使用IOS中的CG旋轉後的文本中心

Question

我必須旋轉圈內的文本，但我無法在弧畫中獲取它。我成功地使用CGAffineTranform進行旋轉，但文本始終不在徑向中心。注意：文本位於弧線內。

float l_angleText=lStartAngle+((lStopAngle-lStartAngle)/2); 
    CGContextSelectFont(p_contex, "Helvetica", 12.5, kCGEncodingMacRoman); 

    CGAffineTransform myTransform=CGAffineTransformRotate(CGAffineTransformMakeScale(1,-1),-l_angleText); 
    CGContextSetTextMatrix(p_contex, myTransform); 

    CGContextSetTextDrawingMode(p_contex, kCGTextFill); 
    CGContextSetRGBFillColor(p_contex, 1.0, 1.0, 1.0, 1.0); //White 

    // get x/y for an the angle. The point in which I start drawing the text. 
    CGPoint lPoints = angleRToPoint(-l_angleText,[self getRadioPoint:l_angleText]);  

    //Move the point according the coords. 
    const double l_pointX = mReference.x+lPoints.x; 
    const double l_pointY = mReference.y-lPoints.y; 
    NSString* lText= [self getText]; 
    CGContextShowTextAtPoint(p_contex, l_pointX, l_pointY, [lText UTF8String] , strlen([lText UTF8String])); 

Answer 1

我剛剛解決了使用象限方法，修正了sin和cos的偏移量。該解決方案提供了良好的體驗。

-(CGPoint) getShiftToCenter:(const float) angle 
{ 
    CGPoint lRet; 
    unsigned int x = getCuadrante(angle); 
    float rx,ry; 
    switch (x) { 
     case 1: 
      rx=-sin(angle); 
      ry=cos(angle); 
      break; 
     case 2: 
      rx=cos(angle); 
      ry=-cos(angle); 
      break; 
     case 3: 
      rx=sin(angle); 
      ry=-cos(angle); 
      break; 
     case 4: 
      rx=-sin(angle); 
      ry=cos(angle); 
      break; 
     default: 
      break; 
    } 


    const float lFactor=4; // this is according to size of the font use it 
    lRet.x=lFactor*rx; 
    lRet.y=lFactor*ry; 

    NSLog(@"Cuad %d , An %f ; x=%f - y=%f", x, angle, lRet.x,lRet.y); 


    return lRet; 
}