这是我的登录系统代码,但 PHP 文件似乎没有登录。我想先解决主要问题,然后我将修改 PHP 文件以从数据库中获取用户。
登录.java:
public class login extends Activity implements OnClickListener {
private EditText usernameEditText;
private EditText passwordEditText;
private Button sendPostReqButton;
private Button clearButton;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.login);
usernameEditText = (EditText) findViewById(R.id.username);
passwordEditText = (EditText) findViewById(R.id.password);
sendPostReqButton = (Button) findViewById(R.id.login);
sendPostReqButton.setOnClickListener(this);
}
@Override
public void onClick(View v) {
if(v.getId() == R.id.login){
String givenUsername = usernameEditText.getEditableText().toString();
String givenPassword = passwordEditText.getEditableText().toString();
System.out.println("Given username :" + givenUsername + " Given password :" + givenPassword);
sendPostRequest(givenUsername, givenPassword);
}
}
private void sendPostRequest(String givenUsername, String givenPassword) {
class SendPostReqAsyncTask extends AsyncTask<String, Void, String>{
@Override
protected String doInBackground(String... params) {
String paramUsername = params[0];
String paramPassword = params[1];
System.out.println("*** doInBackground ** paramUsername " + paramUsername + " paramPassword :" + paramPassword);
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://www.nirmana.lk/hec/android/postLogin.php");
BasicNameValuePair usernameBasicNameValuePair = new BasicNameValuePair("paramUsername", paramUsername);
BasicNameValuePair passwordBasicNameValuePAir = new BasicNameValuePair("paramPassword", paramPassword);
List<NameValuePair> nameValuePairList = new ArrayList<NameValuePair>();
nameValuePairList.add(usernameBasicNameValuePair);
nameValuePairList.add(passwordBasicNameValuePAir);
try {
UrlEncodedFormEntity urlEncodedFormEntity = new UrlEncodedFormEntity(nameValuePairList);
// setEntity() hands the entity (here it is urlEncodedFormEntity) to the request.
httpPost.setEntity(urlEncodedFormEntity);
try {
HttpResponse httpResponse = httpClient.execute(httpPost);
InputStream inputStream = httpResponse.getEntity().getContent();
InputStreamReader inputStreamReader = new InputStreamReader(inputStream);
BufferedReader bufferedReader = new BufferedReader(inputStreamReader);
StringBuilder stringBuilder = new StringBuilder();
String bufferedStrChunk = null;
while((bufferedStrChunk = bufferedReader.readLine()) != null){
stringBuilder.append(bufferedStrChunk);
}
return stringBuilder.toString();
} catch (ClientProtocolException cpe) {
System.out.println("First Exception caz of HttpResponese :" + cpe);
cpe.printStackTrace();
} catch (IOException ioe) {
System.out.println("Second Exception caz of HttpResponse :" + ioe);
ioe.printStackTrace();
}
} catch (UnsupportedEncodingException uee) {
System.out.println("An Exception given because of UrlEncodedFormEntity argument :" + uee);
uee.printStackTrace();
}
return null;
}
@Override
protected void onPostExecute(String result) {
super.onPostExecute(result);
if(result.equals("working")){
Toast.makeText(getApplicationContext(), "HTTP POST is working...", Toast.LENGTH_LONG).show();
}else{
Toast.makeText(getApplicationContext(), "Invalid POST req...", Toast.LENGTH_LONG).show();
}
}
}
SendPostReqAsyncTask sendPostReqAsyncTask = new SendPostReqAsyncTask();
sendPostReqAsyncTask.execute(givenUsername, givenPassword);
}
}
PHP 文件:
<?php
$varUsername = $_POST['paramUsername'];
$varPassword = $_POST['paramPassword'];
if($varUsername == "anuja" && $varPassword == "123"){
echo 'working';
}else{
echo 'invalid';
}
?>
不知道为什么停了?