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java - 使用 JSON 和 PHP 从 Mysql DB 获取数据时出错

转载 作者:行者123 更新时间:2023-11-29 12:36:58 25 4
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我正在学习构建 Android 应用程序,我需要一些帮助来从 MySQL 获取数据。我关注了this tutorial这样做,但我收到此错误:

Error org.json.JSONEXception: Value of type java.lang.String cannot be converted to JSONObject

在我的 MainActivity 上,我有一个按钮,单击该按钮将打开第二个 Activity 。第二个 Activity 必须从数据库获取并显示数据。这是第二个 Activity Restaurant.java

的代码
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;
}
}

这是餐厅.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>

如果需要发布更多代码,请告诉我。非常感谢您的帮助。

我还发现了this answer到了几乎相似的帖子,并试图做出相同的事情,但对我不起作用。相同的错误和空白页。

最佳答案

// 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;
}

关于java - 使用 JSON 和 PHP 从 Mysql DB 获取数据时出错,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26631267/

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