ADT说不幸的是App在用于开始新活动的点击事件后停止了

时间:2013-12-18 07:53:41

标签: java android eclipse android-intent

我的第一个活动代码如下

public class MainActivity extends Activity {

Button mainActLoginBtn, mainActCreateBtn;
EditText mainActUserEdt, mainActPswEdt;
TextView tv;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    mainActLoginBtn = (Button) findViewById(R.id.btn1);
    mainActCreateBtn = (Button) findViewById(R.id.btn2);
    mainActUserEdt = (EditText) findViewById(R.id.edt1);
    mainActPswEdt = (EditText) findViewById(R.id.edt2);
    tv = (TextView) findViewById(R.id.tview);
    mainActLoginBtn.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg0) {
            String userString = mainActUserEdt.getText().toString();
            String pswString = mainActPswEdt.getText().toString();
            if (userString.contentEquals("Dev")
                    && pswString.contentEquals("1234")) {
                tv.setText("Welcome " + userString
                        + "Click Proceed to further process");
                Intent successAct = new Intent("com.example.a_m_s.INTRO");
                startActivityForResult(successAct, 0);

            }
        }

    });

}

@Override
public boolean onCreateOptionsMenu(Menu menu) {
    // Inflate the menu; this adds items to the action bar if it is present.
    getMenuInflater().inflate(R.menu.main, menu);
    return true;
}

应用程序无法成功运行第一个窗口显示,这是一个登录窗口,输入登录凭据并单击登录按钮,Android Emulator中生成错误 如果需要更多参考代码,我无法弄清楚可能出现什么问题请向我们索取。

2 个答案:

答案 0 :(得分:0)

试试这个..

Intent successAct = new Intent(MainActivity.this,INTRO.class);
                startActivity(successAct);

示例:

mainActLoginBtn.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg0) {
            String userString = mainActUserEdt.getText().toString();
            String pswString = mainActPswEdt.getText().toString();
            if (userString.contentEquals("Dev")
                    && pswString.contentEquals("1234")) {
                tv.setText("Welcome " + userString
                        + "Click Proceed to further process");
                 Intent successAct = new Intent(MainActivity.this,INTRO.class);
                    startActivity(successAct);

            }
        }

    });

答案 1 :(得分:0)

除了撰写startActivityForResult(successAct, 0);之外,您还需要撰写startActivity(successAct);

mainActLoginBtn.setOnClickListener(new View.OnClickListener() {

    @Override
    public void onClick(View arg0) {
        String userString = mainActUserEdt.getText().toString();
        String pswString = mainActPswEdt.getText().toString();
        if (userString.contentEquals("Dev")
                && pswString.contentEquals("1234")) {
            tv.setText("Welcome " + userString
                    + "Click Proceed to further process");
             Intent successAct = new Intent("com.example.a_m_s.INTRO");
                startActivity(successAct);

        }
    }

});