我需要帮助才能从表中获取记录并查看单选按钮并插入其中一条记录取决于用户选择的另一个表,我在单选按钮中查看记录但是当我尝试选择一个单选按钮时插入另一张表没有任何反应。
<form method="post" action="">
<?php
$link=mysql_connect("localhost","root","")or die("Can't Connect...");
mysql_query("SET NAMES 'utf8'");
mysql_query("SET CHARACTER 'utf8'");
mysql_select_db("project",$link) or die("Can't Connect to Database...");
$query=mysql_query("SELECT * FROM mrsh7on WHERE `major`='$major'");
if (mysql_num_rows($query) > 0)
{
while($row = mysql_fetch_assoc($query))
{
echo"<div>";
echo "<input name=\"radio\" type=\"radio\" required value=".$row['vip_id'].">
".$row['name']."<br />";
echo"</div>";
}
$name = $row['name'];
$vip_id = $row['vip_id'];
}
if(isset($_POST['submit']) and !empty($_POST['submit']))
{
if(!empty($_POST['radio']))
{
$id = $_SESSION['sess_id'];
$major=$_SESSION['sess_major'];
$vip_id=$_SESSION['sess_vip_id'];
$name=$_SESSION['sess_name'];
$link=mysql_connect("localhost","root","")or die("Can't Connect...");
mysql_query("SET NAMES 'utf8'");
mysql_query("SET CHARACTER 'utf8'");
mysql_select_db("project",$link) or die("Can't Connect to Database...");
$query=mysql_query("SELECT * FROM vote WHERE `id`='$id'");
if (mysql_num_rows($query) > 0)
{
echo"this user voted before";
}
else
{
$sql = mysql_query("INSERT INTO vote (id, major, vip_id, name)
VALUES ('$id', '$major', '$vip_id', '$name')");
mysql_query($sql,$link) or die("mysql_error()");
echo"Successfully Inserted !";
}
}
}?>
<input type="button" name="submit" value="submit">
</form>
答案 0 :(得分:1)
1)如果这是您的所有代码。从一百万美元的问题开始,你在哪里开始会议?在代码的开头缺少session_start();
。
2) "SELECT * FROM mrsh7on WHERE
专业='$major'"
您从哪里获得$major
?试试echo $major;
,看看你得到了什么。
3) mysql_query($sql,$link) or die("mysql_error()");
这不是正确的方法,而是:
mysql_query($sql) or die(mysql_error());
4)更改此内容:<input type="button" name="submit" value="submit">
到此:
<input type="submit" name="submit" value="submit">
注意:如果可以解决所有问题,请避免使用已弃用的 mysql函数并移至
mysqli
或PDO
。