blocks|key|2794726|text|对我来说似乎没问题。你在合成按钮吗？试一试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2794727|self.myBtn.enabled+=+NO;|code-block|syntax|javascript|2794728|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|$]]

It seems ok to me. Are you synthesizing the button? Try

<pre><code>self.myBtn.enabled = NO;
</code></pre>

blocks|key|4907706|text|一定要试试这个..|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4907707|在.h中：|offset|length|style|CODE|4907708|@property(nonatomic,retain)UIButton+*myBtn;|code-block|syntax|javascript|4907709|在.m中：|4907710|@synthesize+myBtn;|4907711|然后，用[self.myBtn+setEnabled:+NO];代码替换您的[myBtn+setEnabled:+NO];代码。|4907712|entityMap^0|0|1|2|0|0|1|2|0|0|4|S|12|N|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$D|10|E|11|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|12|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|13|8|@$D|14|E|15|F|G]|$D|16|E|17|F|G]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|T|$]]

Do try this ..

In <code>.h</code> :

<pre><code>@property(nonatomic,retain)UIButton *myBtn;
</code></pre>

In <code>.m</code> :

<pre><code>@synthesize myBtn;
</code></pre>

And then,replace your <code>[myBtn setEnabled: NO];</code> code with <code>[self.myBtn setEnabled: NO];</code> code.

blocks|key|4907721|text|您应该将按钮设置为IBOutlet。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4907722|@property+(weak,+nonatomic)+IBOutlet+UIButton+*myBtn;|code-block|syntax|javascript|4907723|这样，您就可以在故事板中连接该按钮。这可能是你的问题。|4907724|然后调用此函数以禁用。|4907725|[_myBtn+setEnabled:NO];|4907726|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You should be setting up your button as a IBOutlet.

<pre><code>@property (weak, nonatomic) IBOutlet UIButton *myBtn;
</code></pre>

That way you can connect that button in storyboards. Which could be your issue.

Then call this to disable.

<pre><code>[_myBtn setEnabled:NO];
</code></pre>

blocks|key|2794756|text|如果您正在寻找快速代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2794757|var+button+=+UIButton()
button.enabled+=+false+//disable+the+button|code-block|syntax|javascript|2794758|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|$]]

If you are looking for swift code: 

<pre><code>var button = UIButton()
button.enabled = false //disable the button
</code></pre>

blocks|key|2794760|text|如果您正在寻找Swift3解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2794761|var+myBtn+=+UIButton()
myBtn.isEnabled+=+true|code-block|syntax|javascript|2794762|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|$]]

If you're looking for Swift3 solution

<pre><code>var myBtn = UIButton()
myBtn.isEnabled = true
</code></pre>

I'm stumped at why this code isn't working. I have a login button that I want to disable when the user isn't logged in.

I have a UIButton delared in my .h file like so:

<pre><code> IBOutlet UIButton *myBtn;
</code></pre>

I've set up a referencing outlet in Interface Builder to this button.

In my .m file, I've tried:

<pre><code>[myBtn setEnabled: NO];
</code></pre>

and 

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

But neither of these disable the button in the conditional statement I'm in. (I want to disable the login button when the user successfully logs in)

I'm able to do this with two other buttons on the same screen, so I know the code is correct. I don't throw any errors, so I think the object exists. The references to myBtn change color in XCode, too, so it appears to be a valid reference.

I must be missing something. What am I doing wrong here? (I'm a Windows developer, relatively new at Objective-C)

Disable UIButton programmatically

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我对为什么这段代码不能工作感到困惑。我有一个登录按钮，当用户没有登录时，我想禁用它。我在我的.h文件中有一个UIButton，如下所示：   IBOutlet UIButton *myBtn;我在Interface Builder中设置了一个指向此按钮的引用插座。在我的.m文件中，我尝试过：[myBtn setEnab...

问以编程方式禁用UIButton
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问以编程方式禁用UIButtonEN