ios - 求解决一个的四叶草布局效果

Question

效果如图6和6p 看起来很容易，我实现了后比较复杂，并且在3.5 上有问题

最好用autolayout xib 实现。
masonry 也可以。

需求

1. margin 边缘间距均一致，按钮整体位于屏幕的位置要舒适（视觉均分）

2. 高宽比例要保持

3. 4个矩形，尽量位于一个View中

4. 6p上不能拉伸图片

2x 素材

自己实现了，效果如图，期待简单思路中。

Answer 1

    UIView *containerView = [UIView new];

    UIView *subView0 = [[UIImageView alloc] initWithImage:[UIImage imageNamed:@"four"]];
    UIView *subView1 = [[UIImageView alloc] initWithImage:[UIImage imageNamed:@"four"]];
    UIView *subView2 = [[UIImageView alloc] initWithImage:[UIImage imageNamed:@"four"]];
    UIView *subView3 = [[UIImageView alloc] initWithImage:[UIImage imageNamed:@"four"]];

    [containerView addSubview:subView0];
    [containerView addSubview:subView1];
    [containerView addSubview:subView2];
    [containerView addSubview:subView3];


    [self.view addSubview:containerView];

#define kPadding 16

    [containerView mas_makeConstraints:^(MASConstraintMaker *make) {
         make.top.left.greaterThanOrEqualTo(@50).priorityHigh();
         make.right.bottom.greaterThanOrEqualTo(@-50).priorityHigh();
         make.center.equalTo(self.view);
     }];
    for(UIView *view in containerView.subviews) {
        [view setContentCompressionResistancePriority:10 forAxis:UILayoutConstraintAxisVertical];
        [view setContentCompressionResistancePriority:10 forAxis:UILayoutConstraintAxisHorizontal];
    }
    [subView0 mas_makeConstraints:^(MASConstraintMaker *make) {
         make.left.top.equalTo(@0);
         UIImage *image = [UIImage imageNamed:@"four"];
         if((image.size.width > 0) && (image.size.height > 0)) {
             make.height.equalTo(subView0.mas_width).multipliedBy(image.size.height/image.size.width).priorityHigh();
         }
     }];

    [subView1 mas_makeConstraints:^(MASConstraintMaker *make) {
         make.right.top.equalTo(@0);
         make.left.equalTo(subView0.mas_right).offset(kPadding);
     }];

    [subView2 mas_makeConstraints:^(MASConstraintMaker *make) {
         make.left.bottom.equalTo(@0);
         make.top.equalTo(subView0.mas_bottom).offset(kPadding);
     }];

    [subView3 mas_makeConstraints:^(MASConstraintMaker *make) {
         make.top.equalTo(subView1.mas_bottom).offset(kPadding);
         make.left.equalTo(subView2.mas_right).offset(kPadding);

         make.right.bottom.equalTo(@0);

         make.height.equalTo(@[subView0, subView1, subView2]);
         make.width.equalTo(@[subView0, subView1, subView2]);
     }];

Answer 2

Menggunakan batu adalah mudah,
1 Buat Paparan besar bernama bigView
2. Buat empat butang dan tambahkannya pada bigView

buttonOne的约束｛
    距top的距离;
    距left的距离;
    据right的距离;
    宽度写成比例（100/375.0*KScreenWidth）;
    高度写成比例（150/667.0*KScreenHeight）;
｝
buttonTwo的约束｛
    距top的距离;
    距left的距离;
    据right的距离;
    宽度等于buttonOne;
    高度等于buttonOne;
｝
buttonThree的约束｛
    距top的距离;
    距left的距离;
    据right的距离;
    宽度等于buttonOne;
    高度等于buttonOne;
｝
buttonFour的约束｛
    距top的距离;
    距left的距离;
    据right的距离;
    宽度等于buttonOne;
    高度等于buttonOne;
｝

Answer 3

Saya tidak begitu memahami keperluan anda Adakah anda bermaksud untuk membuat perkadaran reka letak empat butang ini sama pada 6 dan 6P?
Anda boleh mendapatkan lebar nilai lebar skrin dan memuatkan lebar, ketinggian, jarak dan lain-lain mengikut lebar
lebar biarkan = UIScreen.mainScreen().bounds.width

Answer 4

Maksud utama ialah nisbah kepada paparan induk, seperti yang ditunjukkan di bawah: