Question

我正在使用一个拥有'Staff'列的MySQL数据库。例如：

ID    BUSINESS    STAFF
1     Business 1  Bob
2     Business 2  Bill
3     Business 3  Paul, Bill
4     Business 4  Bob

我的目标是使用Google Charts API创建一个饼图，显示每位员工有多少家企业，而且我在计算总金额方面遇到了麻烦。

图表使用格式：

var data = google.visualization.arrayToDataTable([
['Staff', 'Business'],
['Bob',     2],
['Bill',     2],
['Paul',      1]
]);

我如何回应这些线的数量？我花了40分钟搞砸了COUNT个查询，绝对没有快乐。

Answer 1

我根据您的问题开发此代码，您应该根据您的表和字段更改服务器配置服务器名称，用户，密码，数据库

    $server="server";
    $user="user";
    $password="password";
    $database="database";
    $cid=mysql_connect($server,$user,$password);
    mysql_select_db($database,$cid);
    $query="select * from table";
    $rs=mysql_query($query,$conn);
    $val=0;
    while($row=mysql_fetch_array($rs))
    {
    $data=$row['business'];
    $val=$val+trim(str_replace("Business","", "$data"));
    }
    $total=$val;

我使用了这段代码，我找到了总数;

Answer 2

此处的评论中有一个共同主题，即normalization。通常，将多个值表示为单个列中的逗号分隔列表是一件坏事。

以下是备用数据库设计的一个工作示例，它可以帮助您朝着正确的方向前进：

create table staff
(
id int unsigned not null primary key auto_increment,
staffName varchar(250),
unique key `staffUIdx1` (staffName)
) ENGINE=InnoDB;

create table business
(
id int unsigned not null primary key auto_increment,
businessName varchar(250),
unique key `staffUIdx1` (businessName)
) ENGINE=InnoDB;

CREATE TABLE staffBusiness
(
id int unsigned not null primary key auto_increment,
staffId int unsigned not null,
businessId int unsigned not null,
unique key `staffBusinessUIdx1` (staffId,businessId),
CONSTRAINT `fk_staffBusiness_staff1` FOREIGN KEY (`staffId`) REFERENCES `staff` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION,
CONSTRAINT `fk_staffBusiness_business1` FOREIGN KEY (`businessId`) REFERENCES `business` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB;

insert into staff (staffName) values ('Bob');
insert into staff (staffName) values ('Bill');
insert into staff (staffName) values ('Paul');
insert into staff (staffName) values ('George');


insert into business (businessName) values ('Business 1');
insert into business (businessName) values ('Business 2');
insert into business (businessName) values ('Business 3');
insert into business (businessName) values ('Business 4');

insert into staffBusiness (staffId,businessId) 
select s.id,b.id from staff s join business b 
where s.staffName = 'Bob' and b.businessName in ('Business 1','Business 4'); 

insert into staffBusiness (staffId,businessId) 
select s.id,b.id from staff s join business b 
where s.staffName = 'Bill' and b.businessName in ('Business 2','Business 3'); 

insert into staffBusiness (staffId,businessId) 
select s.id,b.id from staff s join business b 
where s.staffName = 'Paul' and b.businessName in ('Business 3');

...然后查询将如下所示：

select staffName as Staff,count(sb.id) as Business
from staff s
left outer join staffBusiness sb on s.id = sb.staffId
group by staffName;

我已经包括了一名名为“乔治”的第四名员工，表明规范化的方法可以让您的员工也没有业务。

如何累计数据库中的条目

2 个答案: