2008-10-28 289 views
1

因爲我需要顯示一個huge number of labelsmove independently,我需要在pyglet中渲染一個標籤爲紋理(否則更新每個字形的頂點列表太慢)。opengl設置頂點顏色的紋理顏色

我有一個解決方案來做到這一點,但我的問題是包含字形的紋理是黑色的,但我希望它是紅色的。看下面的例子:

from pyglet.gl import * 

def label2texture(label): 
    vertex_list = label._vertex_lists[0].vertices[:] 
    xpos = map(int, vertex_list[::8]) 
    ypos = map(int, vertex_list[1::8]) 
    glyphs = label._get_glyphs() 

    xstart = xpos[0] 
    xend = xpos[-1] + glyphs[-1].width 
    width = xend - xstart 

    ystart = min(ypos) 
    yend = max(ystart+glyph.height for glyph in glyphs) 
    height = yend - ystart 

    texture = pyglet.image.Texture.create(width, height, pyglet.gl.GL_RGBA) 

    for glyph, x, y in zip(glyphs, xpos, ypos): 
     data = glyph.get_image_data() 
     x = x - xstart 
     y = height - glyph.height - y + ystart 
     texture.blit_into(data, x, y, 0) 

    return texture.get_transform(flip_y=True) 

window = pyglet.window.Window() 
label = pyglet.text.Label('Hello World!', font_size = 36) 
texture = label2texture(label) 

@window.event 
def on_draw(): 
    hoff = (window.width/2) - (texture.width/2) 
    voff = (window.height/2) - (texture.height/2) 

    glClear(GL_COLOR_BUFFER_BIT) 
    glEnable(GL_BLEND) 
    glBlendFunc(GL_SRC_ALPHA, GL_ONE_MINUS_SRC_ALPHA) 
    glClearColor(0.0, 1.0, 0.0, 1.0) 
    window.clear() 
    glEnable(GL_TEXTURE_2D); 

    glBindTexture(GL_TEXTURE_2D, texture.id) 
    glColor4f(1.0, 0.0, 0.0, 1.0) #I'd like the font to be red 
    glBegin(GL_QUADS); 
    glTexCoord2d(0.0,1.0); glVertex2d(hoff,voff); 
    glTexCoord2d(1.0,1.0); glVertex2d(hoff+texture.width,voff); 
    glTexCoord2d(1.0,0.0); glVertex2d(hoff+texture.width,voff+texture.height); 
    glTexCoord2d(0.0,0.0); glVertex2d(hoff, voff+texture.height); 
    glEnd(); 

pyglet.app.run() 

任何想法,我怎麼能顏色嗎?

回答

2

您想設置glEnable(GL_COLOR_MATERIAL)。這使得紋理顏色與當前的OpenGL顏色混合。您也可以使用glColorMaterial函數來指定是否應影響每個多邊形的正面/背面/兩者。 Docs here