Question

我尝试通过从后台线程同步运行来实现 - [NSString stringWithContentsOfURL：encoding：error：]异步：

__block NSString *result;
dispatch_queue_t currentQueue = dispatch_get_current_queue();

void (^doneBlock)(void) = ^{
    printf("done! %s",[result UTF8String]);
};

dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT,
                                         (unsigned long)NULL), ^(void) {
    result = [NSString stringWithContentsOfURL:[NSURL URLWithString:@"http://www.google.com/"] encoding:NSUTF8StringEncoding error:nil];
    dispatch_sync(currentQueue, ^{
        doneBlock();
    });
});

它的工作正常，最重要的是它的异步。

我的问题是这样做是否安全，或者是否存在任何线程问题等？

提前致谢：）

Answer 1

这应该是安全的，但为什么要重新发明轮子？

NSURLRequest *req = [NSURLRequest requestWithURL:[NSURL URLWithString:@"http://www.google.com"]];
[NSURLConnection sendAsynchronousRequest:req queue:[NSOperationQueue mainQueue] completionHandler:^(NSURLResponse *response, NSData *data, NSError *error) {
    NSString *result = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
    // etc
}];

Answer 2

您也可以使用：

dispatch_queue_t queue = dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0);

dispatch_async(queue, ^{
        NSError *error = nil;
        NSString *searchResultString = [NSString stringWithContentsOfURL:[NSURL URLWithString:searchURL]
                                                           encoding:NSUTF8StringEncoding
                                                              error:&error];
        if (error != nil) {
            completionBlock(term,nil,error);
        }
        else
        {
            // Parse the JSON Response
            NSData *jsonData = [searchResultString dataUsingEncoding:NSUTF8StringEncoding];
            NSDictionary *searchResultsDict = [NSJSONSerialization JSONObjectWithData:jsonData
                                                                              options:kNilOptions
                                                                                error:&error];
            if(error != nil)
            {
                completionBlock(term,nil,error);
            }
            else
            {

                //Other Work here
            }
        }
    });

但是，它应该是安全的。我被告知虽然使用NSURLConnection而是由于错误调用等通过互联网进行通信。我还在研究这个问题。

Answer 3

-(void)loadappdetails:(NSString*)appid {
    NSString* searchurl = [@"https://itunes.apple.com/lookup?id=" stringByAppendingString:appid];

    [self performSelectorInBackground:@selector(asyncload:) withObject:searchurl];

}
-(void)asyncload:(NSString*)searchurl {
    NSURL* url = [NSURL URLWithString:searchurl];
    NSError* error = nil;
    NSString* str = [NSString stringWithContentsOfURL:url encoding:NSUTF8StringEncoding error:&error];
    if (error != nil) {
        NSLog(@"Error: %@", error);
    }
    NSLog(@"str: %@", str);
}

使stringWithContentsOfURL异步 - 它安全吗？

3 个答案: