Question

标题可能听起来很奇怪，但我会尝试尽可能简单地解释问题。让我们以示例开始吧。我有一个有2列的表--id，ip。我们说我有3行，内容为2,3,5。现在，我需要在id 1和5之间得到任何行，这显然是1和4.目前我仍然坚持这个查询：

// ModelClass.h
@property(retain,nonatomic) NSMutableArray *productListArray;
-(void) getProductInfoWs:(WebService *) webServiceReference :(GlobalValues *) valueReference :(NSString *) categoryId :(UIView *) overlayView;

//  ModelClass.m

-(void) getProductInfoWs:(WebService *) webServiceReference :(GlobalValues *) valueReference :(NSString *) categoryId :(UIView *) overlayView{
productListArray=[localstoreModel getProductInfo:1];
if([[ResponseData objectForKey:@"status"] boolValue]==1){
   productListArray=[[ResponseData objectForKey:@"result"][@"category_products"] mutableCopy];
} 
else{
if([httpResponse statusCode]==200 && [[ResponseData objectForKey:@"status"] boolValue]==0){
[webServiceReference AlertMessage:[ResponseData objectForKey:@"errormsg"]];
}
else{
  [webServiceReference AlertMessage:@"2"];
}
}

//CustomAlert.m
-(void) AlertMessage:(NSString *) message{
UIAlertView *alertBox = [[UIAlertView alloc] initWithTitle:LocalizedString(@"Information!")
                                           message:message
                                           delegate:nil
                                           cancelButtonTitle:LocalizedString(@"OK")
                                           otherButtonTitles:nil];
[alertBox show];
}

Answer 1

听起来很奇怪，但这就是很多人所做的。

创建一个帮助表。将它用于左连接

create table amfn
(   -- All My Favorite Numbers  
    id int auto_increment primary key,
    theWhat char(1) null
)engine=MyIsam;   --  <----- somewhat important

insert amfn(theWhat) values (null),(null),(null),(null),(null),(null),(null),(null),(null),(null); -- 10
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;
insert amfn(theWhat) select theWhat from amfn;

select count(*),min(id),max(id) from amfn;
+----------+---------+---------+
| count(*) | min(id) | max(id) |
+----------+---------+---------+
|  1310720 |       1 | 1310720 |
+----------+---------+---------+
1 row in set (0.00 sec)

您的架构：

create table votes
(   question_id int not null,
    ip varchar(20) not null
);
insert votes (question_id,ip) values (1,'xxxx'),(2,'1.1.1.1'),(3,'1.1.1.1'),(4,'1.6.1.1'),(5,'1.1.1.1');

查询：

select a.id,v.question_id,v.ip
from amfn a
left join votes v
on v.question_id=a.id and v.ip='1.1.1.1'
where a.id between 1 and 5 and v.question_id is null;
+----+-------------+------+
| id | question_id | ip   |
+----+-------------+------+
|  1 |        NULL | NULL |
|  4 |        NULL | NULL |
+----+-------------+------+
2 rows in set (0.00 sec)     <------------- boy that is fast

编辑（显示Conrad Frix的时差）。

我上面创建5242880行的方法，23.5秒。康拉德的方法，168.5秒。我会坚持我的方法：＆gt;

Answer 2

好的，我找到了解决方案 -

我为每个question_id和所有填充了NULL的ip字段创建了新列，之后我在SQL查询中使用了这个逻辑：

SELECT question_id
FROM  `votes` 
WHERE (ip =  'NULL' OR ip = ?)
GROUP BY question_id
HAVING COUNT(*) - COUNT(DISTINCT question_id) = 0
ORDER BY RAND()
LIMIT 1

结果我得到一个尚未使用的随机行，因为它搜索重复项，如果找到任何重复项，则会从搜索中删除它们。我希望我写下这个可以理解的

SQL - 获取那些不存在的行

2 个答案:

您的架构：

查询：

编辑（显示Conrad Frix的时差）。