表 business_circle
字段 id name parent_id
表 goods_area
字段 goods_id city area
$sql = mysql_query("select * from goods_area as a INNER JOIN business_circle as b on a.area=b.name order by a.goods_id asc");
while($rs = mysql_fetch_assoc($sql)){
echo $rs['area'].$rs['parent_id'].'
';
}
这样能调用处parent_id的值但我想调用出条件parent_id = id表business_circle里的name值
不知道讲的清楚吗?高手直接给完整代码,谢谢了
回复讨论(解决方案)
至少我没明白,能不能把运行前的表与运行后的结果表贴出来
不知道你要的是不是这个select * from goods_area as a INNER JOIN business_circle as b on a.area=b.name where a.parent_id =b.id order by a.goods_id asc
”这样能调用处parent_id的值但我想调用出条件parent_id = id表business_circle里的name值“
这句话让人理解不了,举例说明一下吧。
因为name包括父类和子类parent_id里的值就是子类的父类id
我想在列表里即显示子类的name名称也显示父类的name名称
如果他是父类的话,parent_id的值是0其他parent_id都父类的id
因为name包括父类和子类的名称parent_id里的值就是该子类的父类id
我想在列表里即显示子类的name名称也显示这个子类的父类的name名称
还是没有明白,能不能直接把想要的结果做成表贴出来

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