Make fragment shader output (location=1) transparent?

OpenGL
1

Hello, I am using an integer texture attachment to perform picking. My texture is defined as:

        texPickerMain = GLUtil.glGenTexture(gl);
    gl.glBindTexture(gl.GL_TEXTURE_2D, texPickerMain);
    gl.glTexImage2D(gl.GL_TEXTURE_2D, 0, gl.GL_R32I, canvasWidth, canvasHeight, 0, gl.GL_RED_INTEGER, gl.GL_INT, null);
    gl.glTexParameteri(gl.GL_TEXTURE_2D, gl.GL_TEXTURE_MIN_FILTER, gl.GL_NEAREST);
    gl.glTexParameteri(gl.GL_TEXTURE_2D, gl.GL_TEXTURE_MAG_FILTER, gl.GL_NEAREST);
    gl.glFramebufferTexture2D(gl.GL_FRAMEBUFFER, gl.GL_COLOR_ATTACHMENT1, gl.GL_TEXTURE_2D, texPickerMain, 0);

In my fragment shader I have

layout(location = 1) out int pickerId;

During my draw I am drawing to attachments 0 and 1 and then blitting this to my default renderer.

I would like to be able to set pickerId such that it is transparent(?). Basically, I want to draw on the screen but have the users mouse pass through when I readpixels to what is behind it.

My readPixels call is:

        gl.glReadBuffer(gl.GL_COLOR_ATTACHMENT1);
    gl.glReadPixels(mouseX, canvasHeight - mouseY, 1, 1, gl.GL_RED_INTEGER, gl.GL_INT, 0);

Note: I have tried making out pickerId an ivec4 and setting alpha to 0 and that did not seem to make a difference. I also need 32 bits in my pickerId as I need to pack as much information as possible in to the pickerId to map it back afterwards.

2

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