所以我试图编辑我用PFQuery(parse.com)创建的数组。我需要检查它,然后从中删除对象,然后在UITableView中显示它,但是当我运行它时,我收到此错误
- (PFQuery *)queryForParse
{
PFQuery *query1 = [PFQuery queryWithClassName:self.parseClassName];
//[query1 selectKeys: [NSArray arrayWithContentsOfFile:[ NSString @"checkMark"]]];
[query1 whereKey:@"location" nearGeoPoint:[PFGeoPoint geoPointWithLatitude:[LocationController sharedSingleton].locationManager.location.coordinate.latitude longitude:[LocationController sharedSingleton].locationManager.location.coordinate.longitude] withinMiles:50];
[query1 findObjectsInBackgroundWithBlock:^(NSArray *objects, NSError *error) {
if (!error){
placesArray = [[NSMutableArray alloc] initWithArray:objects];
NSMutableArray *toDelete = [NSMutableArray array];
for (PFObject *tempObject in placesArray){
if ([tempObject objectForKey:@"checkMark"] == nil){
[toDelete addObject:tempObject];
}
}
[placesArray removeObjectsInArray:toDelete];
NSLog(@"%@", placesArray);
}
[placesTable reloadData];
}];
return query1;
}
- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{
static NSString *CellIdentifier = @"Cell";
PlacesTableCell *cell = (PlacesTableCell *)[tableView dequeueReusableCellWithIdentifier:CellIdentifier];
if (cell == nil){
cell = [[PlacesTableCell alloc] init];
}
PFObject *tempObject = [placesArray objectAtIndex:indexPath.row];
cell.message.text = [tempObject objectForKey:@"message"];
return cell;
}
答案 0 :(得分:0)
听起来你可能只需要添加“numberOfRowsInSection:
”方法。
例如:
- (NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section
{
// assuming only one section and one table here
return([placesArray count]);
}