Question

我正在开发一个应用程序，我需要同时更新2个PHP数据库。从第一个PHP DB我得到需要在主屏幕上显示的数据，因此它在onPostExecute中处理。但对于第二个PHP DB我需要检查并更新其中的记录，我无法弄清楚如何在一个Java程序中执行多个PHP请求。代码如下：

Java代码

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

    String json_url;
    Context context;

    public BackgroundTask (Context c) { context = c;}

    public BackgroundTask() {

    }

    @Override
    protected String doInBackground(String... params) {
        String type = params[0];
        String login_url = "... my link ..."; // to get data of the user
        if (type.equals("login")) {
            try {
                String user_name = params[1];

                URL url = new URL(login_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 post_data = URLEncoder.encode("user_name", "UTF-8") + "=" + URLEncoder.encode(user_name, "UTF-8");

                bufferedWriter.write(post_data);
                bufferedWriter.flush();
                bufferedWriter.close();
                outputStream.close();
                InputStream inputStream = httpURLConnection.getInputStream();
                BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream, "iso-8859-1"));
                StringBuilder stringBuilder = new StringBuilder();
                String results = "";
                String line = "";
                while ((STRING_JSON = bufferedReader.readLine()) != null) {
                    results += line;
                    stringBuilder.append(STRING_JSON + "\n");

                }
                bufferedReader.close();
                inputStream.close();
                httpURLConnection.disconnect();
                //connectionStatus = true;
                Log.i("doInBackgroundTry->try", "stringbuilder in the end is " + stringBuilder.toString());
                master_string = stringBuilder.toString().trim();
                return stringBuilder.toString().trim();
            } catch (MalformedURLException e) {
                Toast.makeText(context, "malformed exception error is " + e.toString(), Toast.LENGTH_LONG).show();
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return null;
    }

    @Override
    public void onPreExecute() {
        json_url = " ... my link ...";
    }

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

    @Override
    public void onPostExecute(String result) {

    // random updation of the UI

    }
}

所有第二个PHP DB都有一个ID，如果它已经在表中，将被检查，如果是，那么什么都不会发生，但如果不是那么它将被添加到数据库中。

Answer 1

首先，在调用execute时，你必须通过某个函数传递类型。您可以再次执行此操作，但使用其他类型来处理第二个查询。例如：

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

    String json_url;
    Context context;

    public BackgroundTask (Context c) { context = c;}

    public BackgroundTask() {

    }

    @Override
    protected String doInBackground(String... params) {
        String type = params[0];
        String login_url = "... my link ..."; // to get data of the user
        if (type.equals("login")) {

        return null;
    }

    else if (type.equals("second_query_type")
    {
        // bla bla bla 
        // random code bla bla bla

        // if you are not going to update your UI, then just return null here
    }

    @Override
    public void onPreExecute() {
        json_url = " ... my link ...";
    }

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

    @Override
    public void onPostExecute(String result) {

    // random updation of the UI

    }
}

现在，您可以使用三种方法执行第二个查询。

方法1：通过函数执行执行

public void second_query_type(String type)
{
 new BackgroundTask(type);
}

方法2：创建后台任务对象并调用执行函数

BackgroundTask bt = new BackgroundTask();
bt.execute("second_query")'

方法3：如果此类在另一个类中，请使用这种方式

new BackgroundTask.execute(type);

如果您有任何其他问题，请随时提出......

通过一个Java异步任务

1 个答案: