SELECT * FROM city WHERE city.pname=$row_a['pname']$row_a['pname']为另一个表格的变量,求问这条句语可行否,如果出错,错在哪
回复讨论(解决方案)
MYSQL中根本没有 =$row_a['pname'] 这种东西。
把这个sql打印出来 放数据库里执行下 看看能不能执行
你那个$row_a['pname']是从PHP程序中传来的吧,改成这样就能执行了,大概
$sql="SELECT * FROM city WHERE city.pname='$row_a['pname']";$result=mysql_query($sql);
如果单纯用MYSQL,则需要把$row_a['pname']换成特定的值,如
SELECT * FROM city WHERE city.pname='A'
单纯sql能不能执行的。
你这个应该是多表交叉查询
数据表 row_a
数据表 city
表 row_a中的字段pname 与city 中的字段pname 相等
所以语句应该这样写:
select c.* from city as c inner join row_a as r on c.pname=r.pname
你可以搜一下 inner join , left join , right join 基本就知道了

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