根据不同表中的ID计算所有行

时间:2019-01-18 07:25:30

标签: php mysql prepared-statement

我想基于photo计算product id。我认为循环计数中有一些事情要做。所以我编码如下,但是我得到了

致命错误:在非对象上调用成员函数fetch_assoc()

<?
include_once($_SERVER['DOCUMENT_ROOT']."/cgi-bin/connect.php");

$stmt=$mysqli->prepare("select * from products_db order by prd_id");
$stmt->execute();
$result=$stmt->get_result();
$data=array();
while($row=$result->fetch_assoc()){
    //count no. of img in loop
    $stmt=$mysqli->prepare("select pic_id from photo_db where pic_sid=?");
    $stmt->bind_param('s',$row['prd_sid']);
    $result=$stmt->get_result();
    $nPic=$result->num_rows;

    $data[]=array($row['prd_code'],$row['prd_en'],$row['cat_id'],
number_format($row['prd_cost']),$row['prd_stts'],$nPic,$row['prd_rcm']);
}
echo json_encode(array('data' => $data));
?>

然后我去做:

$stmt=$mysqli->prepare("select *,count(pic.pic_id) as npic 
from products_db prd 
inner join 
photo_db pic 
on prd.prd_sid=pic.pic_sid 
order by prd.prd_id");

它正在工作但由于与pic_sid中的photo_db无关联而未显示某些产品

我希望所有产品都显示照片计数,即使数据库中不存在这些照片也是如此。

2 个答案:

答案 0 :(得分:0)

尝试此查询,希望它能工作。

select count(pi.pic_id),pr.* from photo_db pi where pi.pic_sid in (select pr.prd_sid from products_db pr ) group by pi.pic_id

答案 1 :(得分:0)

感谢@Sayed Mohd Ali的建议。我想出了解决方案。

select *,
(select count(pic_id) as npic from photo_db where pic.pic_sid=prd.prd_sid) as npic
from products_db prd
left join photo_db pic on pic.pic_sid=prd.prd_sid
group by prd.prd_sid