Question

id  || week_id || user_id || catch_id(0,1,2)
1   || 2       || 6       || 0
1   || 3       || 6       || 1
1   || 4       || 6       || 1
1   || 5       || 6       || 1
1   || 6       || 6       || 0
1   || 7       || 6       || 0
1   || 8       || 6       || 2
1   || 9       || 6       || 0
1   || 10      || 6       || 0
1   || 11      || 6       || 1

我需要为每个用户找出 catch = 1 连续最多一周和 catch = 0 连续一周（查找全部）。我希望我能说清楚。

在上表中

用户6的最大连续捕获= 1 3 （周3,4,5）

用户6的最大连续周数= 0 = 2 （第6,7周和/或第9,10周）

我如何去做。我可以在纯粹的SQL中执行此操作。也欢迎使用php解决方案

Answer 1

这适用于SQL解决方案。虽然它只会给你一个有问题的catch_id的week_id。我不知道你的桌子叫什么，所以我在下面的答案中称它为consecutive：

drop table if exists consecutive;

create table consecutive
(id int,week_id int,user_id int,catch_id int);

insert into consecutive values (1,2,6,0);
insert into consecutive values (1,3,6,1);
insert into consecutive values (1,4,6,1);
insert into consecutive values (1,5,6,1);
insert into consecutive values (1,6,6,0);
insert into consecutive values (1,7,6,0);
insert into consecutive values (1,8,6,2);
insert into consecutive values (1,9,6,0);
insert into consecutive values (1,10,6,0);
insert into consecutive values (1,11,6,1);

select w,count(*) as max_consecutive_weeks
from
(
select
case when @cur_catch_id != catch_id then @cur_week_id := week_id else @cur_week_id end as w,
@cur_catch_id := catch_id as catchChange,
c.*
from consecutive c
cross join (select @cur_catch_id := -1,@cur_week_id := -1) t
where user_id = 6
order by week_id asc
) t
where catch_id = 1
group by w
order by max_consecutive_weeks desc,w asc
limit 1;

通过将where catch_id = 1更改为where catch_id = 0，您可以使用相同的查询来获取catch_id = 0的最大连续week_id。

祝你好运！

Answer 2

如果PHP没问题，我会直接做到：

以ASC顺序week_id检索具有catch = x（x为0或1，具体取决于您要计算的内容）的所有项目
遍历这些项目：
- 检查week_id是否存在差距
- 更新最高

Answer 3

编写一个查询并获取一个名为数据的数组，周数= catch（键是周，catch是值）

$streak = array();
$WeekId = 0;
$prev   = 0;
$count  = 1;
foreach ($data as $week => $catch) 
{
    if($week == ++$WeekId && $prev == $catch)
    {
        $count ++;
        $WeekId = $week;
    }
    else 
    {
        if($prev !== 0)
        {
            $streak[$prev][$count]      = $count;
        }
        $WeekId                     = $week;    
        $count                      = 1;
        $prev                       = $catch;

    }
}
$streak[$prev][$count]      = $count;

现在计算每个 $ streak [0] 和 $ streak [1]

的max（）

Answer 4

我没有尝试过代码，但应该可以使用;可能需要稍微调整一下。

使用SQL并获取按week_id

排序的所有行

$currentcatch = '';
$run = 0;
$results = array();

while($record) {
    if ($record['catch_id'] == $currentcatch) {
        $run++;
    } else {
        if (!empty($currentcatch)) {
        if (empty($results[$currentcatch]) {
           $results[$currentcatch] = $run;
        } else {
           if ($results[$currentcatch] < $run) {
              $results[$currentcatch] = $run;
           }
        }
        }

        $run = 1;
        $currentcatch = $record['catch_id'];
    }
}

print_r($results);

Answer 5

这是一个PHP解决方案。我已经在我的灯上测试了它，它应该可以工作。

/*
  steps:
  1.sort the week_ids
  2.iterate the sorted week_ids and try to get all possible max numbers of consecutive records
  3.show the greatest one.

*/
$numstr = array(4,2,5,6,7,1);  //sample week_ids,should be fetched from db by catch_id
sort($numstr); 
$int_max = 1;  
$maxs_array = array();
for($i=0;$i<sizeof($numstr);$i++)
{
    $k = $i;
    while($numstr[$k])
    {
         if($numstr[$k+1] && $numstr[$k+1] == $numstr[$k]+1)  //duplicate week_ids not considered yet
         {
            $int_max ++;
         }
         else
         {
            array_push($maxs_array,$int_max);
            $int_max = 1;
             continue 2;
          }
          $k++;
     }
}
sort($maxs_array);
echo array_pop($maxs_array);  //output 4

查找连续记录的最大数量

5 个答案: