如何选择SQL

Question

$sql=mysql_query("SELECT a.ssmid, a.image,b.ssmid,b.image,c.ssmid,c.image FROM a user_photos_offline INNER JOIN user_profile_pic b on a.ssmid = b.ssmid INNER JOIN user_family_photos c on a.ssmid = c.ssmid"); 

    while($row=mysql_fetch_assoc($sql)){ 
     echo $row['image']; 
    } 

我想选择数据库中的所有图像3个表，我有3个表，这些表名user_photos_offline，user_profile_pic，user_family_photos和列名是image所有表。

而且在3代表我有ssmid同一列，现在我要选择图像的这3个表

Answer 1

来指代图像seperately，用AS：

SELECT 
    a.ssmid, a.image AS a_image, 
    b.ssmid, b.image AS b_image, 
    c.ssmid, c.image AS c_image 
FROM user_photos_offline a 
INNER JOIN user_profile_pic b on a.ssmid = b.ssmid 
INNER JOIN user_family_photos c on a.ssmid = c.ssmid 

在PHP中，你的指数将改变这些标识符：

while($row=mysql_fetch_assoc($sql)){ 
    echo $row['a_image']; 
    echo $row['b_image']; 
    echo $row['c_image']; 
} 

Answer 2

如果在每个表的ssmid之间没有关系,user_profile_pic和user_family_photos，不需要使用INNER JOIN。使用UNION代替：

$sql=mysql_query(
    "SELECT ssmid, image, CONCAT('folder1/', image) as image_filename FROM user_photos_offline 
    UNION 
    SELECT ssmid, image, CONCAT('folder2/', image) as image_filename FROM user_profile_pic 
    UNION 
    SELECT ssmid, image, CONCAT('folder3/', image) as image_filename FROM user_family_photos" 
    ); 

while($row=mysql_fetch_assoc($sql)){ 
    echo $row['image']; 
    echo $row['image_filename']; 
}