InkRecognizerGuide.DrawnBoxRight - свойство
Обновлен: Ноябрь 2007
Gets or sets the right coordinate of the rectangle that is physically drawn on the tablet screen and in which writing can occur.
Пространство имен: System.Windows.Ink
Сборка: IAWinFX (в IAWinFX.dll)
Синтаксис
'Декларация
Public Property DrawnBoxRight As Double
'Применение
Dim instance As InkRecognizerGuide
Dim value As Double
value = instance.DrawnBoxRight
instance.DrawnBoxRight = value
public double DrawnBoxRight { get; set; }
public:
property double DrawnBoxRight {
double get ();
void set (double value);
}
/** @property */
public double get_DrawnBoxRight()
/** @property */
public void set_DrawnBoxRight(double value)
public function get DrawnBoxRight () : double
public function set DrawnBoxRight (value : double)
Значение свойства
Тип: System.Double
The right coordinate of the rectangle that is physically drawn on the tablet screen and in which writing can occur.
Заметки
The boundaries of the drawn box are visual cues that specify where writing can occur. The user normally writes within the boundaries of the lines. The drawn box is specified by the DrawnBoxTop, DrawnBoxLeft, DrawnBoxBottom, and DrawnBoxRight properties. These properties are for the recognizer's use only. Setting them does not, by itself, draw visual clues on the display. You must draw the visual clues by using the application or the control.
Another rectangle, the writing box, is the invisible box in which writing can actually take place. It is larger than the drawn box and provides the users a margin of error if they draw ink outside the lines of the drawn box. The writing box is specified by the WritingBoxTop, WritingBoxLeft, WritingBoxBottom, and WritingBoxRight properties.
The DrawnBoxRight property throws an ArgumentOutOfRangeException if you try to set it to a value less that AnalysisRegion.MinXY or greater than AnalysisRegion.MaxXY.
Примеры
The following sets all of the values in the InkRecognizerGuide simultaneously.
Dim guideBoxTop As Double = 0
Dim guideBoxBottom As Double = 50
Dim guideBoxLeft As Double = 0
Dim guideBoxRight As Double = 600
Dim WRITINGBOXMARGIN As Integer = 10
' Find the midline of the guide box.
Dim midline As Double = (guideBoxBottom - guideBoxTop) / 2 + guideBoxTop
theGuide.Rows = 1
theGuide.Columns = 0
theGuide.Midline = midline
theGuide.WritingBoxLeft = guideBoxLeft - WRITINGBOXMARGIN
theGuide.WritingBoxTop = guideBoxTop - WRITINGBOXMARGIN
theGuide.WritingBoxRight = guideBoxRight + WRITINGBOXMARGIN
theGuide.WritingBoxBottom = guideBoxBottom + WRITINGBOXMARGIN
theGuide.DrawnBoxLeft = guideBoxLeft
theGuide.DrawnBoxTop = guideBoxTop
theGuide.DrawnBoxRight = guideBoxRight
theGuide.DrawnBoxBottom = guideBoxBottom
double guideBoxTop = 0;
double guideBoxBottom = 50;
double guideBoxLeft = 0;
double guideBoxRight = 600;
const int WRITINGBOXMARGIN = 10;
// Find the midline of the guide box.
double midline = (guideBoxBottom - guideBoxTop) / 2 + guideBoxTop;
theGuide.Rows = 1;
theGuide.Columns = 0;
theGuide.Midline = midline;
theGuide.WritingBoxLeft = guideBoxLeft - WRITINGBOXMARGIN;
theGuide.WritingBoxTop = guideBoxTop - WRITINGBOXMARGIN;
theGuide.WritingBoxRight = guideBoxRight + WRITINGBOXMARGIN;
theGuide.WritingBoxBottom = guideBoxBottom + WRITINGBOXMARGIN;
theGuide.DrawnBoxLeft = guideBoxLeft;
theGuide.DrawnBoxTop = guideBoxTop;
theGuide.DrawnBoxRight = guideBoxRight;
theGuide.DrawnBoxBottom = guideBoxBottom;
Платформы
Windows Vista
Среды .NET Framework и .NET Compact Framework поддерживают не все версии каждой платформы. Поддерживаемые версии перечислены в разделе Требования к системе для .NET Framework.
Сведения о версии
.NET Framework
Поддерживается в версии: 3.0