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尽管我知道这个问题已经多次发布在堆栈溢出上了，但我已经尝试了所有的解决方案，并没有为我工作。Android数据从数据库中取数据库不工作（php + json + mysql）
我试图在列表视图中显示数据。数据以json格式存储，在读取数据时，似乎android并未读取JSON数组，因此数据不会在列表视图中显示，而是保持为空。
的Java文件：
public class ReadResult extends ListActivity {
private ProgressDialog pDialog;
JSONParser jParser = new JSONParser();
ArrayList<HashMap<String, String>> ResultFetch;
private static String url_readResult = "http://10.0.2.2/Result-Viewer/php/ReadData.php";
private static final String TAG_SUCCESS = "success";
private static final String TAG_SCORE = "Score";
private static final String TAG_SEMESTER = "Semester";
private static final String TAG_PRODUCTS = "products";
JSONArray products = null;
ListView list;
@Override
protected void onCreate(Bundle savedInstanceState) {
// TODO Auto-generated method stub
super.onCreate(savedInstanceState);
setContentView(R.layout.read_result);
list = (ListView) getListView();
ResultFetch = new ArrayList<HashMap<String, String>>();
new LoadResult().execute();
}
class LoadResult extends AsyncTask<String, String, String> {
protected void onPreExecute() {
super.onPreExecute();
pDialog = new ProgressDialog(ReadResult.this);
pDialog.setMessage("Loading Result. Please wait...");
pDialog.setIndeterminate(false);
pDialog.setCancelable(false);
pDialog.show();
}
@Override
protected String doInBackground(String... params) {
// TODO Auto-generated method stub
List<NameValuePair> param = new ArrayList<NameValuePair>();
JSONObject json = jParser.makeHttpRequest(url_readResult, "GET",
param);
Log.d("Result: ", json.toString());
try {
int success = json.getInt(TAG_SUCCESS);
if (success == 1) {
JSONObject object = new JSONObject(json.toString());
products = object.getJSONArray(TAG_PRODUCTS);
for (int i = 0; i < products.length(); i++) {
JSONObject c = products.getJSONObject(i);
String Semester = c.getString(TAG_SEMESTER);
String Score = c.getString(TAG_SCORE);
HashMap<String, String> map = new HashMap<String, String>();
map.put(TAG_SEMESTER, Semester);
map.put(TAG_SCORE, Score);
ResultFetch.add(map);
}
} else {
Toast toast = Toast.makeText(getApplicationContext(),
"No Result Found", Toast.LENGTH_SHORT);
toast.show();
}
} catch (JSONException e) {
e.printStackTrace();
}
return null;
}
protected void onPostExecute(String file_url) {
pDialog.dismiss();
runOnUiThread(new Runnable() {
public void run() {
ListAdapter adapter = new SimpleAdapter(ReadResult.this,
ResultFetch, R.layout.list_item, new String[] {
TAG_SEMESTER, TAG_SCORE }, new int[] {
R.id.semester, R.id.score });
list.setAdapter(adapter);
}
});
}
}
}
php文件：
<?php
session_start();
$cid = $_SESSION["cid"];
$rno = $_SESSION["rno"];
$response = array();
$conn=new PDO('mysql:host=localhost;dbname=result','root' ,'');
$result=$conn->query("Select * from $cid where RegistrationNumber = '$rno'");
if($result->rowcount()>0)
{
$response["products"] = array();
foreach($result as $row)
{
$product["Semester"] = $row["Semester"];
$product["Score"] = $row["Score"];
$response["success"] = 1;
array_push($response["products"], $product);
echo json_encode($response);
}
}
else
{
$response["success"] = 0;
$response["message"] = "No record found.";
echo json_encode($response);
}
?>
我已经通过了通过手的CID，RNO和密码页面，以下是生成的JSON输出通过运行读取数据页面。
JSON输出：
{"products":[{"Semester":"0","Score":"0"}],"success":1}
所以你做了什么来调试呢？你检查什么是在Android中到达？也许你的PHP正在吐出警告，腐蚀JSON。 –
我正在读取结果页面中的进度条，但之后没有任何工作。我在结果页面的日志文件中获得成功= 1。之后，我得到一个警告/错误：对于产品没有价值。 –
您是否考虑过使用Gson将您的json映射到Java对象？它只会在几行内实现你想要做的。 –