blocks|key|926236|text|尝试为UIControlStateDisabled+(禁用的灰色图像)和UIControlStateNormal(正常图像)设置不同的图像，以便按钮为您生成禁用状态。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|926237|entityMap^0|3|M|10|K|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

Try to set the different images for <code>UIControlStateDisabled</code> (disabled gray image) and <code>UIControlStateNormal</code>(Normal image) so the button generate the disabled state for you.

blocks|key|710053|text|您可以使用以下代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|710054|sendButton.enabled+=+YES;
sendButton.alpha+=+1.0;

or

sendButton.enabled+=+NO;
sendButton.alpha+=+0.5;|code-block|syntax|javascript|710055|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 can use following code: 

<pre><code>sendButton.enabled = YES;
sendButton.alpha = 1.0;

or

sendButton.enabled = NO;
sendButton.alpha = 0.5;
</code></pre>

blocks|key|710100|text|另一个选项是更改禁用状态的文本颜色(例如，为浅灰色)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|710101|在情节提要编辑器中，从State+Config弹出按钮中选择Disabled。使用Text+Color弹出按钮更改文本颜色。|offset|length|style|CODE|710102|在代码中，使用-setTitleColor:forState:消息。|710103|entityMap^0|0|B|C|U|8|15|A|0|7|O|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]|$D|P|E|Q|F|G]|$D|R|E|S|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|K|$]]

Another option is to change the text color (to light gray for example) for the disabled state.

In the storyboard editor, choose <code>Disabled</code> from the <code>State Config</code> popup button. Use the <code>Text Color</code> popup button to change the text color.

In code, use the <code>-setTitleColor:forState:</code> message.

blocks|key|926409|text|你可以使用前两个答案，这样会得到更好的结果。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|926410|在属性检查器中(当您选择按钮时)，将State+Config更改为Disabled，以设置将在禁用时显示的新图像(删除Content->Enabled检查中的标记以使其禁用)。|926411|当您将状态更改为enabled时，映像将从该状态加载该状态。|926412|将alpha逻辑添加到其中是一个很好的细节。|926413|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

You can use the both first answers and is going to be a better result. 

In the attribute inspector (when you're selecting the button), change the State Config to Disabled to set the new Image that is going to appear when it is disabled (remove the marker in the Content->Enabled check to made it disabled). 

And when you change the state to enabled, the image will load the one from this state. 

Adding the alpha logic to this is a good detail.

blocks|key|710282|text|如果使用文本按钮，则可以将实例方法放入viewDidLoad|type|unstyled|depth|inlineStyleRanges|entityRanges|data|710283|-+(void)setTitleColor:(UIColor+*)color+forState:(UIControlState)state|code-block|syntax|javascript|710284|示例：|710285|[self.syncImagesButton+setTitleColor:[UIColor+grayColor]+forState:UIControlStateDisabled];|710286|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|$]]

If you use a text button, you can put into viewDidLoad the instance method

<pre><code>- (void)setTitleColor:(UIColor *)color forState:(UIControlState)state
</code></pre>

example:

<pre><code>[self.syncImagesButton setTitleColor:[UIColor grayColor] forState:UIControlStateDisabled];
</code></pre>

blocks|key|710314|text|#import+"UIButton%2BMy.h"
#import+<QuartzCore/QuartzCore.h>
@implementation+UIButton+(My)


-(void)fade+:(BOOL)enable{
++++self.enabled=enable;//
++++self.alpha=enable?1.0:0.5;
}

@end|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|710315|.h：|unstyled|710316|#import+<UIKit/UIKit.h>
@interface+UIButton+(My)

-(void)fade+:(BOOL)enable;

@end|710317|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>#import "UIButton+My.h"
#import &lt;QuartzCore/QuartzCore.h&gt;
@implementation UIButton (My)


-(void)fade :(BOOL)enable{
 self.enabled=enable;//
 self.alpha=enable?1.0:0.5;
}

@end
</code></pre>

.h:

<pre><code>#import &lt;UIKit/UIKit.h&gt;
@interface UIButton (My)

-(void)fade :(BOOL)enable;

@end
</code></pre>

blocks|key|926631|text|只需将状态配置更改为禁用，并选择您想要的，禁用状态的背景图像|type|unstyled|depth|inlineStyleRanges|entityRanges|data|926632|​|926633|📷|atomic|offset|length|926634|926635|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/997a1daf5a60c10c258df36fa46cd89c.png|imageAlt^0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@$G|V|H|W|1|X]]|A|$]]|$1|I|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q|R|-4]]]]

just change state config to disable and choose what you want, background Image for disabled state

<a href="https://i.stack.imgur.com/ZklpC.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ZklpC.png" alt="enter image description here"></a>

blocks|key|710527|text|这是一个相当古老的问题，但有一个简单得多的方法来实现这一点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|710528|myButton.userInteractionEnabled+=+false|offset|length|style|CODE|710529|这将仅禁用任何触摸手势，而不会更改按钮的外观|710530|entityMap^0|0|0|13|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

This is quite an old question but there's a much simple way of achieving this.

<code>myButton.userInteractionEnabled = false</code>

This will only disable any touch gestures without changing the appearance of the button

blocks|key|710613|text|在Swift+>+3.0中创建一个扩展，而不是每个按钮本身|type|unstyled|depth|inlineStyleRanges|entityRanges|data|710614|extension+UIButton+{
++++override+open+var+isEnabled+:+Bool+{
++++++++willSet{
++++++++++++if+newValue+==+false+{
++++++++++++++++self.setTitleColor(UIColor.gray,+for:+UIControlState.disabled)
++++++++++++}
++++++++}
++++}
}|code-block|syntax|javascript|710615|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|$]]

Create an extension in Swift > 3.0 rather than each button by itself

<pre><code>extension UIButton {
 override open var isEnabled : Bool {
 willSet{
 if newValue == false {
 self.setTitleColor(UIColor.gray, for: UIControlState.disabled)
 }
 }
 }
}
</code></pre>

blocks|key|710729|text|这个问题有很多答案，但如果你真的想使用backgroundColor来设计按钮样式，这些答案看起来都不是很有用。UIButton有很好的选项，可以为不同的控件状态设置不同的图像，但背景色没有相同的功能。因此，其中一个解决方案是添加扩展，它将从颜色生成图像，并将它们应用到按钮。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|710730|extension+UIButton+{
++private+func+image(withColor+color:+UIColor)+->+UIImage?+{
++++let+rect+=+CGRect(x:+0.0,+y:+0.0,+width:+1.0,+height:+1.0)
++++UIGraphicsBeginImageContext(rect.size)
++++let+context+=+UIGraphicsGetCurrentContext()

++++context?.setFillColor(color.cgColor)
++++context?.fill(rect)

++++let+image+=+UIGraphicsGetImageFromCurrentImageContext()
++++UIGraphicsEndImageContext()

++++return+image
++}

++func+setBackgroundColor(_+color:+UIColor,+for+state:+UIControlState)+{
++++self.setBackgroundImage(image(withColor:+color),+for:+state)
++}
}|code-block|syntax|javascript|710731|此解决方案只有一个问题--此更改不会应用于在storyboard中创建的按钮。对我来说，这不是问题，因为我更喜欢从代码中设计UI样式。如果你想使用故事板，那么@IBInspectable需要一些额外的魔力。|offset|length|style|CODE|710732|第二种选择是子类化，但我更倾向于避免这种情况。|710733|entityMap^0|0|0|27|E|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

This question has a lot of answers but all they looks not very useful in case if you really want to use backgroundColor to style your buttons.
UIButton has nice option to set different images for different control states but there is not same feature for background colors.
So one of solutions is to add extension which will generate images from color and apply them to button.

<pre><code>extension UIButton {
 private func image(withColor color: UIColor) -&gt; UIImage? {
 let rect = CGRect(x: 0.0, y: 0.0, width: 1.0, height: 1.0)
 UIGraphicsBeginImageContext(rect.size)
 let context = UIGraphicsGetCurrentContext()

 context?.setFillColor(color.cgColor)
 context?.fill(rect)

 let image = UIGraphicsGetImageFromCurrentImageContext()
 UIGraphicsEndImageContext()

 return image
 }

 func setBackgroundColor(_ color: UIColor, for state: UIControlState) {
 self.setBackgroundImage(image(withColor: color), for: state)
 }
}
</code></pre>

Only one issue with this solution -- this change won't be applied to buttons created in storyboard. As for me it's not an issue because I prefer to style UI from code. If you want to use storyboards then some additional magic with <code>@IBInspectable</code> needed.

Second option is subclassing but I prefer to avoid this.

blocks|key|927038|text|为不同的状态设置title+color：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|927039|@IBOutlet+weak+var+loginButton:+UIButton!+{
++++++++didSet+{
++++++++++++loginButton.setTitleColor(UIColor.init(white:+1,+alpha:+0.3),+for:+.disabled)
++++++++++++loginButton.setTitleColor(UIColor.init(white:+1,+alpha:+1),+for:+.normal)
++++++++}
++++}|code-block|syntax|javascript|927040|用法：(文本颜色会自动改变)|927041|loginButton.isEnabled+=+false|927042|​|927043|📷|atomic|927044|927045|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/79c5c410ba1fe864669d4774a77a68e3.png|imageAlt^0|8|B|0|0|0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|19|8|@]|D|@]|E|$]]|$1|Q|3|R|5|S|7|1A|8|@]|D|@$9|1B|A|1C|1|1D]]|E|$]]|$1|T|3|P|5|6|7|1E|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11|12|-4]]]]

Set <code>title color</code> for different states:

<pre><code>@IBOutlet weak var loginButton: UIButton! {
 didSet {
 loginButton.setTitleColor(UIColor.init(white: 1, alpha: 0.3), for: .disabled)
 loginButton.setTitleColor(UIColor.init(white: 1, alpha: 1), for: .normal)
 }
 }
</code></pre>

Usage: (text color will get change automatically)

<pre><code>loginButton.isEnabled = false
</code></pre>

<a href="https://i.stack.imgur.com/DaLGD.png" rel="noreferrer"><img src="https://i.stack.imgur.com/DaLGD.png" alt="enter image description here"></a>

blocks|key|927158|text|看起来下面的代码运行得很好。但在某些情况下，这是行不通的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|927159|sendButton.enabled+=+NO;
sendButton.alpha+=+0.5;|code-block|syntax|javascript|927160|如果上面的代码不起作用，请将其封装在主线程中。所以|927161|dispatch_async(dispatch_get_main_queue(),+%5E{
++++sendButton.enabled+=+NO;
++++sendButton.alpha+=+0.5
});|927162|如果你选择swift，就像这样|927163|DispatchQueue.main.async+{
++++sendButton.isEnabled+=+false
++++sendButton.alpha+=+0.5
}|927164|此外，我在这里编写了UIButton扩展函数。|927165|extension+UIButton+{
++++func+enable()+{
++++++++DispatchQueue.main.async+{
++++++++++++self.isEnabled+=+true
++++++++++++self.alpha+=+1.0
++++++++}
++++}

++++func+disable()+{
++++++++DispatchQueue.main.async+{
++++++++++++self.isEnabled+=+false
++++++++++++self.alpha+=+0.5
++++++++}
++++}
}|927166|谢谢你，祝你编码愉快！|927167|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

It looks like the following code works fine. But in some cases, this doesn't work.
<pre><code>sendButton.enabled = NO;
sendButton.alpha = 0.5;
</code></pre>
If the above code doesn't work, please wrap it in Main Thread.
so
<pre><code>dispatch_async(dispatch_get_main_queue(), ^{
 sendButton.enabled = NO;
 sendButton.alpha = 0.5
});
</code></pre>
if you go with swift, like this
<pre><code>DispatchQueue.main.async {
 sendButton.isEnabled = false
 sendButton.alpha = 0.5
}
</code></pre>
In addition, here I write UIButton extension functions.
<pre><code>extension UIButton {
 func enable() {
 DispatchQueue.main.async {
 self.isEnabled = true
 self.alpha = 1.0
 }
 }

 func disable() {
 DispatchQueue.main.async {
 self.isEnabled = false
 self.alpha = 0.5
 }
 }
}
</code></pre>
Thank you and enjoy coding!!!

blocks|key|711044|text|也许你可以继承你的按钮并覆盖你的isEnabled变量。优点是您可以在多个地方重用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|711045|override+var+isEnabled:+Bool+{
+++++didSet+{
+++++++++if+isEnabled+{
+++++++++++++self.alpha+=+1
+++++++++}+else+{
+++++++++++++self.alpha+=+0.2
+++++++++}
+++++}
+}|code-block|syntax|javascript|711046|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|$]]

Maybe you can subclass your button and override your isEnabled variable. The advantage is that you can reuse in multiple places. 

<pre><code>override var isEnabled: Bool {
 didSet {
 if isEnabled {
 self.alpha = 1
 } else {
 self.alpha = 0.2
 }
 }
 }
</code></pre>

In my iPhone app, I have a UIButton which I have created in Interface Builder. I can successfully enable and disable it like this in my code ...

<pre><code>sendButton.enabled = YES;
</code></pre>

or

<pre><code>sendButton.enabled = NO;
</code></pre>

However, the visual look of the button is always the same! It is not faded or grey. If I attempt to click it though, it is enabled or disabled as expected. Am I missing something? Shouldn't it look faded or grey?

Disabled UIButton not faded or grey

Swift

在我的iPhone应用程序中，我有一个在界面生成器中创建的UIButton。我可以在我的代码中成功地启用和禁用它...sendButton.enabled = YES;或sendButton.enabled = NO;但是，按钮的视觉外观始终是相同的！它不是褪色的或灰色的。但是，如果我尝试单击它，它会按预期启用或禁用。...

问已禁用UIButton未褪色或呈灰色
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