如何在ID中获得php mysql中的所有值？

Question

我有一个样本数据

product (ID, name)
         1 | 'iPhone'
         2 | 'iPad'
         3 | 'iWatch'
product_meta (ID, product_id, meta_key, meta_value) 
               1      1         image   iPhone.png       
               2      2         view        123

并使用查询：

SELECT p.*, m.*
FROM product AS p
LEFT JOIN product_meta AS m ON m.product_id = p.ID
WHERE p.ID = 1
GROUP BY p.ID 

如何获得所有价值的结果是

product(ID, name, image, view) =>  1 | iPhone | iPhone.png | 123

Answer 1

假设您的样本数据不正确，并且您要尝试对结果进行PIVOT ，则可以将MAX与CASE使用：

select p.id, 
   p.name, 
   max(case when pm.meta_key = 'image' then pm.meta_value end) image,   
   max(case when pm.meta_key = 'view' then pm.meta_value end) view
from product AS p
    left join product_meta AS pm ON pm.product_id = p.ID
where p.ID = 1
group by p.ID 

Answer 2

您可以像从其余查询中选择table1.column1，table2.column2一样。

Answer 3

基于与sgeddes相同的假设：

SELECT p.id, p.name, mimage.meta_value, mview.meta_value
FROM product AS p
LEFT JOIN product_meta AS mimage
    ON mimage.product_id = p.id AND mimage.meta_key = 'image'
LEFT JOIN product_meta AS mview
    ON mview.product_id = p.id AND mview.meta_key = 'view'
WHERE p.id = 1