iOS tableview 多个部分

Question

我在 iOS 上使用以下代码时遇到问题，我有 3 个部分，每个部分都有一定数量的行，这些行由函数定义:countcells。

但是每次我运行它都会出现同样的错误:2014-07-29 22:16:32.710 TYM-APP[5598:60b] * 由于未捕获的异常 'NSRangeException' 而终止应用程序，原因:'* -[__NSArrayI objectAtIndex:]: index 2 beyond边界 [0 .. 0]'

代码有什么问题？？

  - (NSInteger)numberOfSectionsInTableView:(UITableView *)tableView
{
   return 3;
}

- (NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section
{
   if (section == 0)
   {
       NSString *querySQL = [NSString stringWithFormat: @"SELECT count(*) FROM homedata where module_ID= 77"];
       return [self countcells:querySQL];
   }
   else if(section == 1)
   {
       NSString *querySQL = [NSString stringWithFormat: @"SELECT count(*) FROM homedata where module_ID= 78"];
       return  [self countcells:querySQL];
   }

   else if (section == 2)
   {
       return[self countcells:@"SELECT count(*) FROM homedata where module_ID= 81"];
   }
   else return 0;
}


- (int) countcells: (NSString *)query
{
   int e;

   NSString* text;
   NSString * paths=[self getWritableDBPath];

   const char *dbpath =  [paths UTF8String];
   sqlite3_stmt    *statement;
   static sqlite3 *database = nil;

   if (sqlite3_open(dbpath, &database) == SQLITE_OK)
   {
      NSString *querySQL = [NSString stringWithFormat:query];

      const char *query_stmt = [querySQL UTF8String];

      //  NSLog(@"Databasae opened = %@", userN);

      if (sqlite3_prepare_v2(database,
                           query_stmt, -1, &statement, NULL) == SQLITE_OK)
      {

         while(sqlite3_step(statement) == SQLITE_ROW)
         {
            text = [[NSString alloc]initWithUTF8String:(const char *)     sqlite3_column_text(statement, 0)];
            e = sqlite3_column_int(statement, 0);
            NSLog( @"Valeur directe: %d", e);
         }
         //  username.text=@"No Username";


         sqlite3_finalize(statement);
      }
      //  NSLog( @"Save Error: %s", sqlite3_errmsg(database) );
      NSLog( @"Count Value: %@", text);
      sqlite3_close(database);
   }

   return e;
}

Answer 1

int e 具有垃圾值。确保它应该 >= 0 值。