我终于学会了怎么做,但问题是,当我点击lol.php?id=2时,它仍然会显示此处的所有内容: mysql_select_db("blog");
$result = mysql_query("SELECT post_id, title, body FROM posts ORDER BY post_id DESC&qu
我刚刚学会了如何从表单的POST中获取值: echo "POST parameter '$key' has '$value'";但我的问题是如何插入这些数据:$key $value to the mysql database.Can我只是这样做:
mysqli_query($con,"INSERT INTO tableName