如果语句代码

Question

好的，我正在尝试创建一个应用程序，它只接受任何文本的1个单词。我使用edittext让用户放入文本和按钮为他们提供一个祝词，不正确或不正确

问题是，当我使用if语句，我不能让它的工作能有一个人请认这里的问题是我的代码

public class IfActivity extends Activity { 
Button GO; 
EditText TEXT; 

/** Called when the activity is first created. */ 
@Override 
public void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.main); 

    GO = (Button) findViewById(R.id.go); 
    TEXT = (EditText) findViewById(R.id.Textin); 

    GO.setOnClickListener(new View.OnClickListener() { 

     public void onClick(View arg0) { 
      // TODO Auto-generated method stub 

      if ("hello".equals(TEXT)) { 
       Toast andEggs = Toast.makeText(IfActivity.this, 
         "PASSWORD IS CORRECT", Toast.LENGTH_SHORT); 
       andEggs.show(); 
      } else { 
       Toast andEggs = Toast.makeText(IfActivity.this, 
         "PASSWORD IS INCORRECT", Toast.LENGTH_SHORT); 
       andEggs.show(); 
      } 

     } 
    }); 

} 

}

Answer 1

使用此：

public void onClick(View arg0) {    
// TODO Auto-generated method stub    
String tempTxt = ((EditText) findViewById(R.id.Textin)).getText().toString(); 
if ("hello".equals(tempTxt)) { 
    Toast andEggs = Toast.makeText(IfActivity.this,"PASSWORD IS CORRECT",Toast.LENGTH_SHORT); 
    andEggs.show(); 
    } 
else{ 
    Toast andEggs = Toast.makeText(IfActivity.this,"PASSWORD IS INCORRECT",Toast.LENGTH_SHORT); 
    andEggs.show(); 
    } 
} 

Answer 2

使用此：

if("hello".equals((findViewById(R.id.Textin)).getText().toString())) { // show toast} 

Answer 3

TEXT不是字符串。它是EditText的引用变量。

if (TEXT.getText().toString().equals("hello")) 
{ 

}