计算mysql中值为1的行之间的行数

Question

I have the following data:

client, visit page 2
1,0
2,0
3,1
4,0
5,1
6,0
7,0
8,1

now I want to know how many clients have been since the last client visited page 2. So the outcome would be:

client 3, 3; client 5, 2; client 8,3

eventually I only need to numbers of occurrences of the gaps so here I would have as an ultimate answer:

gap,# occurrences
1,0    
2,1
3,2
4,0

Is this possible in mysql and best would be without a for loop?

The data is to large to load it completely in python/php. One way would be to use python/php and select all rows where 'visit page 2' is 1 and then do a for loop with a select query with count(rows) between succes n-1 and succes n. But this takes to much time.

Answer 1

Creates the effect of a row_number with the @rn variable, and maintains the prior row number where vp2=1 occurred in variable @lastWas .

The cross-join facilitates not having to do set @rn:= etc stuff at the top. And that is the sole purpose of it.

Note the derived table named inr . All derived tables require an alias name even if not used explicitly by name. One can run the innards of the derived table separately. This is used at the beginning of development for a sanity check, then several of the columns there can be ignored in the query on the outside that wraps it. Certainly as seen here. See inr-only at bottom of answer for a view of just that derived table.

Schema

create table x
(   id int auto_increment primary key, 
    client int not null,
    vp2 int not null
);

insert x(client,vp2) values 
(1,0),(2,0),(3,1),(4,0),
(5,1),(6,0),(7,0),(8,1);

Query

select client,rn-lastWas as gap from
(   select id,
    client,
    vp2,
    @lastWas as lastWas,
    greatest(@rn:=@rn+1,@lastWas:=if(vp2=1,@rn,@lastWas)) as junk, -- forces recalc immediacy inside greatest() function, caring not at all of results
    @rn as rn
    from x
    cross join (select @rn:=0 as r,@lastWas:=0 as l) as blahJunk
    order by id
) inr
where vp2=1

Results

+--------+------+
| client | gap  |
+--------+------+
|      3 |    3 |
|      5 |    2 |
|      8 |    3 |
+--------+------+

Advanced Mysql User Variable Techniques

---

inr-only:

This is a view of just the inr derived table, to help visualize the above.

+----+--------+-----+---------+------+------+
| id | client | vp2 | lastWas | junk | rn   |
+----+--------+-----+---------+------+------+
|  1 |      1 |   0 | 0       | 1    |    1 |
|  2 |      2 |   0 | 0       | 2    |    2 |
|  3 |      3 |   1 | 0       | 3    |    3 |
|  4 |      4 |   0 | 3       | 4    |    4 |
|  5 |      5 |   1 | 3       | 5    |    5 |
|  6 |      6 |   0 | 5       | 6    |    6 |
|  7 |      7 |   0 | 5       | 7    |    7 |
|  8 |      8 |   1 | 5       | 8    |    8 |
+----+--------+-----+---------+------+------+

---

brownie points section below

More schema

create table bon
(   -- bunch of numbers
    i int not null
);
insert bon(i) values (1),(2),(3),(4);

Brownie points query

select bon.i as gap,ifnull(qry.rowCount,0) as occurrences
from bon
left join
(   select gap,count(*) as rowCount from
    (   select client,rn-lastWas as gap from
        (   select id,
            client,
            vp2,
            @lastWas as lastWas,
            greatest(@rn:=@rn+1,@lastWas:=if(vp2=1,@rn,@lastWas)) as junk, -- forces recalc
            @rn as rn
            from x
            cross join (select @rn:=0 as r,@lastWas:=0 as l) as blahJunk
            order by id
        ) inr -- derived table alias
        where vp2=1
    ) xxx -- every derived table requires an alias
    group by gap
) qry -- derived table alias
on qry.gap=bon.i

results

+-----+-------------+
| gap | occurrences |
+-----+-------------+
|   1 |           0 |
|   2 |           1 |
|   3 |           2 |
|   4 |           0 |
+-----+-------------+

Answer 2

One approach, keep a total of the current gap and reset on each visit.

select v, count(*) from(
    select client,
           case when visit = 0 then @v := @v + 1
                else @v := @v + 1 end v,
           case when visit = 1 then @v := 0 end gap
    from table join (select @v := 0) v
    order by client
) q
where gap is not null
group by v
;