16

我刚刚编写了一些代码来缩放字体以适应矩形（的长度）。它从 18 宽度开始并向下迭代直到适合。

这似乎非常低效，但我找不到非循环的方式来做到这一点。这条线用于可缩放的游戏网格中的标签，因此我看不到解决方案（包装、切断和延伸超出矩形都是不可接受的）。

它实际上非常快，我正在为数百个矩形执行此操作，它的速度足以让它减慢一点。

如果没有人想出更好的方法，我会从表中加载起始猜测（这样它就比 18 更接近）并使用它——除了滞后它工作得很好。

public Font scaleFont(String text, Rectangle rect, Graphics g, Font pFont) {
    float nextTry=18.0f;
    Font font=pFont;

    while(x > 4) {                             
            font=g.getFont().deriveFont(nextTry);
            FontMetrics fm=g.getFontMetrics(font);
            int width=fm.stringWidth(text);
            if(width <= rect.width)
                return font;
            nextTry*=.9;            
    }
    return font;
}
4

6 回答 6

22

半伪代码：

public Font scaleFont(
    String text, Rectangle rect, Graphics g, Font font) {
    float fontSize = 20.0f;

    font = g.getFont().deriveFont(fontSize);
    int width = g.getFontMetrics(font).stringWidth(text);
    fontSize = (rect.width / width ) * fontSize;
    return g.getFont().deriveFont(fontSize);
}

迭代的推导：

/**
 * Adjusts the given {@link Font}/{@link String} size such that it fits
 * within the bounds of the given {@link Rectangle}.
 *
 * @param label    Contains the text and font to scale.
 * @param dst      The bounds for fitting the string.
 * @param graphics The context for rendering the string.
 * @return A new {@link Font} instance that is guaranteed to write the given
 * string within the bounds of the given {@link Rectangle}.
 */
public Font scaleFont(
    final JLabel label, final Rectangle dst, final Graphics graphics ) {
  assert label != null;
  assert dst != null;
  assert graphics != null;

  final var font = label.getFont();
  final var text = label.getText();

  final var frc = ((Graphics2D) graphics).getFontRenderContext();

  final var dstWidthPx = dst.getWidth();
  final var dstHeightPx = dst.getHeight();

  var minSizePt = 1f;
  var maxSizePt = 1000f;
  var scaledFont = font;
  float scaledPt = scaledFont.getSize();

  while( maxSizePt - minSizePt > 1f ) {
    scaledFont = scaledFont.deriveFont( scaledPt );

    final var layout = new TextLayout( text, scaledFont, frc );
    final var fontWidthPx = layout.getVisibleAdvance();

    final var metrics = scaledFont.getLineMetrics( text, frc );
    final var fontHeightPx = metrics.getHeight();

    if( (fontWidthPx > dstWidthPx) || (fontHeightPx > dstHeightPx) ) {
      maxSizePt = scaledPt;
    }
    else {
      minSizePt = scaledPt;
    }

    scaledPt = (minSizePt + maxSizePt) / 2;
  }

  return scaledFont.deriveFont( (float) Math.floor( scaledPt ) );
}

想象一下，您想向具有矩形边界的组件添加标签，以r使标签完全填满组件的区域。可以写：

final Font DEFAULT_FONT = new Font( "DejaVu Sans", BOLD, 12 );
final Color COLOUR_LABEL = new Color( 33, 33, 33 );

// TODO: Return a valid container component instance.
final var r = getComponent().getBounds();
final var graphics = getComponent().getGraphics();

final int width = (int) r.getWidth();
final int height = (int) r.getHeight();

final var label = new JLabel( text );
label.setFont( DEFAULT_FONT );
label.setSize( width, height );
label.setForeground( COLOUR_LABEL );

final var scaledFont = scaleFont( label, r, graphics );
label.setFont( scaledFont );
于 2009-05-18T04:41:44.213 回答
4

将所有宽度变量更改为 float 而不是 int 以获得更好的结果。

public static Font scaleFontToFit(String text, int width, Graphics g, Font pFont)
{
    float fontSize = pFont.getSize();
    float fWidth = g.getFontMetrics(pFont).stringWidth(text);
    if(fWidth <= width)
        return pFont;
    fontSize = ((float)width / fWidth) * fontSize;
    return pFont.deriveFont(fontSize);
}
于 2011-05-06T23:22:06.913 回答
3

您可以使用插值搜索：

public static Font scaleFont(String text, Rectangle rect, Graphics g, Font pFont) {
    float min=0.1f;
    float max=72f;
    float size=18.0f;
    Font font=pFont;

    while(max - min <= 0.1) {
        font = g.getFont().deriveFont(size);
        FontMetrics fm = g.getFontMetrics(font);
        int width = fm.stringWidth(text);
        if (width == rect.width) {
            return font;
        } else {
            if (width < rect.width) {
                min = size;
            } else {
                max = size;
            }
            size = Math.min(max, Math.max(min, size * (float)rect.width / (float)width));
        }
    }
    return font;
}
于 2009-05-18T07:13:33.003 回答
2 
private Font scaleFont ( String text, Rectangle rect, Graphics gc )
{
    final float fMinimumFont = 0.1f;
    float fMaximumFont = 1000f;

    /* Use Point2d.Float to hold ( font, width of font in pixels ) pairs. */
    Point2D.Float lowerPoint = new Point2D.Float ( fMinimumFont, getWidthInPixelsOfString ( text, fMinimumFont, gc ) );
    Point2D.Float upperPoint = new Point2D.Float ( fMaximumFont, getWidthInPixelsOfString ( text, fMaximumFont, gc ) );
    Point2D.Float midPoint = new Point2D.Float ();

    for ( int i = 0; i < 50; i++ )
    {
        float middleFont = ( lowerPoint.x + upperPoint.x ) / 2;

        midPoint.setLocation ( middleFont, getWidthInPixelsOfString ( text, middleFont, gc ) );

        if ( midPoint.y >= rect.getWidth () * .95 && midPoint.y <= rect.getWidth () )
            break;
        else if ( midPoint.y < rect.getWidth () )
            lowerPoint.setLocation ( midPoint );
        else if ( midPoint.y > rect.getWidth () )
            upperPoint.setLocation ( midPoint );
    }

    fMaximumFont = midPoint.x;

    Font font = gc.getFont ().deriveFont ( fMaximumFont );

    /* Now use Point2d.Float to hold ( font, height of font in pixels ) pairs. */
    lowerPoint.setLocation ( fMinimumFont, getHeightInPixelsOfString ( text, fMinimumFont, gc ) );
    upperPoint.setLocation ( fMaximumFont, getHeightInPixelsOfString ( text, fMaximumFont, gc ) );

    if ( upperPoint.y < rect.getHeight () )
        return font;

    for ( int i = 0; i < 50; i++ )
    {
        float middleFont = ( lowerPoint.x + upperPoint.x ) / 2;

        midPoint.setLocation ( middleFont, getHeightInPixelsOfString ( text, middleFont, gc ) );

        if ( midPoint.y >= rect.getHeight () * .95 && midPoint.y <= rect.getHeight () )
            break;
        else if ( midPoint.y < rect.getHeight () )
            lowerPoint.setLocation ( midPoint );
        else if ( midPoint.y > rect.getHeight () )
            upperPoint.setLocation ( midPoint );
    }

    fMaximumFont = midPoint.x;

    font = gc.getFont ().deriveFont ( fMaximumFont );

    return font;
}


private float getWidthInPixelsOfString ( String str, float fontSize, Graphics gc )
{
    Font font = gc.getFont ().deriveFont ( fontSize );

    return getWidthInPixelsOfString ( str, font, gc );
}

private float getWidthInPixelsOfString ( String str, Font font, Graphics gc )
{
    FontMetrics fm = gc.getFontMetrics ( font );
    int nWidthInPixelsOfCurrentFont = fm.stringWidth ( str );

    return (float) nWidthInPixelsOfCurrentFont;
}


private float getHeightInPixelsOfString ( String string, float fontSize, Graphics gc )
{
    Font font = gc.getFont ().deriveFont ( fontSize );

    return getHeightInPixelsOfString ( string, font, gc );
}

private float getHeightInPixelsOfString ( String string, Font font, Graphics gc )
{
    FontMetrics metrics = gc.getFontMetrics ( font );
    int nHeightInPixelsOfCurrentFont = (int) metrics.getStringBounds ( string, gc ).getHeight () - metrics.getDescent () - metrics.getLeading ();

    return (float) nHeightInPixelsOfCurrentFont * .75f;
}
于 2013-06-05T21:32:17.060 回答
1

您可以使用二分搜索模式来提高效率 - 具有一定粒度的高/低 - 1、0.5 或 0.25 点。

比如，猜18，太高了？移动到 9，太低了？13.5，太低了？15.75，太高了？14！

于 2009-05-18T04:30:23.033 回答
0

另一种明显的方法是将文本预先绘制在位图上，然后缩小位图以适应矩形；但是，由于手工制作的字体设计和提示等，找到合适的字体大小会产生最好看的（虽然可能不是最快的）结果。

于 2009-05-18T04:30:19.227 回答