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Start your free trialraja nusantara
Courses Plus Student 2,084 PointsPassing A Block Error : challenge task 3 of 4
The Question is : Now switch to LoginViewController.m. Blocks can also be declared and passed around as variables, as we can see in the existing code below. This code declares a block named 'loginHandler'. In the login method (after the block), add a call to the PFUser class method 'logInWithUsernameInBackground:password:block:'. Use "test" as both the username and password parameters. Then pass in 'loginHandler' as the third parameter. You can use block variables in this manner like how you would use any other variable as a parameter.
Mycode is :
#import "LoginViewController.h"
#import "UIAlertView.h"
#import <Parse/Parse.h>
@implementation LoginViewController
- (void)viewDidLoad {
[super viewDidLoad];
}
- (IBAction)login:(id)sender {
// This is an example of how to declare a block named loginHandler. It starts with a
// return type, then the name of the block with a special syntax for naming. The next
// set of parentheses hold the parameters and their types. Then assigning the block
// on the right side looks just like how we defined it in the video (carat with the
// same parameters in parentheses).
void (^loginHandler)(PFUser *user, NSError *error) = ^(PFUser *user, NSError *error) {
if (error) {
UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:@"Sorry!"
message:@"Error logging in."
delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil];
[alertView show];
}
else {
[self.navigationController popToRootViewControllerAnimated:YES];
}
};
// Add your code below!
[PFUser logInWithUsernameInBackground:@"test" password:@"test" block:loginHandler{}];
}
@end
*** Error said : Compilation Error! With the block defined, this should be like a normal method call. Check your syntax and make sure everything matches the instructions.***
**Please someone help me. I've been stucking on this challenge for one hours now. ** Any help will be much appreciated.
3 Answers
Holger Liesegang
50,595 PointsHi Raja,
try without {} after loginHandler:
[PFUser logInWithUsernameInBackground:@"test" password:@"test" block:loginHandler];
Kind Regards Holger
Holger Liesegang
50,595 PointsYou've got to use the block as a variable hence the loginHandler only: Blocks and Variables
raja nusantara
Courses Plus Student 2,084 PointsThanks For Your Help Holger. Its's working now. Btw, could u please explain why i need to get rid of {} ?