ios - 快速滑动删除按钮的自定义高度
<p><p>我有自定义表格单元格。每个单元格都有不同的高度。我想给滑动删除按钮与单元格高度。</p>
<p> <a href="/image/uZNGY.png" rel="noreferrer noopener nofollow"><img src="/image/uZNGY.png" alt="I have attached screen shot for"/></a> </p>
<p>我在单元格中使用了 View 。 View 高度为单元格高度 - 16。上下边距 8。</p>
<p>所以请帮我做这件事。</p></p>
<br><hr><h1><strong>Best Answer-推荐答案</ strong></h1><br>
<p><p>在您的自定义 Cell Swift 类中使用此代码</p>
<pre><code>override func layoutSubviews() {
let fltHeight:CGFloat = 46
var subviews: = self.subviews
let subview: UIView? = subviews as? UIView
if NSClassFromString("UITableViewCellDeleteConfirmationView") != nil {
if (subview?.isKind(of: NSClassFromString("UITableViewCellDeleteConfirmationView")!))! {
let deleteButtonView: UIView? = (subview?.subviews)
var buttonFrame: CGRect? = deleteButtonView?.frame
buttonFrame?.origin.x = (deleteButtonView?.frame.origin.x)!
buttonFrame?.origin.y = (deleteButtonView?.frame.origin.y)!
buttonFrame?.size.width = (deleteButtonView?.frame.size.width)!
buttonFrame?.size.height = fltHeight
deleteButtonView?.frame = buttonFrame!
// Placing at the center of the cell.
subview?.frame = CGRect(x: CGFloat((subview?.frame.origin.x)!),
y: CGFloat((subview?.frame.origin.y)! + ((subview?.frame.size.height)!-fltHeight)/2),
width: CGFloat((subview?.frame.size.width)!),
height: CGFloat(fltHeight))
deleteButtonView?.clipsToBounds = true
subview?.clipsToBounds = true
}
}
}
</code></pre></p>
<p style="font-size: 20px;">关于ios - 快速滑动删除按钮的自定义高度，我们在Stack Overflow上找到一个类似的问题：
<a href="https://stackoverflow.com/questions/37719670/" rel="noreferrer noopener nofollow" style="color: red;">
https://stackoverflow.com/questions/37719670/
</a>
</p>
页:
[1]