在 Libgdx 中绘制渐变

Drawing a gradient in Libgdx(在 Libgdx 中绘制渐变)

问题描述

好的，我有这个代码

@Override
public void render() {
    // do not update game world when paused
    if (!paused) {
        // Update game world by the time that has passed
        // since last render frame
        worldController.update(Gdx.graphics.getDeltaTime());
    }
    // Sets the clear screen color to: Cornflower Blue
    Gdx.gl.glClearColor(0x64/255.0f, 0x95/255.0f, 0xed/255.0f,
            0xff/255.0f);
    // Clears the screen
    Gdx.gl.glClear(GL10.GL_COLOR_BUFFER_BIT);
    // Render game world to screen
    worldRenderer.render();
}

它会在屏幕上绘制浅蓝色背景.我正在尝试创建一个渐变，从顶部的深蓝色到底部的浅蓝色.有没有一种简单的方法可以做到这一点?我是 Libgdx 和 OpenGL 的新手，所以我想从一本书中学习，但我似乎找不到这个问题的答案.我听说过绘制一个大正方形并让顶点具有不同的颜色，但我不确定如何做到这一点.

And it draws a light blue background onto the screen. I am attempting to create a gradient that goes from a dark blue at the top, to a light blue towards the bottom. Is there a simple way to do this? I'm new to Libgdx, and OpenGL so i'm trying to learn from a book but I can't seem to find the answer to this one. I've heard of drawing a big square and having the vertices different colors, but I'm unsure of how to do this.

推荐答案

在 libGDX 中，ShapeRenderer 对象包含一个 drawRect() 方法，该方法接受其位置和大小以及四种颜色的参数.这些颜色将转换为 4 角渐变.如果您想要垂直渐变，只需将顶角设为一种颜色，将底角设为另一种颜色.像这样的:

In libGDX, the ShapeRenderer object contains a drawRect() method that takes arguments for its position and size as well as four colors. Those colors are converted to a 4-corners gradient. If you want a vertical gradient, just make the top corners one color and the bottom corners another color. Something like this:

shapeRenderer.filledRect(x, y, width, height, lightBlue, lightBlue, darkBlue, darkBlue);

来自 ShapeRenderer 的 API:

From the API for ShapeRenderer:

4个颜色参数指定矩形左下角、右下角、右上角和左上角的颜色，让您可以创建渐变.

The 4 color parameters specify the color for the bottom left, bottom right, top right and top left corner of the rectangle, allowing you to create gradients.

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