我有 3 个表,我试图通过从每个表中放入一些特定数据来获得最终表。表格如下:
第一个表名为tbl_category,如下所示:
+------+---------------+----------------+
| cid | category_name | category_image |
+------+---------------+----------------+
| 1 | cars | 1.jpg |
+------+---------------+----------------+
名为 tbl_wallpaper 的第二个表是这样的:
+------+---------------+----------------+-----------+
| id | cat_id | scat_name | image |
+------+---------------+----------------+-----------+
| 33 | 1 | blue cars | 001.png |
+------+---------------+----------------+-----------+
第三个表叫做tbl_scategory,它是这样的:
+------+---------------+
| scid |scategory_name |
+------+---------------+
| 1 | blue cars |
+------+---------------+
我试图达到的目标是得到这样的东西
"HD_WALLPAPER": [
{
"cid": "1",
"scat_name": "blue cars",
"category_name": "cars",
"category_image": "1.jpg",
"category_image_thumb": "001.png",
"total_wallpaper": "1"
},
表示数据来自表tbl_category和tbl_wallpaper和tbl_scategory。我正在尝试获取存储在表 tbl_scategory(tbl_category 的子类别)
中的壁纸总数我正在使用这段代码,但它带来了整个壁纸列
if(isset($_GET['cat_list']))
{
$jsonObj= array();
$cat_order=API_CAT_ORDER_BY;
$query="SELECT tbl_category.cid, tbl_category.category_name, tbl_category.category_image, tbl_scategory.scid, tbl_wallpaper.scat_name FROM tbl_category ,
tbl_scategory, tbl_wallpaper
ORDER BY tbl_category.".$cat_order."";
$sql = mysqli_query($mysqli,$query)or die(mysql_error());
while($data = mysqli_fetch_assoc($sql))
{
//Wallpaper count
$query_wall = "SELECT COUNT(*) as num FROM tbl_wallpaper WHERE scat_name='".$data['cid']."'";
$total_wall = mysqli_fetch_array(mysqli_query($mysqli,$query_wall));
$total_wall = $total_wall['num'];
$row['cid'] = $data['cid'];
$row['scat_name'] = $data['scat_name'];
$row['category_name'] = $data['category_name'];
$row['category_image'] = $file_path.'images/'.$data['category_image'];
$row['category_image_thumb'] = $file_path.'images/thumbs/'.$data['category_image'];
$row['total_wallpaper'] = $total_wall;
array_push($jsonObj,$row);
}
$set['HD_WALLPAPER'] = $jsonObj;
header( 'Content-Type: application/json; charset=utf-8' );
echo $val= str_replace('\\/', '/', json_encode($set,JSON_UNESCAPED_UNICODE | JSON_PRETTY_PRINT));
die();
}
不知道是什么问题
注意:表格有更多行,我只是把这些部分放在示例中
最佳答案
试试这个:
SELECT t1.cid, t1.category_name, t1.category_image, t2.scat_name, t2.image
FROM tbl_category AS t1, tbl_wallpaper AS t2, tbl_scategory AS t3
WHERE t1.id = t2.cat_id
AND t2.scat_name = t3.scategory_name
希望能解决你的问题
关于php - 从 3 个不同的 mysql 表中选择数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46313044/