我有一个分段的UITableView,其中填充了来自我的数据库的数据。数据采用JSON格式,然后通过一个人的工作变化分成三个MutableArrays。访问BOOL值并从数组中设置表格单元格
NSData *data = [NSData dataWithContentsOfURL:url];
json = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:nil];
ridersInVan = [[NSMutableArray alloc] init];
first = [[NSMutableArray alloc] init];
second = [[NSMutableArray alloc] init];
third = [[NSMutableArray alloc] init];
for (int i=0; i< [json count]; i++)
{
item = json[i];
if ([@"1" isEqual: item[@"watch"]])
{
[first addObject:item];
} else if ([@"2" isEqual: item[@"watch"]])
{
[second addObject:item];
} else if ([@"3" isEqual: item[@"watch"]])
{
[third addObject:item];
}
}
ridersInVan = [NSMutableArray arrayWithObjects:first, second, third, nil];
我创建了tableview中并填充一切,但我试图做的是建立基于某些值在阵列内
{
driver = 0;
expiration = "2013-10-08";
greenCard = 1;
id = 5;
name = "greg smith";
paid = 1;
phoneNumber = "123 345-1234";
showNumber = 1;
watch = 3;
}
驱动程序,支付,shownumber是文本颜色所有BOOL的我怎样才能使用布尔值我有这件事
- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath: (NSIndexPath *)indexPath
{
UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:@"Cell" forIndexPath:indexPath];
cell.textLabel.text = [[[ridersInVan objectAtIndex:indexPath.section] objectAtIndex:indexPath.row] objectForKey:@"name"];
cell.detailTextLabel.text = [[[ridersInVan objectAtIndex:indexPath.section] objectAtIndex:indexPath.row] objectForKey:@"phoneNumber"];
if (paid == NO && currentDate <= 8)
{
cell.textLabel.textColor = START;
} else if (paid == YES) {
cell.textLabel.textColor = PAID;
isPaid = YES;
} else if (paid == NO && currentDate > 8 && currentDate <= 15)
{
cell.textLabel.textColor = LATE;
isLate = YES;
} else if (paid == NO && currentDate > 15 && currentDate <= 28)
{
cell.textLabel.textColor = AFTER_VAN_DATE;
afterDate = YES;
} else if (paid == NO && currentDate > 28)
{
cell.textLabel.textColor = OFF_OF_VAN;
offOfVan = YES;
}
return cell;
}
设置文本颜色210即时尝试设置付款价值的阵列中的付费价值..任何想法?
您已经添加了相当多的代码,但我仍然不明白您的数据模型或您的问题实际是什么。我建议你将问题简化为重点 - 我有这个数据,我怎么能得到这个数据 - 这是我的,但它取而代之。 – Wain