使用JSON和PHP从Mysql数据库获取数据时出错
问题描述:
我正在学习构建android应用程序,我需要一些帮助从MySQL获取数据。我跟着this tutorial做,但我得到这个错误:使用JSON和PHP从Mysql数据库获取数据时出错
Error org.json.JSONEXception: Value of type java.lang.String cannot be converted to JSONObject
在我的MainActivity我有一个按钮,点击打开第二个活动的时候。第二项活动必须从数据库中获取并显示数据。这是第二个活动Restaurant.java
public class Restaurants extends Activity {
private String jsonResult;
private String url = "http://10.0.2.2/app1/GetRestaurants.php";
private ListView listView;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.restaurants);
listView = (ListView) findViewById(R.id.listView1);
accessWebService();
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
// Inflate the menu; this adds items to the action bar if it is present.
getMenuInflater().inflate(R.menu.main, menu);
return true;
}
// Async Task to access the web
private class JsonReadTask extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... params) {
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(params[0]);
try {
HttpResponse response = httpclient.execute(httppost);
jsonResult = inputStreamToString(
response.getEntity().getContent()).toString();
}
catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return null;
}
private StringBuilder inputStreamToString(InputStream is) {
String rLine = "";
StringBuilder answer = new StringBuilder();
BufferedReader rd = new BufferedReader(new InputStreamReader(is));
try {
while ((rLine = rd.readLine()) != null) {
answer.append(rLine);
}
}
catch (IOException e) {
// e.printStackTrace();
Toast.makeText(getApplicationContext(),
"Error..." + e.toString(), Toast.LENGTH_LONG).show();
}
return answer;
}
@Override
protected void onPostExecute(String result) {
ListDrwaer();
}
}// end async task
public void accessWebService() {
JsonReadTask task = new JsonReadTask();
// passes values for the urls string array
task.execute(new String[] { url });
}
// build hash set for list view
public void ListDrwaer() {
List<Map<String, String>> restaurantList = new ArrayList<Map<String, String>>();
try {
JSONObject jsonResponse = new JSONObject(jsonResult);
JSONArray jsonMainNode = jsonResponse.optJSONArray("restoranti");
for (int i = 0; i < jsonMainNode.length(); i++) {
JSONObject jsonChildNode = jsonMainNode.getJSONObject(i);
String name = jsonChildNode.optString("name");
String menu = jsonChildNode.optString("menu");
String outPut = name + "-" + menu;
restaurantList.add(createRestaurants("restoranti", outPut));
}
} catch (JSONException e) {
Toast.makeText(getApplicationContext(), "Error " + e.toString(),
Toast.LENGTH_SHORT).show();
}
SimpleAdapter simpleAdapter = new SimpleAdapter(this, restaurantList,
android.R.layout.simple_list_item_1,
new String[] { "restoranti" }, new int[] { android.R.id.text1 });
listView.setAdapter(simpleAdapter);
}
private HashMap<String, String> createRestaurants(String name, String menu) {
HashMap<String, String> restaurantNameNo = new HashMap<String, String>();
restaurantNameNo.put(name, menu);
return restaurantNameNo;
}
}
此的代码是restaurants.xml
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:paddingBottom="@dimen/activity_vertical_margin"
android:paddingLeft="@dimen/activity_horizontal_margin"
android:paddingRight="@dimen/activity_horizontal_margin"
android:paddingTop="@dimen/activity_vertical_margin"
tools:context=".MainActivity" >
<ListView
android:id="@+id/listView1"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:layout_alignParentTop="true"
android:layout_centerHorizontal="true"
android:layout_marginTop="14dp" >
</ListView>
请让我知道如果需要发布更多的代码。许多感谢的帮助。
我也发现this answer到几乎相似的帖子,并试图使相同,但没有为我工作。相同的错误和空白页面。
答
// Maybe this will help ...
private class JsonReadTask extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... params) {
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(params[0]);
try {
HttpResponse response = httpclient.execute(httppost);
jsonResult = inputStreamToString(
response.getEntity().getContent()).toString();
// LETS RETURN SOMETING ...
return jsonResult;
}
catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return null;
}
response.getEntity()。getContent(),看起来你没有得到一个JSON对象的内容是什么?加上你总是返回null(所以没有结果在相同的方法) – Drejc 2014-10-29 13:14:53