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java - java.lang.String 类型的值数据库无法转换为 JSONObject

转载 作者:太空宇宙 更新时间:2023-11-04 12:37:21 25 4
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使用下面的代码给出的数据插入到数据库中,但不显示是否成功的对话框。

我的目标 API 是 23。

我应该使用gson吗?

请帮忙。

public class BackgroundTask extends AsyncTask<String, Void, String> {

String register_url = "http://192.168.1.101/loginapp/register.php";

Context ctx;

Activity activity;

ProgressDialog progressDialog;

AlertDialog.Builder builder;

public BackgroundTask(Context ctx)
{
this.ctx = ctx;
activity = (Activity)ctx;
}

@Override
protected void onPreExecute() {
builder = new AlertDialog.Builder(activity);
progressDialog = new ProgressDialog(ctx);
progressDialog.setTitle("Please wait...");
progressDialog.setMessage("Connecting to server...");
progressDialog.setIndeterminate(true);
progressDialog.setCancelable(false);
progressDialog.show();

super.onPreExecute();
}

@Override
protected String doInBackground(String... params) {
String method = params[0];
if(method.equals("register"))
{
try {
URL url = new URL(register_url);
HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection();
httpURLConnection.setRequestMethod("POST");
httpURLConnection.setDoOutput(true);
httpURLConnection.setDoInput(true);
OutputStream outputStream = httpURLConnection.getOutputStream();
BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream,"UTF-8"));
String name = params[1];
String email = params[2];
String password = params[3];
String data = URLEncoder.encode("name", "UTF-8")+"="+URLEncoder.encode(name,"UTF-8")+"&"+
URLEncoder.encode("email","UTF-8")+"="+URLEncoder.encode(email,"UTF-8")+"&"+
URLEncoder.encode("password","UTF-8")+"="+URLEncoder.encode(password,"UTF-8");
bufferedWriter.write(data);
bufferedWriter.flush();
bufferedWriter.close();
outputStream.close();
InputStream inputStream = httpURLConnection.getInputStream();
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
StringBuilder stringBuilder = new StringBuilder();
String line = "";
while((line=bufferedReader.readLine()) != null)
{
stringBuilder.append(line+"\n");
}
httpURLConnection.disconnect();
Thread.sleep(5000);
// return string builder as normal string.
return stringBuilder.toString();

} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
return null;
}

@Override
protected void onProgressUpdate(Void... values) {
super.onProgressUpdate(values);
}

@Override
protected void onPostExecute(String json) {
try {
progressDialog.dismiss();
JSONObject jsonObject = new JSONObject(json);
//need to get json array from json object.
JSONArray jsonArray = jsonObject.getJSONArray("server_response");
//now we can get each of the json data from json array.
JSONObject jo = jsonArray.getJSONObject(0);
//now we can read each data from json object;
String code = jo.getString("code");
String message = jo.getString("message");
if(code.equals("reg_true"))
{
showDialog("Registration Success..",message,code);
}
else if(code.equals("reg_false"))
{
showDialog("Registration failed..",message,code);
}
} catch (JSONException e) {
e.printStackTrace();
}


}

public void showDialog(String title, String message, String code)
{
builder.setTitle(title);
if(code.equals("reg_true")||code.equals("reg_false"))
{
builder.setMessage(message);
builder.setPositiveButton("OK", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
dialog.dismiss();
activity.finish();
}
});
AlertDialog alertDialog = builder.create();
alertDialog.show();

}
}
}

最佳答案

doInBackground()方法的返回值被传递给onPostExecute()。在这个方法里面:

    @Override
protected String doInBackground(String... params) {
String method = params[0];
if(method.equals("register"))
{
try {
URL url = new URL(register_url);
HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection();
//....
// return string builder as normal string.
return stringBuilder.toString();

} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
return null; //YOU ARE RETURNING NULL. CHECK WHETHER params[0] is "register"
}
//....
@Override
protected void onPostExecute(String json) { //json might be null. Try to print it using log.debug()
try {
progressDialog.dismiss();
JSONObject jsonObject = new JSONObject(json); //This might throw exception
//...
} catch (JSONException e) {
e.printStackTrace();
}
}

将日志粘贴到pastebin中并在此处附加链接。另请参阅:String cannot be converted to JSONObject exception

关于java - java.lang.String 类型的值数据库无法转换为 JSONObject,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37184835/

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