , "The response is " +resultString);
JSONObject jsonObj = new JSONObject(resultString);
JSONObject sessionJson = jsonObj.getJSONObject("session");
String sessionId = sessionJson.getString("sessionid");
String name = sessionJson.getString("name");
Log.v(TAG,"The session ID is "+sessionId);
Log.v(TAG,"The name is "+name);
regName = name+"-"+sessionId+"-"+URL;
} catch (Exception e){
e.printStackTrace();
}
return regName;
}
private 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();
} finally{
try {
is.close();
} catch (IOException e){
e.printStackTrace();
}
}
return sb.toString();
}
}
我刚刚添加了一些MainActivity中缺少的代码,
String regNameSplit[] = regName.split("-");
一直显示为空
答案 0 :(得分:2)
尝试使用提供的系统EntityUtils.toString(entity)
,而不是convertStreamToString()
方法。
重要提示:不要捕获泛型Exception
,这会隐藏未经检查的(运行时)异常。您可能隐藏JSONObject constructor中发生的JSONException
。
<强>更新强>
您正在调用SendHttpPost
而不是将结果分配给变量:
HandleJSON.SendHttpPost(URL, jsonobj);
应该是:
regName = HandleJSON.SendHttpPost(URL, jsonobj);
答案 1 :(得分:1)
我没有看到这个有什么问题,你能告诉我regname有什么用吗?
在您的主动活动中,只需更改以下内容:
regname = HandleJSON.SendHttpPost(URL, jsonobj);
您没有回拨的regname被分配给您在sendhttppost返回的name和sessionid。