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MySQl里类似Oracle rownum的实现

2013年12月12日 ⁄ 综合 ⁄ 共 1173字 ⁄ 字号 评论关闭
-- dense rank,写法1
set @curr_cut:=0, @prev_cnt:=0, @rank:=0;
select actor_id, 
       @curr_cnt:=cnt as cnt,
       @rank:=if(@prev_cnt<>@curr_cnt,@rank:=@rank+1,@rank) as rank
       @prev_cnt:=@curr_cnt as dummy
from (select actor_id,count(*) as cnt from sakila.film_actor group by actor_id order by cnt desc limit 10) as der;

结果:

actor_id

cnt

rank

dummy

107

42

1

42

102

41

2

41

198

40

3

40

181

39

4

39

23

37

5

37

81

36

6

36

106

35

7

35

158

35

7

35

13

35

7

35

37

35

7

35

-- dense rank,写法2,不使用dummy列
set @curr_cnt:=0, @prev_cnt:=0, @rank:=0;
select actor_id,
       @curr_cnt:=cnt as cnt,
       @rank:=if(@prev_cnt<>@curr_cnt,@rank:=@rank+1,@rank) as rank
from (select actor_id,count(*) as cnt from sakila.film_actor group by actor_id order by cnt desc limit 10) as der
where least(0,@prev_cnt:=@curr_cnt)=0;

 

结果:

actor_id

cnt

rank

107

42

1

102

41

2

198

40

3

181

39

4

23

37

5

81

36

6

158

35

7

13

35

7

37

35

7

144

35

7

 

-- rank
set @curr_cnt:=0, @rank:=0;
select actor_id,@curr_cnt:=cnt as cnt,@rank:=@rank+1 as rank
from (select actor_id,count(*) as cnt from sakila.film_actor group by actor_id oeder by cnt desc limit 10) as der;

结果:

actor_id

cnt

rank

107

42

1

102

41

2

198

40

3

181

39

4

23

37

5

81

36

6

158

35

7

13

35

8

37

35

9

144

35

10

 

-- rownum
set @rownum:=0;
select actor_id,first_name,@rownum:=@rownum+1 as rownum from sakila.actor order by first_name limit 5;

结果:

actor_id

first_name

rownum

132

ADAM

1

71

ADAM

2

165

AL

3

173

ALAN

4

146

ALBERT

5

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