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
------解决方案--------------------
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'