【问题标题】:Values storing 2 times in database在数据库中存储 2 次的值
【发布时间】:2013-11-11 07:23:32
【问题描述】:

我必须在数据库中插入 x 和 y 触摸坐标值。但它在数据库中存储了 2 次。当我长按时,我得到触摸坐标并为 button_tag 插入一些标签。但是我长按一次,为什么这些值存储了 2 次​​p>

表:

button_tag   xcoor    ycoor
2             123.5    320.9
23            123.5    320.9


- (void)tapTest:(UILongPressGestureRecognizer *)sender {
    NSLog(@"coordinate is %f %f", [sender locationInView:wbCont.scrollView].x, [sender locationInView:wbCont.scrollView].y);


     xcor = [sender locationInView:wbCont.scrollView].x;
     ycor = [sender locationInView:wbCont.scrollView].y;


    universal_button_tag = arc4random() % 99;

    NSLog(@"Universal Button tag: %d",universal_button_tag);

    NSArray *paths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory,NSUserDomainMask, YES);
    NSString *documentsDirectory = [paths objectAtIndex:0];


    NSString *path = [documentsDirectory stringByAppendingPathComponent:@"db.sqlite"];
    NSLog(@"filepath %@",path);


    if (sqlite3_open([path UTF8String], &database) == SQLITE_OK) {



        const char *sql = [[NSString stringWithFormat:@"SELECT button_tag,xcoor,ycoor FROM touch where  button_tag='%d' AND xcoor = '%f' AND ycoor = '%f'",universal_button_tag,xcor,ycor] cStringUsingEncoding:NSUTF8StringEncoding];


        NSLog(@"sql is %s",sql);

        BOOL favExist = false;

        sqlite3_stmt *statement, *addStmt;

        if (sqlite3_prepare_v2(database, sql, -1, &statement, NULL) == SQLITE_OK) {
            // We "step" through the results - once for each row.
            while (sqlite3_step(statement) == SQLITE_ROW) {



                favExist = true;
            }
        }


        if(!favExist){


            //Changes



            const char *sqlInsert = [[NSString stringWithFormat:@"insert into touch (button_tag,xcoor,ycoor) values ('%d','%f','%f')", universal_button_tag,xcor,ycor] cStringUsingEncoding:NSUTF8StringEncoding];

            //----

            NSLog(@"sql insert is %s",sqlInsert);


            // [catID release];

            if(sqlite3_prepare_v2(database, sqlInsert, -1, &addStmt, NULL) != SQLITE_OK)
                NSAssert1(0, @"Error while creating add statement. '%s'", sqlite3_errmsg(database));

            NSLog(@"error is %s",sqlite3_errmsg(database));

            if(SQLITE_DONE != sqlite3_step(addStmt))
                NSAssert1(0, @"Error while inserting data. '%s'", sqlite3_errmsg(database));


        }

    }

}

【问题讨论】:

  • 您对favExist 的检测在我看来是错误的。您尝试在给定随机 id 的情况下在数据库中查找现有记录。该记录永远不会存在,或者极不可能。

标签: ios database insert sqlite uigesturerecognizer


【解决方案1】:

针对手势识别器的不同状态调用手势识别器的目标操作。您需要检查方法中的状态。

- (void)tapTest:(UILongPressGestureRecognizer *)sender {
    if (sender.state == UIGestureRecognizerStateEnded) {
        // handle completed gesture
    }
}

有关详细信息,请参阅UIGestureRecognizer 的文档。

【讨论】:

  • 在我的代码工作之前。我从数据库中删除了一列。在它不起作用之后
  • 这与您的原始问题或我的回答有什么关系?
猜你喜欢
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 2013-03-12
  • 1970-01-01
  • 2021-08-06
相关资源
最近更新 更多