PHP:星级评定系统的概念?

发布于 2024-10-09 02:30:16 字数 1539 浏览 1 评论 0原文

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瞎闹 2024-10-16 02:30:16

这是一个非常简单的 mysql 示例:

drop table if exists image;
create table image
(
image_id int unsigned not null auto_increment primary key,
caption varchar(255) not null,
num_votes int unsigned not null default 0,
total_score int unsigned not null default 0,
rating decimal(8,2) not null default 0
)
engine = innodb;

drop table if exists image_vote;
create table image_vote
(
image_id int unsigned not null,
user_id int unsigned not null,
score tinyint unsigned not null default 0,
primary key (image_id, user_id)
)
engine=innodb;

delimiter #

create trigger image_vote_after_ins_trig after insert on image_vote
for each row
begin
 update image set 
    num_votes = num_votes + 1,
    total_score = total_score + new.score,
    rating = total_score / num_votes  
 where 
    image_id = new.image_id;
end#

delimiter ;

insert into image (caption) values ('image 1'),('image 2'), ('image 3');

insert into image_vote (image_id, user_id, score) values
(1,1,5),(1,2,4),(1,3,3),(1,4,2),(1,5,1),
(2,1,2),(2,2,1),(2,3,4),
(3,1,4),(3,5,2);

select * from image;
select * from image_vote;

here's a very simple mysql example:

drop table if exists image;
create table image
(
image_id int unsigned not null auto_increment primary key,
caption varchar(255) not null,
num_votes int unsigned not null default 0,
total_score int unsigned not null default 0,
rating decimal(8,2) not null default 0
)
engine = innodb;

drop table if exists image_vote;
create table image_vote
(
image_id int unsigned not null,
user_id int unsigned not null,
score tinyint unsigned not null default 0,
primary key (image_id, user_id)
)
engine=innodb;

delimiter #

create trigger image_vote_after_ins_trig after insert on image_vote
for each row
begin
 update image set 
    num_votes = num_votes + 1,
    total_score = total_score + new.score,
    rating = total_score / num_votes  
 where 
    image_id = new.image_id;
end#

delimiter ;

insert into image (caption) values ('image 1'),('image 2'), ('image 3');

insert into image_vote (image_id, user_id, score) values
(1,1,5),(1,2,4),(1,3,3),(1,4,2),(1,5,1),
(2,1,2),(2,2,1),(2,3,4),
(3,1,4),(3,5,2);

select * from image;
select * from image_vote;
第几種人 2024-10-16 02:30:16

ColorBox 的制造商制作了一个 jQuery/PHP 评级系统,您可以使用它,称为 ColorRating。 ColorBox 是一个很棒程序,所以我认为 ColorRating 也可以。我会检查一下,因为它可能会为您省去很多麻烦:

http://colorpowered.com/color rating/

The maker of ColorBox has made a jQuery/PHP rating system that you can use called ColorRating. ColorBox is a great program so I think that ColorRating would be as well. I'd check it out as it might save you a lot of trouble:

http://colorpowered.com/colorrating/

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