如何重试基于块的URL请求 [英] How to retry a block based URL Request

问题描述

我使用iOS7的新网址请求方法获取数据，如下所示：

  NSMutableURLRequest * request = [NSMutableURLRequest 
 requestWithURL：[NSURL URLWithString：[self.baseUrl 
 stringByAppendingString：path]]]; 
 
 NSURLSessionDataTask * dataTask = [[NSURLSession sharedSession] dataTaskWithRequest：request completionHandler：^（NSData * data，NSURLResponse * response，NSError * error）{
 NSHTTPURLResponse * httpResponse =（NSHTTPURLResponse *）response ; 
 NSUInteger responseStatusCode = [httpResponse statusCode]; 
 
 if（responseStatusCode！= 200）{
 // RETRY（$）
} else 
 completionBlock（results [@result] [符号]）; 
}]; 
 [dataTask resume];不幸的是，我不时得到HTTP响应，表明服务器不可达（<$ c $ b   
 c> response code！= 200 ），并需要向服务器重新发送相同的请求。 
 
 
 
如何做到这一点？我需要完成上面的代码片段，其中我的注释 // RETRY 是？
 
 
 
成功提取后调用完成块。 
但是如何再次发送同样的请求？
 
 
 
谢谢！
解决方案
将请求代码放入方法中，并在 dispatch_async 块中再次调用它;）
   - （void）requestMethod {
 
 NSMutableURLRequest * request = [NSMutableURLRequest 
 requestWithURL：[NSURL URLWithString：[self.baseUrl 
 stringByAppendingString ：路径]]]; 
 
 __weak typeof（self）weakSelf = self; 
 NSURLSessionDataTask * dataTask = [[NSURLSession sharedSession] dataTaskWithRequest：request completionHandler：^（NSData * data，NSURLResponse * response，NSError * error）{
 NSHTTPURLResponse * httpResponse =（NSHTTPURLResponse *）response; 
 NSUInteger responseStatusCode = [httpResponse statusCode]; 
 
 if（responseStatusCode！= 200）{
 dispatch_async（dispatch_get_global_queue（DISPATCH_QUEUE_PRIORITY_DEFAULT，0ul），^ {
 [weakSelf requestMethod]; 
}）; 
} else 
 completionBlock（results [@result] [symbol]）; 
}]; 
 [dataTask resume]; 
 
} 
  
 
I am fetching data using iOS7's new URL request methods, like so:
NSMutableURLRequest *request = [NSMutableURLRequest
    requestWithURL:[NSURL URLWithString:[self.baseUrl 
    stringByAppendingString:path]]];

NSURLSessionDataTask *dataTask = [[NSURLSession sharedSession] dataTaskWithRequest:request completionHandler:^(NSData *data, NSURLResponse *response, NSError *error) {
    NSHTTPURLResponse* httpResponse = (NSHTTPURLResponse*)response;
    NSUInteger responseStatusCode = [httpResponse statusCode];

    if (responseStatusCode != 200) { 
        // RETRY (??????) 
    } else       
        completionBlock(results[@"result"][symbol]);
}];
[dataTask resume];
Unfortunately, from time to time I get HTTP responses indicating the server is not reachable (response code != 200) and need to resend the same request to the server. 

How can this be done? How would I need to complete my code snippet above where my comment // RETRY is?

In my example I call the completion block after a successful fetch. 
But how can I send the same request again?

Thank you!
解决方案
Put your request code in a method and call it again in a dispatch_async block ;)
- (void)requestMethod {

    NSMutableURLRequest *request = [NSMutableURLRequest
                                    requestWithURL:[NSURL URLWithString:[self.baseUrl
                                                                         stringByAppendingString:path]]];

    __weak typeof (self) weakSelf = self;
    NSURLSessionDataTask *dataTask = [[NSURLSession sharedSession] dataTaskWithRequest:request completionHandler:^(NSData *data, NSURLResponse *response, NSError *error) {
        NSHTTPURLResponse* httpResponse = (NSHTTPURLResponse*)response;
        NSUInteger responseStatusCode = [httpResponse statusCode];

        if (responseStatusCode != 200) {
            dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0ul), ^{
                [weakSelf requestMethod];
            });
        } else       
            completionBlock(results[@"result"][symbol]);
    }];
    [dataTask resume];

}


                        
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