小编Div*_*ers的帖子

我有以下两个班级:

class Animal {
    public static void staticMethod(int i) {
        System.out.println("Animal : static -- " + i);
    }

    public void instanceMethod(int i) {
        System.out.println("Animal : instance -- " + i);
    }
}

class Cat extends Animal {
    public static void staticMethod(int i) {
        System.out.println("Cat : static -- " + i);
    }

    public void instanceMethod(int i) {
        System.out.println("Cat : instance -- " + i);
    }

    public static void main(String[] args) {
        Cat myCat = new Cat();
        myCat.staticMethod(1);                       // Cat : static …

我有代码:

public void AlarmStart() {
    Calendar cal = Calendar.getInstance();
    cal.add(Calendar.MINUTE, 5);
    Intent intent = new Intent(MainNote.this, AlarmReceiver.class);
    intent.putExtra("alarm_message", "MESS");
    PendingIntent sender = PendingIntent.getBroadcast(MainNote.this, 1,
        intent, PendingIntent.FLAG_UPDATE_CURRENT);
    AlarmManager am = (AlarmManager) getSystemService(ALARM_SERVICE);
    am.set(AlarmManager.RTC_WAKEUP, cal.getTimeInMillis(), sender);
}

它按时调用AlarmReceiver类.

public class AlarmReceiver extends BroadcastReceiver {

    @Override
    public void onReceive(Context context, Intent intent) {
        Bundle bundle = intent.getExtras();
        String message = bundle.getString("alarm_message");
        NotifierHelper.sendNotification(?????, MainNote.class, "ba", "baba",
            2, true, true);
    } // Problem here
}

然后NotifierHelper类:

public class NotifierHelper {

    private static final int NOTIFY_1 …

或者也许更好地使用其他类来同步访问数据？

我的自定义视图中的下一个代码扩展了EditText:

protected void onDraw(Canvas canvas) {
    super.onDraw(canvas);

                int count = getLineCount();
                Canvas cvs= new Canvas();
                Drawable dr = this.getBackground();
                Rect r = mRect;
                Paint paint = mPaint;
                mTextPaint.setTextSize(this.getTextSize());
                Paint PaintText = mTextPaint;

                for (int i = 0; i < count; i++) {
                    int baseline = getLineBounds(i, r);
                    cvs.drawText(Integer.toString(i+1), r.left, baseline + 1, PaintText);
                    cvs.drawLine(r.left, baseline + 1, r.right, baseline + 1, paint); 
                }
                cvs.drawLine(PaintText.measureText(Integer.toString(count)), this.getTop(), PaintText.measureText(Integer.toString(count)), this.getBottom(), paint);
                dr.setBounds(0, 0, this.getRight(), this.getBottom());
                dr.draw(cvs);
                this.setBackgroundDrawable(dr);
            }

为什么这个观点的背景上没有任何内容？

我在DialogFragment中有下一个代码:

@Override
public Dialog onCreateDialog(Bundle savedInstanceState) {
    Log.i(getClass().getName(), "onCreateDialog");
    final FileSystemAdapter adapter = getAdapter();
    adapter.showFS(null);

    AlertDialog b = new AlertDialog.Builder(getActivity())
            .setTitle(title)
            .setCancelable(false)
            .setPositiveButton(R.string.ok, new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialogInterface, int i) {
                    //some actions
                }
            })
            .setNegativeButton(R.string.cancel, new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialogInterface, int i) {
                    dismiss();
                }
            })
            .setAdapter(adapter, new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialogInterface, int i) {
                    adapter.showFS(adapter.getItem(i));
                }
            })
            .create();
    return b;
}

单击列表元素后,最后一个onClick方法正在调用,然后对话框正在关闭.

如何防止这种行为并保持对话框打开,直到dismiss方法调用？

我是非常新的PHP,所以我有一个非常棘手的问题.我尝试编写能帮助我使用DB的类.代码在这里:

<?php
    class DbHelper{
        private $databaseURL;
        private $databaseUName;
        private $databasePWord;
        private $databaseName;
        private $nameOfDbWithWorkers;
        private $connection;

        function __construct($dbURL, $dbUserName, $dbPword, $dbName, $nameOfDbWithWorkers){
            $this->databaseURL = $dbURL;
            $this->databaseUName = $dbUserName;
            $this->databasePWord = $dbPword;
            $this->databaseName = $dbName;
            $this->nameOfDbWithWorkers = $nameOfDbWithWorkers;
        }

        function setConnectionToDb(){
            $connection = mysql_connect($this->databaseURL,$this->databaseUName,$this->databasePWord) OR DIE("can't connect to DB"); 
            mysql_select_db($this->databaseName, $connection)or die ("Error while connecting to database");
        }

        function getUser($login, $pass){
            $query = "SELECT type FROM $this->nameOfDbWithWorkers WHERE login = '$login' and password = '$pass';"; 
            $queryResult = $this->getDataFromDbByQuery($query);
            echo $queryResult;
            if …

可能重复:
警告:mysql_fetch_array():提供的参数不是有效的MySQL结果

我有下一堂课:

<?php
    class DbHelper{
        private $databaseURL;
        private $databaseUName;
        private $databasePWord;
        private $databaseName;
        private $nameOfDbWithWorkers;
        private $connection;

        function __construct($dbURL, $dbUserName, $dbPword, $dbName, $nameOfDbWithWorkers){
            $this->databaseURL = $dbURL;
            $this->databaseUName = $dbUserName;
            $this->databasePWord = $dbPword;
            $this->databaseName = $dbName;
            $this->nameOfDbWithWorkers = $nameOfDbWithWorkers;
        }

        function setConnectionToDb(){
            $this->connection = mysql_connect($this->databaseURL,$this->databaseUName,$this->databasePWord) OR DIE("can't connect to DB"); 
            mysql_select_db($this->databaseName, $this->connection)or die ("Error while connecting to database");
        }

        function getUser($login, $pass){
            echo "$login, $pass";
            $query = "SELECT type FROM $this->nameOfDbWithWorkers WHERE login = '$login' and password = '$pass';"; 
            $queryResult …