Question

我有两个表，它们是properties和property_images。 properties和property_images表中的字段如下所示

属性

property_id 
name
address
city
type

property_images

prop_img_id
photoname
property_id

我有一个查询来从两个表中获取结果。但是当我运行查询时，我只为每个属性获得一个图像。但我得到了正确的图像数量。我想在查询中获取属性的所有照片。我的查询是：

SELECT *, photo, COUNT(property_photo.photo) as photocount
FROM properties LEFT JOIN
     property_photo
     ON properties.property_id = property_photo.property_id
GROUP BY properties.property_id
LIMIT 6

Answer 1

您可以使用GROUP_CONCAT()：

SELECT p.*, COUNT(pp.photoname) as photocount,
       GROUP_CONCAT(pp.photoname SEPARATOR '|')
FROM properties p LEFT JOIN
     property_photo pp
     ON p.property_id = pp.property_id
GROUP BY p.property_id
LIMIT 6;

这会将照片的名称放在一个字符串中，每个名称都有一个垂直条。

Answer 2

使用内部查询来计算，然后将其与照片属性

连接

    select tt.photo,t.photocount from property_photo tt inner join 
(SELECT p.property_id, COUNT(pp.photo) as photocount 
  FROM properties p LEFT JOIN property_photo pp
 ON p.property_id = pp.property_id GROUP BY p.property_id ) t on t.property_id=tt.property_id;

MySQL查询获取所有照片

2 个答案: