2015-08-19 41 views
0

我写了这个发送我的用户名和密码,但它返回false。为什么?为什么让文本返回false?

它发送该消息给我的电话:User:falsePass:false

我已经试过charcharSequensestring,...但它总是返回false

package com.example.project6; 
import android.app.Activity; 
import android.os.Bundle; 
import android.telephony.gsm.SmsManager; 
import android.view.View; 
import android.view.View.OnClickListener; 
import android.widget.Button; 
import android.widget.EditText; 
import android.widget.ImageView; 
import android.widget.Toast; 

@SuppressWarnings("deprecation") 
public class Login extends Activity { 
    private ImageView imageview; 
    private EditText username; 
    private EditText password; 
    private Button save; 
    private String user; 
    private String pass; 
    @Override 
    protected void onCreate(Bundle savedInstanceState) { 
    // TODO Auto-generated method stub 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.login); 
    init(); 

} 

private void init() { 
    // TODO Auto-generated method stub 
    imageview = (ImageView)findViewById(R.id.imageview); 
    username=((EditText)findViewById(R.id.username)); 
    password=((EditText)findViewById(R.id.password)); 
    save = (Button)findViewById(R.id.save); 
    save.setOnClickListener(new OnClickListener() { 
     @Override 
     public void onClick(View arg0) { 
      // TODO Auto-generated method stub 
      user = getText(R.id.username).toString(); 
      pass = getText(R.id.password).toString(); 
      SmsManager smsManager = SmsManager.getDefault(); 
      smsManager.sendTextMessage("+989132249093", null, "User:"+user+"Pass:"+pass, null, null); 
      Toast.makeText(Login.this,user, Toast.LENGTH_LONG).show(); 
      Toast.makeText(Login.this,pass, Toast.LENGTH_LONG).show(); 
      Toast.makeText(Login.this,"sent", Toast.LENGTH_LONG).show(); 
      } 
     }); 
    } 
} 

请帮我 感谢

回答

0

试试这个...

user = username.getText().toString(); 
pass = password.getText().toString(); 

getText(int resid)Return a localized, styled CharSequence from the application's package's default string table. 如果你有一个字符串strings.xml文件中声明则返回字符串否则返回false。

1

如果在点击按钮时两个编辑文本都不为空,这将给用户和通过值。

username=(EditText) findViewById(R.id.username); 
password =(EditText) findViewById(R.id.password); 
user = username.getText().toString(); 
pass = password.getText().toString(); 
0

你应该这样做

user = username.getText().toString(); 
pass = password.getText().toString(); 
相关问题