好像做不对。我试图在我的表单中加入两个表,WHERE是从URL (www.websiste.com?reference=38)获取的变量
两个表有相同的commonkey (ProgramCode)
表1:程序:id_program | ProgramName | ReleaseDate | ProgramCode
表2:id_program_genre | ProgramCode | id_genre
我得到这个,但它不工作
$sql_select = "SELECT * FROM program
INNER JOIN ProgramGenre
ON program.ProgramCode = ProgramGenre.ProgramCode
WHERE ProgramCode='$_GET[reference]'";
我做错了什么?
尝试使用
$sql_select="SELECT * from programs p,ProgramGenre g where p.ProgramCode=g.ProgramCode and p.ProgramCode='".$_REQUEST['reference']."'";
或者
$sql_select = "SELECT * FROM programs
INNER JOIN ProgramGenre
ON programs.ProgramCode = ProgramGenre.ProgramCode
WHERE ProgramCode='".$_GET['reference']."'";
SQL不知道ProgramCode
指的是哪一列
假设有一个参数化的查询,
$sql_select = "SELECT * FROM program
INNER JOIN ProgramGenre
ON program.ProgramCode = ProgramGenre.ProgramCode
WHERE program.ProgramCode=?";