2014-12-30 29 views
-3

我想向php url发送电子邮件和密码,但我无法做到这一点..我想从edittext中获取字符串并将其发送到PHP myadmin localhost我尝试了很多,但没有能够做到这一点......请帮助我完整的解释...谢谢:)无法在Android中使用httppost

<?php 
 
//error_reporting(0); 
 

 
$con = mysql_connect('localhost','root',''); 
 
mysql_select_db('sample_app_ws',$con); 
 

 
// Insert Query For Send Value To Data Base 
 
//echo $_REQUEST['act'] ; 
 
if(isset($_REQUEST['act']) && $_REQUEST['act'] == 'Insert_UserData') { 
 
\t $email = $_REQUEST['email']; 
 

 
\t $password = $_REQUEST['password']; 
 
\t 
 
$result= mysql_query("INSERT INTO sample_login(email,password,enterDate) VALUES ('".$email."','".$password."',NOW())"); 
 
\t if($result) \t 
 
\t $sendvale ['DataSubmit'] = "DataSubmit"; 
 
\t else $sendvale ['err'] = "ErrDataSubmit"; 
 
\t echo json_encode($sendvale); 
 
} 
 
elseif(isset($_REQUEST['act']) && $_REQUEST['act'] == 'Select_UserName'){ 
 
\t $email = $_REQUEST['email']; 
 
\t $password = $_REQUEST['password']; 
 
\t 
 
\t $result = mysql_query("Select * from sample_login where email = '".$email."' and password = '".$password."'"); 
 
\t if(mysql_num_rows($result) > 0){ 
 
\t $sendvale ['personaldetails'] = "VaildUsrNameAndPassword"; 
 
\t \t echo json_encode($sendvale); 
 
\t }else{ 
 
\t $sendvale ['personaldetails'] = "InVaildUsrNameAndPassword"; 
 
\t \t echo json_encode($sendvale); 
 
\t } 
 
} 
 
else{ 
 
\t $sendvale ['err'] = "POST_DATA_IS_Empty"; 
 
\t \t echo json_encode($sendvale); 
 
\t } 
 

 
?>

我提供它涉及到这个PHP代码的Android的Java代码..

java代码....

public class LoginActivity extends Activity implements OnClickListener { 

private EditText userName; 
private EditText password; 
private Button submit; 
private Button cancel; 
public static final String URL = " http://127.0.0.1/sample_app_ws/sample_login.php"; 

@Override 
protected void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.activity_login); 

    createObjects(); 
} 

private void createObjects() { 

    userName = (EditText) findViewById(R.id.editTextUserName); 
    password = (EditText) findViewById(R.id.editTextpassword); 
    submit = (Button) findViewById(R.id.buttonLogin); 
    submit.setOnClickListener(this); 

    cancel = (Button) findViewById(R.id.buttonCAncel); 

} 

private void sendData(String userName, String Password) { 

    ArrayList<NameValuePair> arrayList = new ArrayList<NameValuePair>(); 

    arrayList.add(new BasicNameValuePair("email", userName)); 
    arrayList.add(new BasicNameValuePair("password", Password)); 

    try { 

     HttpClient client = new DefaultHttpClient(); 
     HttpPost httpPost = new HttpPost(URL); 
     httpPost.setEntity(new UrlEncodedFormEntity(arrayList)); 
     HttpResponse httpResponse = client.execute(httpPost); 

     // httpResponse.getEntity().equals("0"); 

    } catch (Exception e) { 

     Toast.makeText(this, e.getMessage(), Toast.LENGTH_LONG).show(); 
    } 

} 

@Override 
public void onClick(View v) { 
    new Thread(new Runnable() { 

     @Override 
     public void run() { 
      sendData(userName.getText().toString(), password.getText() 
        .toString()); 
     } 
    }); 

} 

}

+0

什么是错误? – segarci

+0

我正确不知道我使用了一些代码从网络使用Listvalue对和httpclient和发布方法,但没有发生甚至没有发现日志甚至没有打印任何东西......你能提供给我一些代码段的Android java支持我的PHP代码 –

+0

什么是您的Java代码? – segarci

回答

0

试图改变这一点:

public static final String URL = " http://127.0.0.1/sample_app_ws/sample_login.php"; 

到你的IP地址,这样的 -

public static final String URL = " http://your IP address/sample_app_ws/sample_login.php"; 

进入到CMD - 写ipconfig并获得你的IP地址。

+0

我试过了..但没有发生..仍然无法传输数据... –

+0

看看我的答案在这里http://stackoverflow.com/questions/27655786/android-values-upload-database-with-json/ 27656738#27656738 – Josef

+0

我们是否总是需要添加。 jSonparser类,并在您的代码中,您只需添加值...因为我必须转换hp代码,但我想要相同的php代码不会wana更改它... –