如何重复ASIHTTPRequest？

Question

给出以下示例代码:

// 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[exampleModel sendRequest]从a 做某事时UIViewController,它有效!但是,然后我[exampleModel sendRequest]再次从另一个做UIViewController,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？

Answer 1

您不应尝试重用请求对象.它保持状态.真的设计为在请求结束后处理掉.

设计不像NSURLConnection,NSURLRequest,NSURLResponse类那样干净(基本上将三者合并为一个并将下面的低级核心基础类包装起来).如果您需要处理低级别的HTTP内容,它仍然比以一种普通方式使用NSURLConnection要好得多.如果不这样做,高级类有一些优点(比如访问UIWebView使用的相同缓存).