为什么不能显示数据，请高手指点！

Question

<?php
$db = mysql_connect("localhost","root","deskbook";
mysql_select_db("my1",$db);
$sql = "INSERT INTO my set (name) VALUES ('$name')";
$a = mysql_query($sql);
echo "okn";
?>;
连接数据库
连接表
输入数据
发送数据

究竟那里不对啊？请指点一下，谢谢了。

Answer 1

<?php
$db = mysql_connect("localhost","root","deskbook";
mysql_select_db("my",$db);
$sql = mysql_query("insert into my set name = $name",$db);
echo "ok!n";
?>;

<form action="a2.php" method="post">;
姓名<input type="Text" name="name">;<br>;
<input type="Submit" name="Submit" value="确定">;
</form>;
--------------------------------------------------------------------------------
这样就可以了!
$sql = mysql_query("insert into my set name = '$name'",$db);
这句的name = '$name'改为name = $name
--------------------------------------------------------------------------------

Answer 2

<?php
$db = mysql_connect("localhost","root","deskbook";
mysql_select_db("my",$db);
$sql = mysql_query("insert into my set name = '$name'",$db);
echo "ok!n";
?>;

<form action="a2.php" method="post">;
姓名<input type="Text" name="name">;<br>;
<input type="Submit" name="Submit" value="确定">;
</form>;

怎么在同一页面提交数据？有人能修改一下么？谢谢!

Answer 3

使用name＝"aaa"可以显示，
name="$name"就不显示数据了。
可是就算是同一页面一样不显示，既然可以增加记录，应该是变量$name的问题，不过问题出在那里，input的$name变量无法接收数据，这是为什么？