Question

我的应用中有一个按钮，当我点击它时，它会声明方法insertintodatabase。但是，当我点击它时，没有任何事情发生，即使log没有显示任何内容。问题出在哪儿？请建议。

private final OkHttpClient client = new OkHttpClient();
    public void SignUp(View view)
    {
        insertToDatabase();

    }
    private void insertToDatabase(){
        class SendPostReqAsyncTask extends AsyncTask<String, Void, String> {
            @Override
            protected  void onPreExecute()
            {
                name = usernam.getText().toString();
                pass = passw.getText().toString();
                emails = email.getText().toString();
                Log.e("GetText","called");

            }
            @Override
            protected String doInBackground(String... params) {

                String json = "";

                try {
                    JSONObject jsonObject = new JSONObject();
                    jsonObject.accumulate("name", name);
                    jsonObject.accumulate("password", pass);
                    jsonObject.accumulate("email", emails);
                    json = jsonObject.toString();
                    Log.e("MYAPP", "getjson");

                } catch (JSONException e) {
                    Log.e("MYAPP", "unexpected JSON exception", e);
                }
                try{
                    RequestBody formBody = new FormEncodingBuilder()
                            .add("result", json)
                            .build();
                    Request request = new Request.Builder()
                            .url("https://justedhak.comlu.com/receiver.php")
                            .post(formBody)
                            .build();

                    Response response = client.newCall(request).execute();
                    if (!response.isSuccessful()) throw new IOException("Unexpected code " + response);


            } catch (IOException e){
                    Log.e("MYAPP", "unexpected JSON exception", e);
                }

                    return "success";
            }

            @Override
            protected void onPostExecute(String result) {
                super.onPostExecute(result);

                Toast.makeText(getApplicationContext(), result, Toast.LENGTH_LONG).show();

            }

php

$username= $_POST['username'];
$password= $_POST['password'];
$email= $_POST['email'];
$image =$_POST['image'];
$sql = "insert into USERS (username,password,email) values ('$username','$password','$email')";
 if(mysqli_query($con,$sql)){
  echo 'success';
}
else{
echo 'failure';
  }
mysqli_close($con);

Answer 1

某处你必须创建SendPostReqAsyncTask的对象并调用execute方法传递参数......

// Example class
class DownloadFilesTask extends AsyncTask<URL, Integer, Long> {
 // some code
}

// Call the execute method of Async Task
new DownloadFilesTask().execute(url1, url2, url3);

Answer 2

//像这样使用..

public class <Your Root class> extends Activity{

    @Override
    public void onCreate(Bundle savedInstanceState) {
    // some code

         private final OkHttpClient client = new OkHttpClient();
         public void SignUp(View view)
         {
            new SendPostReqAsyncTask().execute();
         }
    }
    class SendPostReqAsyncTask extends AsyncTask<String, Void, String> {
          @Override
          protected  void onPreExecute()
          {
            name = usernam.getText().toString();
            pass = passw.getText().toString();
            emails = email.getText().toString();
            Log.e("GetText","called");

         }
         @Override
         protected String doInBackground(String... params) {

         String json = "";

                        try {
                            JSONObject jsonObject = new JSONObject();
                            jsonObject.accumulate("name", name);
                            jsonObject.accumulate("password", pass);
                            jsonObject.accumulate("email", emails);
                            json = jsonObject.toString();
                            Log.e("MYAPP", "getjson");

                        } catch (JSONException e) {
                            Log.e("MYAPP", "unexpected JSON exception", e);
                        }
                        try{
                            RequestBody formBody = new FormEncodingBuilder()
                                    .add("result", json)
                                    .build();
                            Request request = new Request.Builder()
                                    .url("https://justedhak.comlu.com/receiver.php")
                                    .post(formBody)
                                    .build();

                            Response response = client.newCall(request).execute();
                            if (!response.isSuccessful()) throw new IOException("Unexpected code " + response);


                    } catch (IOException e){
                            Log.e("MYAPP", "unexpected JSON exception", e);
                        }

                            return "success";
                    }

                    @Override
                    protected void onPostExecute(String result) {
                        super.onPostExecute(result);

                        Toast.makeText(getApplicationContext(), result, Toast.LENGTH_LONG).show();

                    }

}

Answer 3

因为OkHttp支持异步，我想你也可以参考以下方式：

我们假设您mHandler = new Handler(Looper.getMainLooper());

内有onCreate

private void updateToDatabase() {
        // POST request
        OkHttpClient client = new OkHttpClient();
        RequestBody requestBody = new FormEncodingBuilder()
                .add("key1", "value1")
                .add("key2", "value2")
                .build();
        Request request = new Request.Builder()
                .url("http://...")
                .post(requestBody)
                .build();
        client.newCall(request).enqueue(new Callback() {
            @Override
            public void onFailure(final Request request, final IOException e) {
                mHandler.post(new Runnable() {
                    @Override
                    public void run() {
                        Toast.makeText(mContext, e.toString(), Toast.LENGTH_SHORT).show();
                    }
                });                                        
            }

            @Override
            public void onResponse(Response response) throws IOException {
                mHandler.post(new Runnable() {
                    @Override
                    public void run() {
                        Toast.makeText(mContext, response.body().string(), Toast.LENGTH_SHORT).show(); 
                    }
                });                                       
            }
        });
    }

如果您仍想使用AsyncTask，请按以下步骤更新您的代码：

public class MainActivity extends AppCompatActivity {
    ...
    public void SignUp(View view)
    {
        new SendPostReqAsyncTask().execute();
    }

    ...
    private class SendPostReqAsyncTask extends AsyncTask<String, Void, String> {
        @Override
        protected void onPreExecute() {
            name = usernam.getText().toString();
            pass = passw.getText().toString();
            emails = email.getText().toString();
            Log.e("GetText", "called");
        }
        @Override
        protected String doInBackground(String... params) {
            String json = "";
            try {
                JSONObject jsonObject = new JSONObject();
                jsonObject.accumulate("name", name);
                jsonObject.accumulate("password", pass);
                jsonObject.accumulate("email", emails);
                json = jsonObject.toString();
                Log.e("MYAPP", "getjson");
            } catch (JSONException e) {
                Log.e("MYAPP", "unexpected JSON exception", e);
            }
            try {
                OkHttpClient client = new OkHttpClient();
                RequestBody formBody = new FormEncodingBuilder()
                        .add("result", json)
                        .build();
                Request request = new Request.Builder()
                        .url("https://justedhak.comlu.com/receiver.php")
                        .post(formBody)
                        .build();
                Response response = client.newCall(request).execute();
                if (!response.isSuccessful()) throw new IOException("Unexpected code " + response);
            } catch (IOException e) {
                Log.e("MYAPP", "unexpected JSON exception", e);
            }
            return "success";
        }
        @Override
        protected void onPostExecute(String result) {
            super.onPostExecute(result);
            Toast.makeText(getApplicationContext(), result, Toast.LENGTH_LONG).show();
        }
    }
}

单击按钮时，async方法未运行

3 个答案: