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java - 如何在android上将JSONObject显示到TextView中

转载 作者:行者123 更新时间:2023-11-30 07:05:46 25 4
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我已经设置了一个 php 脚本来创建 json here ,但是当我尝试显示 JSONobject 时,我在 Android 监视器上遇到了类似这样的错误。

Value (html)(body)(script of type java.lang.String cannot be converted to JSONObject

谁能告诉我如何解决这个问题吗?下面我的代码来尝试

my MainActivity.java

package flix.yudi.okhttp1;
import android.app.ProgressDialog;
import android.os.AsyncTask;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.util.Log;
import android.widget.ListAdapter;
import android.widget.ListView;
import android.widget.SimpleAdapter;
import android.widget.Toast;

import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

import java.util.ArrayList;
import java.util.HashMap;

public class MainActivity extends AppCompatActivity {

private String TAG = MainActivity.class.getSimpleName();

private ProgressDialog pDialog;
private ListView lv;

// URL to get contacts JSON
private static String url = "http://zxccvvv.cuccfree.com/send_data.php";

ArrayList<HashMap<String, String>> contactList;

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);

contactList = new ArrayList<>();

lv = (ListView) findViewById(R.id.list);

new GetContacts().execute();
}

/**
* Async task class to get json by making HTTP call
*/
private class GetContacts extends AsyncTask<Void, Void, Void> {

@Override
protected void onPreExecute() {
super.onPreExecute();
// Showing progress dialog
pDialog = new ProgressDialog(MainActivity.this);
pDialog.setMessage("Please wait...");
pDialog.setCancelable(false);
pDialog.show();

}

@Override
protected Void doInBackground(Void... arg0) {
HttpHandler sh = new HttpHandler();

// Making a request to url and getting response
String jsonStr = sh.makeServiceCall(url);

Log.e(TAG, "Response from url: " + jsonStr);

if (jsonStr != null) {
try {
JSONObject jsonObj = new JSONObject(jsonStr);

// Getting JSON Array node
JSONArray contacts = jsonObj.getJSONArray("");

// looping through All Contacts
for (int i = 0; i < contacts.length(); i++) {
JSONObject c = contacts.getJSONObject(i);

String id = c.getString("id");
String ask = c.getString("ask");

// tmp hash map for single contact
HashMap<String, String> contact = new HashMap<>();

// adding each child node to HashMap key => value
contact.put("id", id);
contact.put("ask", ask);

// adding contact to contact list
contactList.add(contact);
}
} catch (final JSONException e) {
Log.e(TAG, "Json parsing error: " + e.getMessage());
runOnUiThread(new Runnable() {
@Override
public void run() {
Toast.makeText(getApplicationContext(),
"Json parsing error: " + e.getMessage(),
Toast.LENGTH_LONG)
.show();
}
});

}
} else {
Log.e(TAG, "Couldn't get json from server.");
runOnUiThread(new Runnable() {
@Override
public void run() {
Toast.makeText(getApplicationContext(),
"Couldn't get json from server. Check LogCat for possible errors!",
Toast.LENGTH_LONG)
.show();
}
});

}

return null;
}

@Override
protected void onPostExecute(Void result) {
super.onPostExecute(result);
// Dismiss the progress dialog
if (pDialog.isShowing())
pDialog.dismiss();
/**
* Updating parsed JSON data into ListView
* */
ListAdapter adapter = new SimpleAdapter(
MainActivity.this, contactList,
R.layout.list_item, new String[]{"ask"}, new int[]{R.id.ask});

lv.setAdapter(adapter);
}

}
}

my HttpHandler.java

package flix.yudi.okhttp1;

import android.util.Log;

import java.io.BufferedInputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.ProtocolException;
import java.net.URL;

public class HttpHandler {

private static final String TAG = HttpHandler.class.getSimpleName();

public HttpHandler() {
}

public String makeServiceCall(String reqUrl) {
String response = null;
try {
URL url = new URL(reqUrl);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("GET");
// read the response
InputStream in = new BufferedInputStream(conn.getInputStream());
response = convertStreamToString(in);
} catch (MalformedURLException e) {
Log.e(TAG, "MalformedURLException: " + e.getMessage());
} catch (ProtocolException e) {
Log.e(TAG, "ProtocolException: " + e.getMessage());
} catch (IOException e) {
Log.e(TAG, "IOException: " + e.getMessage());
} catch (Exception e) {
Log.e(TAG, "Exception: " + e.getMessage());
}
return response;
}

private String convertStreamToString(InputStream is) {
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();

String line;
try {
while ((line = reader.readLine()) != null) {
sb.append(line).append('\n');
}
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
is.close();
} catch (IOException e) {
e.printStackTrace();
}
}
return sb.toString();
}
}

我从here获取了引用码

有人可以帮我解决这个错误吗?

Edited code

if (jsonStr != null) {
try {
JSONArray jsonObj = new JSONArray(jsonStr);

// Getting JSON Array node
JSONArray pertanyaan = jsonObj.getJSONArray("pertanyaan");

// looping through All Contacts
for (int i = 0; i < pertanyaan.length(); i++) {
JSONArray c = pertanyaan.getJSONArray(i);

String id = c.getString("id");
String ask = c.getString("ask");

Edit

Error
I/OpenGLRenderer: Initialized EGL, version 1.4
E/MainActivity: Response from url: <html><body><script type="text/javascript" src="/aes.js" ></script>
<script>function toNumbers(d){var e=[];d.replace(/(..)/g,function(d){e.push(parseInt(d,16))});return e}function
toHex(){for(var d=[],d=1==arguments.length&&arguments[0].constructor==Array?arguments[0]:arguments,e="",f=0;f<d.length;f++)e+=(16>d[f]?"0":"")+d[f].toString(16);return e.toLowerCase()}
var a=toNumbers("f655ba9d09a112d4968c63579db590b4"),b=toNumbers("98344c2eee86c3994890592585b49f80"),c=toNumbers("455e95bd78dbe99a933749187199f824");
document.cookie="__test="+toHex(slowAES.decrypt(c,2,a,b))+"; expires=Thu, 31-Dec-37 23:55:55 GMT; path=/";
location.href="http://zxccvvv.cuccfree.com/send_data.php?i=1";</script><noscript>This site requires Javascript to work, please enable Javascript in your browser or use a browser with Javascript support</noscript></body></html>
E/MainActivity: Json parsing error: Value <html><body><script of type java.lang.String cannot be converted to JSONArray
V/RenderScript: 0x5598622250 Launching thread(s), CPUs 8

最佳答案

如果您有权访问服务器,则可以使其返回以下格式的字符串:

{"contacts":[{"id":"1","ask":"pertanyaan ke 1"},{"id":"2","ask":"pertanyaan ke 2"},{"id":"3","ask":"pertanyaan ke 3"},{"id":"4","ask":"pertanyaan ke 4"},{"id":"5","ask":"pertanyaan ke 5"}]}

如果您想将其读取为 JSONObject ,其中 JSONObject jsonObj = new JSONObject(jsonStr);

然后使用JSONArray contacts = jsonObj.getJSONArray("contacts");来解析它

<小时/>

编辑后:

将此 JSONArray c = pertanyaan.getJSONArray(i); 更改为 JSONObject c = pertanyaan.getJsonObject(i)

关于java - 如何在android上将JSONObject显示到TextView中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40155347/

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