ios-UITableView從jsonURL加載數據

UITableView從jsonURL加載數據

在UITableview的JSON的URL綁定數據，但是在執行過程中報錯。

NSData *response = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];
NSString *json_string = [[NSString alloc] initWithData:response encoding:NSUTF8StringEncoding];
statuses = [parser objectWithString:json_string error:nil];
[self.dropdownTblView reloadData];
for (NSDictionary *status in statuses) 
{
    _altTitle = [status valueForKey:@"Title"];
    NSLog(@"Title %@",_altTitle);
}

- (NSInteger)numberOfSectionsInTableView:(UITableView *)tableView
{
return 1;
}

- (NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section
{
   NSLog(@"%d",[statuses count]);
   return [statuses count];

}

- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{
static NSString *CellIdentifier = @"Cell";

UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:CellIdentifier];
if (cell == nil)
{
    cell = [[[UITableViewCell alloc] initWithStyle:UITableViewCellStyleDefault reuseIdentifier:CellIdentifier] autorelease];
}
return cell;

 //Here I'm getting an error    
id obj = [statuses objectAtIndex:indexPath.row];
NSString *name = [obj valueForKey:@"Title"];
cell.textLabel.text =name;
return cell;
}

JSON內容：

[
  {
    "Id": 1,
    "Title": "Tamil to English",
    "AltTitle": "த|மி|ழ்| |மூ|ல|ம்| |ஆ|ங்|கி|ல|ம்",
    "Description": "Learn English through Tamil",
    "Code": 1,
    "Version": "1.0",
    "SourceLanguageIndicator": "அ",
    "TargetLanguageIndicator": "A",
    "SourceLanguageCode": "ta",
    "TargetLanguageCode": "en",
    "Updated": "2013-02-21T03:33:19.6601651+00:00"
  }    
]    

最佳回答：

上面的代碼在未執行到你報錯的地方的時候,就返回了吧

return cell; //這一句直接就返回了.不會執行下面的代碼.

 //Here I'm getting an error    
id obj = [statuses objectAtIndex:indexPath.row];