Question

我有一个没有显示错误的代码，但是当我单击该图标时，它显示“审阅已停止”。再次打开应用。这是我的代码。

主要活动

    public class MainActivity extends Activity {
    Spinner spinner;
    Button Submit;
    String text;
    TextView t1, t2;
    EditText name, comment, nam;
    static SQLiteDatabase db;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    name = (EditText) findViewById(R.id.editTextTextPersonName1);
    t1 = (TextView) findViewById(R.id.textView);
    t2 = (TextView) findViewById(R.id.textView2);
    comment = (EditText) findViewById(R.id.editTextTextMultiLine);
    comment.setSelection(0);
    db = openOrCreateDatabase("MainDB", Context.MODE_PRIVATE, null);
    db.execSQL("CREATE TABLE IF NOT EXISTS demo(name VARCHAR(20),spinner VARCHAR(15),comment VARCHAR(100));");

}

public void showmsg(View v)
{

    Spinner spinner = (Spinner) findViewById(R.id.spinner1);
    String text = spinner.getSelectedItem().toString();
    db.execSQL("INSERT INTO demo VALUES('"+name.getText()+"','"+text+"','"+comment.getText()+"');");
    String sql = "SELECT COUNT(*) FROM demo ";
    sql += "WHERE spinner IN ('Excellent','Best','Good','Fair')";
    Cursor c = db.rawQuery(sql, null);
    c.moveToFirst();
    int cnt = c.getInt(0);
    t1.setText(""+cnt);
    c.close();

    String sql1 = "SELECT COUNT(*) FROM demo ";
    sql1 += "WHERE spinner IN ('Costly','Poor','Bad','Worst')";
    Cursor c1 = db.rawQuery(sql1, null);
    c1.moveToFirst();
    int cnt1 = c1.getInt(0);
    t2.setText(""+cnt1);
    c1.close();
    Toast.makeText(MainActivity.this, "Thanks for your comment", Toast.LENGTH_LONG).show();
    name.setText("");
    spinner.setSelection(0);
    comment.setText("");
}

}

我无法在Google中找到答案，所以我尝试在此处发布...
显示运行该应用程序时Review已停止。

致谢

“应用程序停止”再次打开

0 个答案: