我使用AFNetworking
框架來調用Web服務,我已經使用 POST方法從這樣的Web服務響應:如何在iOS中執行block後得到響應?
AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
manager.requestSerializer = [AFJSONRequestSerializer serializer];
[manager.requestSerializer setTimeoutInterval:25];
[manager POST:stringURL parameters:param
success:^(AFHTTPRequestOperation *operation, id responseObject) {
NSLog(@"RESPONCE : %@", responseObject);
}
failure:^(AFHTTPRequestOperation *operation, NSError *error)
{
NSLog(@"Error: %@", error);
NSString *msg = error.localizedDescription;
UIAlertView *alertview = [[UIAlertView alloc]initWithTitle:@"That Shift" message:msg delegate:nil cancelButtonTitle:@"Dissmiss" otherButtonTitles:nil, nil];
[alertview show];
}];
NSLog(@"Right AFter Block Execution RESPONCE : %@", responseObject);
我希望我的回答之後其NSLog(@"Right AFter Block Execution RESPONCE : %@", responseObject);
執行任何一個可以建議我該怎麼做?
原因BEHINDE這樣做:
背後這樣做的原因是,我想我已經知道該代碼後立即使用另一塊,我想使用第二塊性反應,這將是右後那在NSLog(@"Right AFter Block Execution RESPONCE : %@", responseObject);
的plce。
你不能因爲操作是異步的。 – trojanfoe
將此放入成功塊內有什麼問題? – YvesLeBorg