obaro / simplewebapi Goto Github PK
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License: MIT License
I neither get a response nor do I get any progressBar movement.
package com.example.shreyanshu.sample;
import android.os.AsyncTask;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.ProgressBar;
import android.widget.TextView;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;
public class MainActivity extends AppCompatActivity {
EditText emailText;
TextView responseView;
ProgressBar progressBar;
static final String API_KEY = "key added";
static final String API_URL = " https://api.fullcontact.com/v2/person.json?";
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
emailText = (EditText) findViewById(R.id.emailText);
responseView = (TextView) findViewById(R.id.responseView);
progressBar = (ProgressBar) findViewById(R.id.progressBar);
Button queryButton = (Button) findViewById(R.id.queryButton);
queryButton.setOnClickListener(new View.OnClickListener() {
public void onClick(View view) {
new RetrieveFeedTask().execute();
}
});
}
class RetrieveFeedTask extends AsyncTask<Void, Void, String> {
private Exception exception;
String email;
public RetrieveFeedTask() {
this.email = emailText.getText().toString();
}
protected void OnPreExecute() {
progressBar.setVisibility(View.VISIBLE);
responseView.setText("About to start");
}
protected String doInBackground(Void...urls) {
try {
URL url = new URL(API_URL + "email=" + email + "&apiKey=" + API_KEY);
Log.e("URL", url.toString());
HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
try {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(urlConnection.getInputStream()));
StringBuilder stringBuilder = new StringBuilder();
String line;
while ((line = bufferedReader.readLine()) != null) {
stringBuilder.append(line).append("\n");
}
bufferedReader.close();
return stringBuilder.toString();
}
finally {
urlConnection.disconnect();
}
}
catch (Exception e) {
Log.e("Error", e.getMessage(), e);
return null;
}
}
protected void onPostExcecute(String response) {
if (response == null) {
response = "There was an ERROR";
}
progressBar.setVisibility(View.GONE);
Log.e("INFO", response);
responseView.setText(response);
}
}
}
I am passing two values instead one and using my api url since it accepts two parameter.
Just tell me how to api which i created i have my url and it accepts two parameter and return response in Json which i want to print
{
Status=1
message"dewegow523yrinc" any random everytime i am calling
}
requires com.android.application plugin to build
Obaro,
Would you consider adding a license to your code such as MIT license to allow others to use it? That would also probably get more credit for you when people do use your code. There's a short explanation of how to do that here, https://github.com/blog/1530-choosing-an-open-source-license
In particular, I am interested in using part of your SimpleWebAPI as starter code for an Android app and possibly in a YouTube video, but need a license to use it.
Many thanks,
Joe James
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