【问题标题】：selecting from drop list and display the right result using php mysql inner join从下拉列表中选择并使用 php mysql inner join 显示正确的结果
【发布时间】：2013-05-19 16:16:09
【问题描述】：

我有一个包含省区市的搜索查询

我需要的是，当用户选择省时，浏览器必须显示与此选择相关的写入结果，当用户选择省和地区时，浏览器必须显示与该省和地区相关的结果，与村庄相同，但

问题是系统显示所有与省区村选择相关的用户结果都是同一个用户

查询：

 $sql = mysql_query("SELECT user_id,first_name, last_name, birth_date, registered_date, 
    s.specialization_name,
    g.governorate_name,
    d.district_name,
    v.village_name 
          FROM members u
                       INNER JOIN  specialization s 
                        ON u.specialization = s.specialization_id
                        INNER JOIN governorate g
                        ON u.governorate = g.governorate_id
                        INNER JOIN districts d
                        ON u.district = d.district_id
                        INNER JOIN village v
                        ON u.village = v.id
                       where (governorate = '' or governorate = '$bygov') or
                             (district = '' or district = '$bydist') or
                             (village = '' or village = '$byvillage')")
                             or die(mysql_error());

【问题讨论】：

标签： php mysql inner-join

【解决方案1】：

尝试使用 AND 组合 WHERE 选项

$sql = mysql_query("SELECT user_id,first_name, last_name, birth_date, registered_date, 
s.specialization_name,
g.governorate_name,
d.district_name,
v.village_name 
      FROM members u
                   INNER JOIN  specialization s 
                    ON u.specialization = s.specialization_id
                    INNER JOIN governorate g
                    ON u.governorate = g.governorate_id
                    INNER JOIN districts d
                    ON u.district = d.district_id
                    INNER JOIN village v
                    ON u.village = v.id
                   where (governorate = '' or governorate = '$bygov') AND
                         (district = '' or district = '$bydist') AND
                         (village = '' or village = '$byvillage')")
                         or die(mysql_error());

【讨论】：