我有3张彼此相互关联的表格。
store_manufacture -------------- 类别 ----------------- --------- Store_Categories
----------------- -------------------- -----------------------
sm_id | sm_name cat_id | cat_name sc_id|store_id|cat_id
----------------- -------------------- -----------------------
12 | HP 1 | Travel 1 | 12 | 1
2 | Health 2 | 12 | 2
3 | Electronics 3 | 12 | 3
我想用所有类别ID显示storename一次,我必须将其发布到下一页。这是我试过的代码:
PHP
$cat_fetch=mysqli_query($con,"SELECT
sm_id,sm_brand_name,cat_id,sm_image,sm_link FROM `store_manufacture` sm
INNER JOIN store_category sc ON sc.store_id=sm.sm_id");
while($row=mysqli_fetch_array($cat_fetch,MYSQLI_ASSOC){
$id=$row['sm_id'];
echo " <h5> <a href=''>" .$row['sm_brand_name']." ". $row['cat_id']."</a></h5>";
}
输出
所需输出
HP (Travel, Health,Electronics)
答案 0 :(得分:0)
"SELECT
sm.sm_name,GROUP_CONCAT(c.cat_name) as cat_name FROM `store_manufacture` sm , Categories c, Store_Categories sc
where sm.sm_id = sc.store_id AND c.cat_id = sc.cat_id GROUP BY sm.sm_id";
这是查询你应该运行希望这会帮助你