blocks|key|151921|text|你有没有试过这样的东西|type|unstyled|depth|inlineStyleRanges|entityRanges|data|151922|for+o+:+AnyObject+in+self.subviews+{
+++++if+let+v+=+o+as?+NSView+{
+++++++++v.removeFromSuperview()
+++++}
}|code-block|syntax|javascript|151923|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

did you try something like

<pre><code>for o : AnyObject in self.subviews {
 if let v = o as? NSView {
 v.removeFromSuperview()
 }
}
</code></pre>

blocks|key|2738370|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2738371|for+view+in+container_view.subviews+{
++++view.removeFromSuperview()
}|code-block|syntax|javascript|2738372|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this:

<pre><code>for view in container_view.subviews {
 view.removeFromSuperview()
}
</code></pre>

blocks|key|154393|text|你的语法有点不对劲。请确保显式强制转换。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|154394|+let+theSubviews+:+Array<NSView>+=+container_view.subviews+as+Array<NSView>
+for+view+in+theSubviews+{
+++++view.removeFromSuperview()
+}|code-block|syntax|javascript|154395|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Your syntax is slightly off. Make sure you cast explicitly.

<pre><code> let theSubviews : Array&lt;NSView&gt; = container_view.subviews as Array&lt;NSView&gt;
 for view in theSubviews {
 view.removeFromSuperview()
 }
</code></pre>

blocks|key|151950|text|尝试一下，我测试了一下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|151951|++let+theSubviews+=+container_view.subviews
++for+subview+in+theSubviews+{
++++++subview.removeFromSuperview()
++}|code-block|syntax|javascript|151952|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this out , I tested this :

<pre><code> let theSubviews = container_view.subviews
 for subview in theSubviews {
 subview.removeFromSuperview()
 }
</code></pre>

blocks|key|153102|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|153103|var+subViews+=+parentView.subviews+as+Array<UIView>

++++++for+someView+in+subViews
++++++{
++++++++++someView.removeFromSuperview()
++++++}|code-block|syntax|javascript|153104|更新：如果你觉得冒险，那么你可以在UIView上做一个扩展，如下所示：|offset|length|style|BOLD|153105|extension+UIView
{
++++func+removeAllSubViews()
++++{
+++++++for+subView+:AnyObject+in+self.subviews
+++++++{
++++++++++++subView.removeFromSuperview()
+++++++}
++++}

}|153106|然后这样叫它：|153107|parentView.removeAllSubViews()|153108|entityMap^0|0|0|0|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Try this: 

<pre><code>var subViews = parentView.subviews as Array&lt;UIView&gt;

 for someView in subViews
 {
 someView.removeFromSuperview()
 }
</code></pre>

UPDATE: If you are feeling adventurous then you can make an extension on the UIView as shown below: 

<pre><code>extension UIView
{
 func removeAllSubViews()
 {
 for subView :AnyObject in self.subviews
 {
 subView.removeFromSuperview()
 }
 }

}
</code></pre>

And call it like this: 

<pre><code>parentView.removeAllSubViews()
</code></pre>

blocks|key|2738405|text|这应该是最简单的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2738406|let+container_view:+NSView+=+...
container_view.subviews+=+[]|code-block|syntax|javascript|2738407|(其他方法请参见Remove+all+subviews?+)|offset|length|2738408|2738409|注意:这是一个MacOS问题，此答案仅适用于MacOS。它在iOS上不起作用。|style|BOLD|2738410|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/2156015/remove-all-subviews^0|0|0|8|K|0|0|0|7|5|M|5|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Z|8|@]|9|@$I|10|J|11|1|12]]|A|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|14|8|@$I|15|J|16|N|O]|$I|17|J|18|N|O]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|W]]]]

This should be the simplest solution.

<pre><code>let container_view: NSView = ...
container_view.subviews = []
</code></pre>

(see <a href="https://stackoverflow.com/questions/2156015/remove-all-subviews">Remove all subviews?</a> for other methods)

<hr>

Note this is a MacOS question and this answer works only for MacOS. It does not work on iOS.

blocks|key|2738458|text|编辑：(谢谢Jeremiah+/+Rollo)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2738459|到目前为止，在Swift+for+iOS中做到这一点的最好方法是：|2738460|view.subviews.forEach({+$0.removeFromSuperview()+})+//+this+gets+things+done
view.subviews.map({+$0.removeFromSuperview()+})+//+this+returns+modified+array|code-block|syntax|javascript|2738461|%5E%5E这些功能很有趣！|2738462|let+funTimes+=+["Awesome","Crazy","WTF"]
extension+String+{+
++++func+readIt()+{
++++++++print(self)
++++}
}

funTimes.forEach({+$0.readIt()+})|2738463|/结束编辑|2738464|只需执行以下操作：|2738465|for+view+in+self.view.subviews+{
++++view.removeFromSuperview()
}|2738466|或者如果您正在寻找特定的类|2738467|for+view:CustomViewClass!+in+self.view.subviews+{
++++++++if+view.isKindOfClass(CustomViewClass)+{
++++++++++++view.doClassThing()
++++++++}
++++}|2738468|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|15|8|@]|9|@]|A|$G|H]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|17|8|@]|9|@]|A|$G|H]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

EDIT: (thanks Jeremiah / Rollo)

By far the best way to do this in Swift for iOS is:

<pre><code>view.subviews.forEach({ $0.removeFromSuperview() }) // this gets things done
view.subviews.map({ $0.removeFromSuperview() }) // this returns modified array
</code></pre>

^^ These features are fun!

<pre><code>let funTimes = ["Awesome","Crazy","WTF"]
extension String { 
 func readIt() {
 print(self)
 }
}

funTimes.forEach({ $0.readIt() })
</code></pre>

//// END EDIT

Just do this:

<pre><code>for view in self.view.subviews {
 view.removeFromSuperview()
}
</code></pre>

Or if you are looking for a specific class

<pre><code>for view:CustomViewClass! in self.view.subviews {
 if view.isKindOfClass(CustomViewClass) {
 view.doClassThing()
 }
 }
</code></pre>

blocks|key|2951451|text|我不知道您是否设法解决了这个问题，但我最近遇到了一个类似的问题，For循环每次只留下一个视图。我发现这是因为当调用removeFromSuperview()时，self.subviews+(可能)发生了变化。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2951452|为了解决这个问题，我这样做了：|2951453|let+subViews:+Array+=+self.subviews.copy()
for+(var+subview:+NSView!)+in+subViews
{
++++subview.removeFromSuperview()
}|code-block|syntax|javascript|2951454|通过执行.copy()，我可以在修改self.subviews数组的同时删除每个子视图。这是因为复制的数组(subViews)包含对对象的所有引用，并且不会发生变化。|2951455|编辑:在您的情况下，我认为您将使用：|2951456|let+theSubviews:+Array+=+container_view.subviews.copy()
for+(var+view:+NSView!)+in+theSubviews
{
++++view.removeFromSuperview()
}|2951457|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|V|8|@]|9|@]|A|$G|H]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I don't know if you managed to resolve this but I have recently experienced a similar problem where the For loop left one view each time. I found this was because the self.subviews was mutated (maybe) when the removeFromSuperview() was called.

To fix this I did:

<pre><code>let subViews: Array = self.subviews.copy()
for (var subview: NSView!) in subViews
{
 subview.removeFromSuperview()
}
</code></pre>

By doing .copy(), I could perform the removal of each subview while mutating the self.subviews array. This is because the copied array (subViews) contains all of the references to the objects and is not mutated.

EDIT: In your case I think you would use:

<pre><code>let theSubviews: Array = container_view.subviews.copy()
for (var view: NSView!) in theSubviews
{
 view.removeFromSuperview()
}
</code></pre>

blocks|key|154513|text|在xcodebeta6中，这是可行的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|154514|++++var+subViews+=+self.parentView.subviews
++++for+subview+in+subViews+as+[UIView]+++{
++++++++subview.removeFromSuperview()
++++}|code-block|syntax|javascript|154515|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In xcodebeta6 this worked out.

<pre><code> var subViews = self.parentView.subviews
 for subview in subViews as [UIView] {
 subview.removeFromSuperview()
 }
</code></pre>

blocks|key|2951544|text|我写了这个扩展：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2951545|extension+UIView+{
++++func+lf_removeAllSubviews()+{
++++++++for+view+in+self.subviews+{
++++++++++++view.removeFromSuperview()
++++++++}
++++}
}|code-block|syntax|javascript|2951546|以便可以在UIViewController中使用self.view.lf_removeAllSubviews。等我有了更多的使用swift的经验后，我会把它放到我的https://github.com/superarts/LFramework的swift版本中(到目前为止，1天的经验，是的，对于API，我放弃了下划线)。|offset|length|2951547|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/superarts/LFramework^0|0|0|2A|13|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

I wrote this extension:

<pre><code>extension UIView {
 func lf_removeAllSubviews() {
 for view in self.subviews {
 view.removeFromSuperview()
 }
 }
}
</code></pre>

So that you can use self.view.lf_removeAllSubviews in a UIViewController. I'll put this in the swift version of my <a href="https://github.com/superarts/LFramework" rel="nofollow">https://github.com/superarts/LFramework</a> later, when I have more experience in swift (1 day exp so far, and yes, for API I gave up underscore).

blocks|key|155967|text|你一定要试试这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|155968|func+clearAllScrollSubView+()
{
++++let+theSubviews+=+itemsScrollView.subviews

++++for+(var+view)+in+theSubviews
++++{

++++++++if+view+is+UIView
++++++++{
++++++++++++view.removeFromSuperview()
++++++++}

++++}
}|code-block|syntax|javascript|155969|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

you have to try this

<pre><code>func clearAllScrollSubView ()
{
 let theSubviews = itemsScrollView.subviews

 for (var view) in theSubviews
 {

 if view is UIView
 {
 view.removeFromSuperview()
 }

 }
}
</code></pre>

blocks|key|155971|text|对于iOS/Swift，要删除我使用的所有子视图：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|155972|for+v+in+view.subviews{
+++v.removeFromSuperview()
}|code-block|syntax|javascript|155973|为了摆脱特定类(如UILabel)的所有子视图，我使用：|155974|for+v+in+view.subviews{
+++if+v+is+UILabel{
++++++v.removeFromSuperview()
+++}
}|155975|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

For iOS/Swift, to get rid of all subviews I use:

<pre><code>for v in view.subviews{
 v.removeFromSuperview()
}
</code></pre>

to get rid of all subviews of a particular class (like UILabel) I use:

<pre><code>for v in view.subviews{
 if v is UILabel{
 v.removeFromSuperview()
 }
}
</code></pre>

blocks|key|2951571|text|用于删除特定类的子视图-这是我在Xcode6.1.1中唯一能工作的Swift代码。假设您要删除的唯一子视图的类型为UIButton...|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2951572|for+subView+in+nameofmysuperview.subviews+{
++++if+subView.isKindOfClass(UIButton)+{
++++++++subView.removeFromSuperview()
++++}
}|code-block|syntax|javascript|2951573|entityMap^0|1L|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

For removing just subviews of a specific class - this was the only Swift code that worked for me in Xcode6.1.1.
Assuming the only subviews you want to remove are of type UIButton...

<pre><code>for subView in nameofmysuperview.subviews {
 if subView.isKindOfClass(UIButton) {
 subView.removeFromSuperview()
 }
}
</code></pre>

blocks|key|2951578|text|用于删除所有子视图的扩展，它将被快速删除。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2951579|import+Foundation
import+UIKit

extension+UIView+{
++++///+Remove+allSubView+in+view
++++func+removeAllSubViews()+{
++++++++self.subviews.forEach({+$0.removeFromSuperview()+})
++++}

}|code-block|syntax|javascript|2951580|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Extension for remove all subviews, it is quickly removed.

<pre><code>import Foundation
import UIKit

extension UIView {
 /// Remove allSubView in view
 func removeAllSubViews() {
 self.subviews.forEach({ $0.removeFromSuperview() })
 }

}
</code></pre>

blocks|key|2951588|text|代码可以写得更简单，如下所示。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2951589|view.subviews.forEach+{+$0.removeFromSuperview()+}|code-block|syntax|javascript|2951590|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The code can be written simpler as following.

<pre><code>view.subviews.forEach { $0.removeFromSuperview() }
</code></pre>

blocks|key|156017|text|一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|156018|while+view.subviews.count+>+0++{+(view.subviews[0]+as?+NSView)?.removeFromSuperview()+}+|code-block|syntax|javascript|156019|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<h2>One-liner:</h2>
<pre><code>while view.subviews.count &gt; 0 { (view.subviews[0] as? NSView)?.removeFromSuperview() } 
</code></pre>

blocks|key|156025|text|Swift+3|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|156026|如果将标记添加到视图，则可以删除特定视图。|156027|for+v+in+(view?.subviews)!
{
++++if+v.tag+==+321
++++{
+++++++++v.removeFromSuperview()
++++}
+}|code-block|syntax|javascript|156028|entityMap^0|0|7|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|S|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

Swift 3

If you add a tag to your view you can remove a specific view.

<pre><code>for v in (view?.subviews)!
{
 if v.tag == 321
 {
 v.removeFromSuperview()
 }
 }
</code></pre>

blocks|key|2738835|text|对于Swift+3|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2738836|我这样做是因为仅仅从superview中移除并不会擦除数组中的按钮。|2738837|++++for+k+in+0..<buttons.count+{

++++++buttons[k].removeFromSuperview()

++++}


++++buttons.removeAll()|code-block|syntax|javascript|2738838|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

For Swift 3

I did as following because just removing from superview did not erase the buttons from array.

<pre><code> for k in 0..&lt;buttons.count {

 buttons[k].removeFromSuperview()

 }


 buttons.removeAll()
</code></pre>

blocks|key|2951686|text|对于Swift+4.%2B|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2951687|extension+UIView+{
+++++public+func+removeAllSubViews()+{
++++++++++self.subviews.forEach({+$0.removeFromSuperview()+})

}|code-block|syntax|javascript|2951688|我希望这对你来说是完全有用的。|2951689|entityMap^0|2|9|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

For <code>Swift 4.+</code>

<pre><code>extension UIView {
 public func removeAllSubViews() {
 self.subviews.forEach({ $0.removeFromSuperview() })

}
</code></pre>

i hope this is use full for you.

blocks|key|154688|text|Swift+5|type|unstyled|depth|inlineStyleRanges|entityRanges|data|154689|我创建了两个不同的方法来删除子视图。如果我们把它们放在extension中，就更容易使用了|offset|length|style|CODE|154690|extension+UIView+{
++++///+Remove+all+subview
++++func+removeAllSubviews()+{
++++++++subviews.forEach+{+$0.removeFromSuperview()+}
++++}

++++///+Remove+all+subview+with+specific+type
++++func+removeAllSubviews<T:+UIView>(type:+T.Type)+{
++++++++subviews
++++++++++++.filter+{+$0.isMember(of:+type)+}
++++++++++++.forEach+{+$0.removeFromSuperview()+}
++++}
}|code-block|syntax|javascript|154691|entityMap^0|0|R|9|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

<h1>Swift 5</h1>
I create 2 different methods to remove subview. And it's much easier to use if we put them in <code>extension</code>
<pre><code>extension UIView {
 /// Remove all subview
 func removeAllSubviews() {
 subviews.forEach { $0.removeFromSuperview() }
 }

 /// Remove all subview with specific type
 func removeAllSubviews&lt;T: UIView&gt;(type: T.Type) {
 subviews
 .filter { $0.isMember(of: type) }
 .forEach { $0.removeFromSuperview() }
 }
}
</code></pre>

blocks|key|2951808|text|这是另一种方法，允许您在任何UIView实例(或UIView子类)集合上调用该操作。这使得在.subviews之后插入像filter这样的东西变得更容易，这样您就可以更有选择性，或者在其他UIView集合上调用它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2951809|extension+Array+where+Element:+UIView+{
++func+removeEachFromSuperview()+{
++++forEach+{
++++++$0.removeFromSuperview()
++++}
++}
}|code-block|syntax|javascript|2951810|示例用法：|2951811|myView.subviews.removeEachFromSuperview()

//+or,+for+example:

myView.subivews.filter+{+$0+is+UIImageView+}.removeEachFromSuperview()|2951812|或者，您可以使用UIView扩展来完成相同的任务(尽管不能在UIView实例的任意数组上调用它)：|2951813|extension+UIView+{
++++func+removeSubviews(predicate:+((UIView)+->+Bool)?+=+nil)+
++++++++subviews.filter(
++++++++++++predicate+??+{+_+in+true+}
++++++++).forEach+{
++++++++++++$0.removeFromSuperview()
++++++++}
++++}
}|2951814|2951815|myView.removeSubviews()
myView.removeSubviews+{+$0+is+UIImageView+}|2951816|entityMap^0|E|6|O|6|1A|9|1O|6|2M|6|0|0|0|0|8|6|U|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|19|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|S|3|L|5|6|7|1H|8|@]|D|@]|E|$]]|$1|T|3|U|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|V|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|W|$]]

Here's another approach that allows you call the operation on any collection of <code>UIView</code> instances (or <code>UIView</code> subclasses). This makes it easy to insert things like <code>filter</code> after <code>.subviews</code> so you can be more selective, or to call this on other collections of <code>UIView</code>s.
<pre><code>extension Array where Element: UIView {
 func removeEachFromSuperview() {
 forEach {
 $0.removeFromSuperview()
 }
 }
}
</code></pre>
Example usage:
<pre><code>myView.subviews.removeEachFromSuperview()

// or, for example:

myView.subivews.filter { $0 is UIImageView }.removeEachFromSuperview()
</code></pre>
Alternatively you can accomplish the same thing with a <code>UIView</code> extension (though this can't be called on some arbitrary array of <code>UIView</code> instances):
<pre><code>extension UIView {
 func removeSubviews(predicate: ((UIView) -&gt; Bool)? = nil) 
 subviews.filter(
 predicate ?? { _ in true }
 ).forEach {
 $0.removeFromSuperview()
 }
 }
}
</code></pre>
Example usage:
<pre><code>myView.removeSubviews()
myView.removeSubviews { $0 is UIImageView }
</code></pre>

blocks|key|153377|text|这是你一直在寻找的一行代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|153378|view.subviews.filter({$0.tag+==+321}).forEach({$0.removeFromSuperview()})|code-block|syntax|javascript|153379|在使用这段代码删除视图(someView.tag+=+321)之前，不要忘记标记它。|offset|length|style|CODE|153380|entityMap^0|0|0|C|I|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Here's a one-liner you've been looking for:
<pre><code>view.subviews.filter({$0.tag == 321}).forEach({$0.removeFromSuperview()})
</code></pre>
Don't forget to Tag your view (<code>someView.tag = 321</code>) before you remove it with this code.

I'm looking for a simple method to remove at once all subviews from a superview instead of removing them one by one.

<pre><code>//I'm trying something like this, but is not working
let theSubviews : Array = container_view.subviews
for (view : NSView) in theSubviews {
 view.removeFromSuperview(container_view)
}
</code></pre>

What I am missing?

UPDATE

My app has a main <code>container_view</code>. I have to add different other views as subviews to <code>container_view</code> in order to provide a sort of navigation.

So, when clicking the button to "open" a particular page, I need to remove allsubviews and add the new one.

UPDATE 2 - A working solution (OS X)

I guess Apple fixed it.

Now it is more easy than ever, just call: 

<pre><code>for view in containerView.subviews{
 view.removeFromSuperview()
}
</code></pre>

How to remove all subviews of a view in Swift?

Swift

MacOS

我正在寻找一种简单的方法来立即从superview中删除所有的子视图，而不是逐个删除它们。//I'm trying something like this, but is not workinglet theSubviews : Array = container_view.subviewsfor (view : NS...

问如何在Swift中删除视图的所有子视图？
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