小编tes*_*123的帖子

ClassCastException:android.widget.EditText

我总是得到错误:

10-03 09:55:44.517: ERROR/AndroidRuntime(819): Caused by: java.lang.ClassCastException: android.widget.EditText
Run Code Online (Sandbox Code Playgroud)

在我的代码中,我没有这个词EditText,为什么会出现这种错误?

MultiAutoCompleteTextView之前是EditText.请帮忙

码:

public class AddPizza extends Activity {
    private ImageView iv;
    private MultiAutoCompleteTextView name;
    private RatingBar rating;
    private SQLiteDatabase myDB;
    final String MY_DB_NAME = "PizzaCounter";
    final String MY_DB_TABLE = "Pizza";

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        // TODO Auto-generated method stub
        super.onCreate(savedInstanceState);
        setContentView(R.layout.addpizza);

        iv = (ImageView) findViewById(R.id.imageViewPizza);
        name = (MultiAutoCompleteTextView) findViewById(R.id.multiAutoCompleteTextView1);
        rating = (RatingBar) findViewById(R.id.ratingBar1);
        Button bt = (Button) findViewById(R.id.bt_addform);

        iv.setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View v) { …
Run Code Online (Sandbox Code Playgroud)

android

9
推荐指数
1
解决办法
6431
查看次数

CommandText属性尚未正确初始化(MySqlDataReader)

读者总是null,我不明白为什么.

在使用一种方法连接之前,一切正常.

码:

private MySqlConnection connection;
private MySqlCommand command;
private MySqlDataReader Reader;

public Form1()
{
    InitializeComponent();
    DBint();
}
private void DBint()
{
    string myConnectionString = "SERVER=xxx;PORT=3306;" +
                                "DATABASE=xxx;" +
                                "UID=root;" +
                                "PASSWORD=xxx;";
    connection = new MySqlConnection(myConnectionString);
    command = connection.CreateCommand();
    MySqlDataReader Reader;//error occures her
    connection.Open();
    Reader = command.ExecuteReader();               
    //MessageBox.Show(tmp);                
    //connection.Close();    
}
Run Code Online (Sandbox Code Playgroud)

c# mysql database

8
推荐指数
1
解决办法
4万
查看次数

ListView中的资源$ NotFoundException

我总是得到一个Resources $ NotFoundException,但我不知道为什么.请帮忙

码:

@Override
    public View getView(int position, View convertView, ViewGroup parent) {
        ViewHolder holder = new ViewHolder();
        LayoutInflater li = (LayoutInflater)CustomAdapter.this.getContext().getSystemService(Context.LAYOUT_INFLATER_SERVICE);
        if (convertView == null) {
            convertView = li.inflate(R.layout.customlistitem,null);
            holder.mTexttitle = (TextView) convertView.findViewById(R.id.item_title);
            holder.mTextrate = (TextView) convertView.findViewById(R.id.item_rate);
            convertView.setTag(holder);
        }
        else{
            holder = (ViewHolder) convertView.getTag();
        }
        holder.mTexttitle.setText(pizzen.get(position).title);
        holder.mTextrate.setText(pizzen.get(position).rate);
        //holder.mTextdate.setText(pizzen.get(position).date.toString());
        return convertView;

    }


public class ViewHolder {

    public TextView mTexttitle;
    public TextView mTextrate;
    public TextView mTextdate;
    public ImageView mImageView;

}
Run Code Online (Sandbox Code Playgroud)

堆栈跟踪:

09-30 10:11:07.001: ERROR/XXX List View(1032): android.widget.ListView@43d133c0
09-30 10:11:07.021: ERROR/XXX(1032): …
Run Code Online (Sandbox Code Playgroud)

android

6
推荐指数
1
解决办法
4108
查看次数

在基于服务的类中设置监听器

我有一个问题是在服务中设置ServiceUpdateUIListener以更新UI.制作一个新的Service对象并在那里设置监听器并将其置于intent中是错误的.

代码源是在http://developerlife.com/tutorials/?p=356那里我找不到如何设置监听器并启动服务.

呼叫:

TimerService service = new TimerService();
                TimerService.setUpdateListener(new ServiceUpdateUIListener() {

                    @Override
                    public void updateUI(String time) {
                        clock.setText(time);

                    }
                });

                Intent i  = new Intent(Timer.this,service.class); //service cannot be resolved to a type
                i.putExtra("ms", ms);
                startService(i);  
Run Code Online (Sandbox Code Playgroud)

服务:

 public class TimerService extends Service{

        CountDownTimer timer;
        Chronometer clock;
        public static ServiceUpdateUIListener UI_UPDATE_LISTENER;

        @Override
        public IBinder onBind(Intent intent) {

            return null;
        }
        @Override
        public void onStart(Intent intent, int startId) {
            // TODO Auto-generated method stub
            int ms = intent.getIntExtra("ms", 0);

            timer = …
Run Code Online (Sandbox Code Playgroud)

android android-service

4
推荐指数
2
解决办法
2万
查看次数

垂直拆分布局Android

我想要一个分为两部分垂直的布局.我的应用程序始终处于横向模式.

请帮忙.

我用tablelayout试了一下.

 <TableLayout android:id="@+id/tableLayout1" 
 android:stretchColumns="*" android:layout_width="fill_parent" android:layout_height="fill_parent" xmlns:android="http://schemas.android.com/apk/res/android">
 <TableRow android:id="@+id/tableRow1"
 android:layout_width="fill_parent" android:layout_height="wrap_content">
 </TableRow>
 </TableLayout>
Run Code Online (Sandbox Code Playgroud)

android android-layout

2
推荐指数
1
解决办法
3288
查看次数

ListView中的NullPointerException

海兰!

我有一个listview,应该填充数据库中的数据

码:

private void onCreateDBAndDBTabled() {
        myDB = this.openOrCreateDatabase(MY_DB_NAME, MODE_PRIVATE, null);
        myDB.execSQL("CREATE TABLE IF NOT EXISTS " + MY_DB_TABLE
                + " (_id integer primary key autoincrement, name varchar(100), rate integer(1), eattime datetime)"
                +";");
    ArrayList<Pizza> list = new ArrayList<Pizza>();
Cursor cursor = this.myDB.query(MY_DB_TABLE, new String[] { "name" },null,null,null,null,null,null);
      if (cursor.moveToFirst()) {
         do {
             Log.e("XXX", "Courser Enter: " + cursor.getString(0));
             Pizza pizza = new Pizza();
             pizza.title= cursor.getString(0);
            list.add(pizza); 
            } 
         while (cursor.moveToNext());
      }
      if (cursor != null && !cursor.isClosed()) {
         cursor.close();
      } …
Run Code Online (Sandbox Code Playgroud)

android nullpointerexception android-layout

2
推荐指数
1
解决办法
2996
查看次数

查看无法找到.

我的错误如下:

'/'应用程序中的服务器错误.

未找到视图"登录"或其主控,或者没有视图引擎支持搜索的位置.搜索了以下位置:

~/Views/User/Login.aspx
~/Views/User/Login.ascx
~/Views/Shared/Login.aspx
~/Views/Shared/Login.ascx
~/Views/User/Login.cshtml
~/Views/User/Login.vbhtml
~/Views/Shared/Login.cshtml
~/Views/Shared/Login.vbhtml^
Run Code Online (Sandbox Code Playgroud)

查看/登录/ Login.cshtml:

@{
    ViewBag.Title = "Login";
}
@{
    if (!ViewData["uname"].Equals(null) )
    {
            <div>
            Username: @ViewData["uname"]
            @Session["uname"] = @ViewData["uname"]
            </div>
    }
}
<h2>Login</h2>
<h4>Username:</h4>
<input type="text" name="uname" value=" " />
<button />
Run Code Online (Sandbox Code Playgroud)

控制器/ Usercontroller.cs:

 public class UserController : Controller
    {
        //
        // GET: /User/

        public ActionResult Index()
        {
            return View();
        }
        public ActionResult Login()
        {
            return View();
        }

    }
Run Code Online (Sandbox Code Playgroud)

Global.asax.cs(仅限方法):

public static void RegisterRoutes(RouteCollection routes)
        {
            routes.IgnoreRoute("{resource}.axd/{*pathInfo}");

            routes.MapRoute(
                "Default", // …
Run Code Online (Sandbox Code Playgroud)

c# asp.net-mvc-3

1
推荐指数
1
解决办法
262
查看次数

PHP中的变音符号编码错误

我有食谱表:

CREATE TABLE IF NOT EXISTS `RECIPES` (
  `recipes_id` int(11) NOT NULL AUTO_INCREMENT,
  `title` varchar(50) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
  `text` varchar(2000) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
  `count_persons` int(11) NOT NULL,
  `duration` int(11) NOT NULL,
  `user_id` int(11) NOT NULL,
  `date` datetime NOT NULL,
  `accepted` int(11) NOT NULL DEFAULT '0',
  PRIMARY KEY (`recipes_id`),
  KEY `recipes_user_fk` (`user_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=44 ;
Run Code Online (Sandbox Code Playgroud)

插入:

(20, 'Hundefutter', 'Dose öffnen', 4, 10, 1, '2011-12-31 23:59:5', 0)
Run Code Online (Sandbox Code Playgroud)

使用php我将其解析为json(getByID):

<?php
header('Content-Type: text/html; …
Run Code Online (Sandbox Code Playgroud)

php mysql json character-encoding

1
推荐指数
1
解决办法
545
查看次数