圆形旋转五个按钮保持相等的距离 [英] Circular Rotation Of five buttons maintaining equal distance

问题描述

我想在圆周上旋转五个按钮（圆心与中心（120,120），实际上是方形视图的中心，即240 * 240），半径为100.是否可以这样做，有交互旋转的按钮和正确的外观。

I want to rotate five buttons on the circumference of an circle (circle with the center (120,120), which is actually center of square view i.e. 240*240) with radius 100. IS it possible to do so, with having interaction with the buttons which are rotating and a proper look.

我试过'

x =  round(cx + redious * cos(angl));
y =  round(cy - redious * sin(angl));   

NSString *X=[NSString stringWithFormat:@"%f",x];
[xval addObject:[NSString stringWithFormat:@"%@",X]];
NSString *Y=[NSString stringWithFormat:@"%f",y];
[yval addObject:[NSString stringWithFormat:@"%@",Y]];'

计算点数和：

map.frame= CGRectMake([[fqx objectAtIndex:counter1]intValue],[[fqy objectAtIndex:counter1]intValue], 72, 37);   
website.frame= CGRectMake([[fqx objectAtIndex:counter2]intValue],[[fqy objectAtIndex:counter2]intValue], 72, 37);   
share.frame= CGRectMake([[fqx objectAtIndex:counter3]intValue],[[fqy objectAtIndex:counter3]intValue], 72, 37); 
slideShow.frame= CGRectMake([[fqx objectAtIndex:counter4]intValue],[[fqy objectAtIndex:counter4]intValue], 72, 37);' 

旋转，但它产生奇怪的路径......以三角形方式..
（'map'，'share'，'slideShow'，'website'）我的按钮..：P

to rotate but it generate weird path.. in triangular way.. ('map','share','slideShow','website') ar my buttons.. :P

推荐答案

这太可爱了，不能放弃，所以这里就是。此代码仅限于一个按钮旋转，但我怀疑你将它扩展为多个（提示：使用一个CADisplayLink并在其中一次旋转所有按钮）。

This was too cute to pass up so here it is. This code is limited to one button rotating but I doubt you'll have trouble extending it to take multiple (hint: use ONE CADisplayLink and rotate all buttons at once in that).

- (void)setupRotatingButtons
{
    // call this method once; make sure "self.view" is not nil or the button 
    // won't appear. the below variables are needed in the @interface.
    // center: the center of rotation
    // radius: the radius
    // time:   a CGFloat that determines where in the cycle the button is located at
    //         (note: it will keep increasing indefinitely; you need to use 
    //         modulus to find a meaningful value for the current position, if
    //         needed)
    // speed:  the speed of the rotation, where 2 * M_PI is 1 lap a second
    // b:      the UIButton
    center = CGPointMake(100, 100);
    radius = 100;
    time = 0;
    speed = 2 * M_PI; // <-- will rotate CW 360 degrees per second (1 "lap"/s)

    b = [[UIButton buttonWithType:UIButtonTypeRoundedRect] retain];
    b.titleLabel.text = @"Hi";
    b.frame = CGRectMake(0.f, 0.f, 100, 50);
    // we get the center set right before we add subview, to avoid glitch at start
    [self continueCircling:nil]; 
    [self.view addSubview:b];
    [self.view bringSubviewToFront:b];
    CADisplayLink *dl = [CADisplayLink displayLinkWithTarget:self 
        selector:@selector(continueCircling:)];
    [dl addToRunLoop:[NSRunLoop currentRunLoop] forMode:NSDefaultRunLoopMode];
}

我们还需要实际的continueCircling：方法，它只是：

We also need the actual "continueCircling:" method, which is simply:

- (void)continueCircling:(CADisplayLink *)dl
{
    time += speed * dl.duration;
    b.center = CGPointMake(center.x + radius * cosf(time), 
                           center.y + radius * sinf(time));
}

我确信还有其他更好的方法可以解决这个问题，但是以上工作至少。 ：）

I'm sure there are other more nifty ways to solve this, but the above works at least. :)

编辑：我忘了提及，你需要添加QuartzCore框架和

I forgot to mention, you will need to add QuartzCore framework and

#import <QuartzCore/QuartzCore.h> 

。

编辑2：找到PI常量（ M_PI ），因此用3.1415代替。

Edit 2: Found the PI constant (M_PI), so replaced 3.1415 with that.

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