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我在一个更大的 CAShapelayer 内有 3 个 CAShapelayers。比如说矩形，一个大矩形内的三角形，所有这些 CAShapelayers 都绘制在一个图像上。我需要用图案图像填充更大的矩形，但不需要填充所有其他 2 个形状。我试过屏蔽，偶数规则。请帮忙。目前我得到的完全相反。填充内层而不是外层这就是我正在做的

// mask layer is the outer layer
         CAShapeLayer *maskLayer = [CAShapeLayer layer];
                            maskLayer.fillRule = kCAFillRuleEvenOdd;
                            maskLayer.frame = self.view.frame;


//inner layer1                       
     CAShapeLayer *maskSubLayer1 = [CAShapeLayer layer];
                            maskSubLayer1.fillRule = kCAFillRuleEvenOdd;
                            maskSubLayer1.frame = self.view.frame;


 //inner layer 2              
              CAShapeLayer *maskSubLayer2 = [CAShapeLayer layer];
                            maskSubLayer2.fillRule = kCAFillRuleEvenOdd;
                            maskSubLayer2.frame = self.view.frame;


 //shapelayer and shapelayer 2 are the exact layer that are drawn inside         
 // add the paths to sublayers of the mask
                            maskSubLayer1.path = shapeLayer.path;
                            maskSubLayer2.path = shapeLayer2.path;

                            [maskLayer addSublayer:maskSubLayer1];
                            [maskLayer addSublayer:maskSubLayer2];

  // masking the outer layer                 
           biggershapeLayer.mask = maskLayer;
            [biggershapeLayer setFillColor:[UIColor colorWithPatternImage:[UIImage imageNamed:[imagesArray objectAtIndex:indexPath.row]]].CGColor];
                            [self.view.layer addSublayer:biggershapeLayer];


   *******************************************
   *                                         *
   *                                         *
   *    *******             ******           *
   *    ********            ******           * 
   *     *******            ******           * 
   *      *******           ******           * 
   *                                         *
   *******************************************

This is the current result I want the exact opposite.
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