I'm using a MySQL database which has a 'Staff' column. Example:
ID BUSINESS STAFF
1 Business 1 Bob
2 Business 2 Bill
3 Business 3 Paul, Bill
4 Business 4 Bob
I'm aiming to create a pie chart showing how many businesses each member of staff has, using the Google Charts API, and I'm having trouble counting total amounts.
The chart uses the formatting:
var data = google.visualization.arrayToDataTable([
['Staff', 'Business'],
['Bob', 2],
['Bill', 2],
['Paul', 1]
]);
How would I go about echoing a count of these lines? I've spent a fun 40 minutes messing up COUNT queries, with absolutely no joy.
I develop this code according your question ,you should change according your table and fields also change server configuration server name,user,password,database
$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;
I used this code and i found total ;
There is a common theme in the comments here and that is normalization . As a rule it is A Bad Thing to represent multiple values as a comma-separated list in a single column.
Here is a working example of an alternate DB design that should get you going in the right direction:
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');
...and then the query would look like this:
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;
I've included a 4th member of staff called 'George' to show that a normalized approach allows you to have members of staff with no business too.
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