iphone - UIButton在按下时保持突出显示

Question

这真令人沮丧，我搜索了很多都无济于事。

我有一个按钮。按下该按钮时，它将调用执行网络操作的方法(NSURLRequest)。

该按钮应执行以下操作:

按下时:显示按下的状态图像，触发请求方法

按下后:显示按钮的禁用状态，直到方法完成请求

问题是在整个请求中，按钮均处于突出显示/按下状态。我已附上我目前在下面的代码。

对于按钮:

[myButton setBackgroundImage:[UIImage imageNamed:@"defaultbutton"] forState:UIControlStateNormal];
[myButton setBackgroundImage:[UIImage imageNamed:@"pressedbutton"] forState:UIControlStateHighlighted];
[myButton setBackgroundImage:[UIImage imageNamed:@"disabledbutton"] forState:(UIControlStateDisabled|UIControlStateSelected)];
[squishButton addTarget:self action:@selector(reqMethod) forControlEvents:UIControlEventTouchUpInside];

在请求开始的方法中:

-(void)reqMethod {
NSLog(@"Starting request..");
[myButton setHighlighted:NO];
[myButton setEnabled:NO];
[myButton setSelected:YES];

请求完成后，它会隐藏普通按钮并显示可以正常工作的重置按钮。

Answer 1

您应该对多线程文档有所了解。
http://developer.apple.com/library/ios/#documentation/Cocoa/Conceptual/Multithreading/Introduction/Introduction.html

如果您需要一段代码，我认为这可能很有用

[myButton setBackgroundImage:[UIImage imageNamed:@"defaultbutton"] forState:UIControlStateNormal];
[myButton setBackgroundImage:[UIImage imageNamed:@"pressedbutton"] forState:UIControlStateHighlighted];
[myButton setBackgroundImage:[UIImage imageNamed:@"disabledbutton"] forState:(UIControlStateDisabled|UIControlStateSelected)];
[squishButton addTarget:self action:@selector(reqMethod) forControlEvents:UIControlEventTouchUpInside];

    dispatch_async(dispatch_get_global_queue(0, 0), 
                       ^{
                           //Your request
    dispatch_async(dispatch_get_main_queue(), 
                           ^{
                   NSLog(@"Starting request..");
                   [myButton setHighlighted:NO];
                   [myButton setEnabled:NO];
                   [myButton setSelected:YES];  
                     });
                });