Hi
我正在尝试连接两个数据库表,并将行提取到php变量中。但这是我的第一个联接函数,我不会将任何值返回到表单中
SELECT s.supplierid,
s.name,
s.address,
s.zipCode,
s.cityName,
s.region,
s.country,
s.phone,
s.fax,
s.contactMame,
s.contactTitle,
s.contactPhone,
s.contactEmail,
s.notes,
s.employeeid,
e.employeeid as emp_id,
e.name as emp_name
FROM suppliers s
LEFT JOIN employees e
ON s.employeeid=e.employeeid
WHERE supplierid=? limit 0,1
ORDER BY name DESC
LIMIT :from_record_num, :records_per_page";
然后我把它们分配给像这样的变量
$name = $row['name'];
$address = $row['address'];
$emp_name = $row['emp_name'];
我需要在代码中再次使用它,但要从表中提取所有行。有没有办法使用
SELECT *
加入左边?
试试这样的方法:-(假定分页变量为$from_record_num
和$records_per_page
)
$sql="select
s.supplierid,
s.name,
s.address,
s.zipcode,
s.cityname,
s.region,
s.country,
s.phone,
s.fax,
s.contactmame,
s.contacttitle,
s.contactphone,
s.contactemail,
s.notes,
s.employeeid,
e.employeeid as 'emp_id',
e.name as 'emp_name'
from suppliers s
left join employees e on s.employeeid=e.employeeid
where s.supplierid=?
order by s.name desc
limit {$from_record_num}, {$records_per_page}";