ios - 为什么不是我的所有代码都在我的 findObjectsInBackgroundWithBlock :^ being executed?

Question

这是我的 ViewDidLoad 中的代码

我正在尝试在我拥有未执行的 NSLog 的代码部分执行某些操作。我没能找到有同样问题的人？

我哪里错了？提前致谢!

PFRelation *relation = [staffMember relationForKey:@"completedTraining"];
PFQuery *query = [relation query];
[query includeKey:@"trainingRecordPointer"];
[query findObjectsInBackgroundWithBlock:^(NSArray *completedTrainingRecords, NSError *error){
    if(!error){
        for (PFObject* completedTrainingRecord in completedTrainingRecords) {
            PFObject * recordWtihTypeAndName = [completedTrainingRecord objectForKey:@"trainingRecordPointer"];
            PFObject *outputObject = [[PFObject alloc]initWithClassName:@"NewTrainingRecord"];
            NSString *recordName = [recordWtihTypeAndName valueForKey:@"RecordName"];
            [completeRecordsWithType addObject:recordName];
            [outputObject addObject:recordName forKey:@"recordName"];
            [outputObject addObject:completedTrainingRecord.createdAt forKey:@"date"];
            [[trainingRecordsDictionary objectForKey:[recordWtihTypeAndName objectForKey:@"Type"]] addObject:outputObject];
            [self.tableView reloadData]; //it works up to this point but if I move this line outside
            //the for-loop nothing happens
            NSLog(@"this will execute"); // does execute

        }
        NSLog(@"this wont execute"); // doesn't execute

    } else {
        NSLog(@"Error: %@ %@", error, [error userInfo]);
    }
    NSLog(@"this wont execute"); // doesn't execute

}];

Answer 1

您应该将 [self.tableView reloadData]; 移动到您的 for 循环之外。

您还应该确保在主线程中重新加载 TableView ，而不是在后台线程中。

可能是这样的:

[query findObjectsInBackgroundWithBlock:^(NSArray *completedTrainingRecords, NSError *error){
    if(!error){
        for (PFObject* completedTrainingRecord in completedTrainingRecords) {

                 ... do your stuff ...

        }
         __weak typeof(self) weakSelf = self;
            dispatch_async(dispatch_get_main_queue(), ^ {
                  [weakSelf.tableView reloadData];
        });
    }
}];

您可能会遇到麻烦，因为您试图在后台线程上修改您的 UI。