Question

好的我正在尝试创建一个只接受任何文本的1个单词的应用程序。我使用edittext让用户输入文本和按钮，让他们给你一个包含错误或正确单词的祝酒词

问题是当我使用if语句我不能让它工作时可以请一些人在这里找出问题是我的代码

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

TEXT不是字符串。它是EditText的引用变量。

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

}

Answer 2

使用此：

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

Answer 3

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();
   }
 }

如果声明代码

3 个答案: