我想从MySQL检索数据到android listView。我已遵循此tutorial中的每个步骤,但是当我尝试调用JSON对象时它崩溃了。请帮忙。非常感谢..
private ListView listView;
String myJSON;
JSONArray information = null;
ArrayList<HashMap<String, String>> infoList;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
infoList = new ArrayList<HashMap<String, String>>();
listView = (ListView) findViewById(R.id.listView2);
getData();
}
public void getData() {
class GetDataJSON extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... params) {
DefaultHttpClient httpclient = new DefaultHttpClient(new BasicHttpParams());
HttpPost httppost = new HttpPost("http://192.168.107.115:80/Android/CRUD/retrieveInformation.php");
// Depends on your web service
httppost.setHeader("Content-type", "application/json");
InputStream inputStream = null;
String result = null;
try {
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
inputStream = entity.getContent();
// json is UTF-8 by default
BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream, "UTF-8"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null)
{
sb.append(line + "\n");
}
result = sb.toString();
} catch (Exception e) {
// Oops
}
finally {
try{if(inputStream != null)inputStream.close();}catch(Exception squish){}
}
return result;
}
@Override
protected void onPostExecute(String result){
myJSON=result;
showList();
}
}
GetDataJSON g = new GetDataJSON();
g.execute();
}
protected void showList(){
try {
JSONObject jsonObj = new JSONObject(myJSON);
information = jsonObj.getJSONArray(Config.TAG_RESULTS);
for(int i=0;i<information.length();i++){
JSONObject c = information.getJSONObject(i);
String date = c.getString(Config.TAG_DATE);
Toast.makeText(getApplicationContext(),date,Toast.LENGTH_LONG).show();
String timeIn = c.getString(Config.TAG_TiME_IN);
Toast.makeText(getApplicationContext(),timeIn,Toast.LENGTH_LONG).show();
String timeOut = c.getString(Config.TAG_TIME_OUT);
HashMap<String,String> info = new HashMap<String,String>();
info.put(Config.TAG_DATE, date);
info.put(Config.TAG_TiME_IN, timeIn);
info.put(Config.TAG_TIME_OUT,timeOut);
infoList.add(info);
}
ListAdapter adapter = new SimpleAdapter(
HomePage.this, infoList, R.layout.retrieve_data,
new String[]{Config.TAG_DATE,Config.TAG_TiME_IN,Config.TAG_TIME_OUT},
new int[]{R.id.date,R.id.timeIn,R.id.timeOut}
);
listView.setAdapter(adapter);
} catch (JSONException e) {
e.printStackTrace();
}
}
retrieveInformation.php
<?php
define('HOST','127.0.0.1:3307');
define('USER','root');
define('PASS','');
define('DB','androiddb');
$con = mysqli_connect(HOST,USER,PASS,DB) or die('unable to connect');
$sql = "select * from information";
$res = mysqli_query($con,$sql);
$result=array();
while($row=mysqli_fetch_array($res)){
array_push($result,array('id'=>$row[0],'name'=>$row[1],'weather'=>$row[2],'date'=>$row[3],'status'=>$row[4],
'time_in'=>$row[5], 'time_out'=>$row[6]));
}
print(json_encode(array("result"=>$result)));
mysqli_close($con);
?>
logcat的
Process: com.example.project.myapplication, PID: 31622
java.lang.NullPointerException
at org.json.JSONTokener.nextCleanInternal(JSONTokener.java:116)
at org.json.JSONTokener.nextValue(JSONTokener.java:94)
at org.json.JSONObject.<init>(JSONObject.java:155)
at org.json.JSONObject.<init>(JSONObject.java:172)
at com.example.project.myapplication.GUI.HomePage.showList(HomePage.java:185)
at com.example.project.myapplication.GUI.HomePage$1GetDataJSON.onPostExecute(HomePage.java:176)
at com.example.project.myapplication.GUI.HomePage$1GetDataJSON.onPostExecute(HomePage.java:137)
at android.os.AsyncTask.finish(AsyncTask.java:632)
代码
`JSONObject jsonObj = new JSONObject(myJSON);` ,
class GetDataJSON extends AsyncTask<String, Void, String> {
和showList();
。
答案 0 :(得分:0)
再犯一个愚蠢的错误。忘记打开wifi连接