iphone - 如何重复ASIHTTPRequest？

给定以下示例代码：

// ExampleModel.h

@interface ExampleModel : NSObject <ASIHTTPRequestDelegate> {

}

@property (nonatomic, retain) ASIFormDataRequest *request;
@property (nonatomic, copy) NSString *iVar;

- (void)sendRequest;


// ExampleModel.m

@implementation ExampleModel

@synthesize request;
@synthesize iVar;

# pragma mark NSObject

- (void)dealloc {
    [request clearDelegatesAndCancel];
    [request release];
    [iVar release];
    [super dealloc];
}

- (id)init {
    if ((self = [super init])) {
        // These parts of the request are always the same.
        NSURL *url = [[NSURL alloc] initWithString:@"https://example.com/"];
        request = [[ASIFormDataRequest alloc] initWithURL:url];
        [url release];
        request.delegate = self;
        [request setPostValue:@"value1" forKey:@"key1"];
        [request setPostValue:@"value2" forKey:@"key2"];
    }
    return self;
}

# pragma mark ExampleModel

- (void)sendRequest {
    // Reset iVar for each repeat request because it might've changed.
    [request setPostValue:iVar forKey:@"iVarKey"];
    [request startAsynchronous];
}

@end

# pragma mark ASIHTTPRequestDelegate

- (void)requestFinished:(ASIHTTPRequest *)request {
    // Handle response.
}

- (void)requestFailed:(ASIHTTPRequest *)request {
    // Handle error.
}

当我从[exampleModel sendRequest]执行类似UIViewController的操作时，它起作用了！但是，然后我再次从另一个[exampleModel sendRequest]执行UIViewController并得到：

Terminating app due to uncaught exception 'NSInvalidArgumentException',
reason: '*** -[NSOperationQueue addOperation:]:
operation is finished and cannot be enqueued`

我怎样才能解决这个问题？

最佳答案

您不应该尝试重用请求对象。它保持状态。真正设计为在请求结束后进行处理。

该设计不如NSURLConnection，NSURLRequest，NSURLResponse类那么干净（基本上将这三个类全部融合为一个，并在其下包装了低层的核心基础类）。如果您需要处理低级的HTTP内容，它仍然比以香草方式使用NSURLConnection更好。如果您不这样做，那么高级类具有一些优点（例如访问UIWebView使用的相同缓存）。