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从下拉列表中选择并使用php mysql内部连接显示正确的结果_Php_Mysql_Inner Join - Fatal编程技术网

从下拉列表中选择并使用php mysql内部连接显示正确的结果

从下拉列表中选择并使用php mysql内部连接显示正确的结果,php,mysql,inner-join,Php,Mysql,Inner Join,我有一个搜索查询,包括省区城市 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“区”时,浏览器必须显示与此省和区以及与村庄相关的结果,但 问题是,系统显示与省、区和村选择相关的所有用户,结果相同 查询: 尝试使用和组合您的WHERE选项 $sql = mysql_query("SELECT user_id,first_name, last_name, birth_date, registered_date, s.specialization_name,

我有一个搜索查询,包括省区城市

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“区”时,浏览器必须显示与此省和区以及与村庄相关的结果,但

问题是,系统显示与省、区和村选择相关的所有用户,结果相同

查询:
尝试使用和组合您的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());

尝试使用和组合您的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());