在sqlite3 xcode 6中更新表时出错

Question

我是iOS新手。 目前我正在尝试使用xcode6 obj C和sqlite3构建我自己的应用程序。

我能够打开数据库，创建表，但无法更新。你能帮帮我吗？代码如下

    @interface ViewController : UIViewController
    {
        sqlite3 *db;
    }

- (void)viewDidLoad {
    [self openDB];
    [self createTable:@"custadd" withField1:@"ID" withField2:@"Name" withField3:@"Phone"];
    [super viewDidLoad];
    // Do any additional setup after loading the view, typically from a nib.
}

    //to create table    
    -(void) createTable:(NSString *) tableName
             withField1: (NSString*) field1
             withField2: (NSString*) field2
             withField3: (NSString*) field3
    {
        char *err;
        NSString *sql = [NSString stringWithFormat:@"Create table if not exists '%@' ('%@' char primary key, '%@' char, '%@' char not null );",tableName,field1,field2,field3];
        if (sqlite3_exec(db, [sql UTF8String], NULL, NULL, &err) != SQLITE_OK) {
            sqlite3_close(db);
            NSAssert(0, @"Could not create table");
        }else{
            NSLog(@"Table created");
        }
    }

    - (IBAction)saveButton:(id)sender {
    //    NSString *sql = [NSString stringWithFormat:@"INSERT INTO custadd('ID','Name','Phone') Values ('%@','%@','%@']",custid,name,phone];

    //    NSString *sql = @"INSERT INTO custadd(1,ose,012345)";
        NSString *sql = @".table";

        char *err;
        if (sqlite3_exec(db, [sql UTF8String], NULL, NULL, &err) != SQLITE_OK ){
    //        sqlite3_close(custadd);
    //        NSAssert(0, @"Could not update table");
            NSLog(@"table not updated");

        } else{
            NSLog(@"table updated");
        }

    }