尝试在php中执行SELECT，但我得到

所以当我给它一个特定的KittenID时，我有一个漂亮的代码集来运行并找到我的信息，但它根本不起作用，我很难过。哦，太累了，谁能告诉我哪里出了问题？是的，我有：

<?php

date_default_timezone_set('America/New_York');
//If statements:

//find:
date_default_timezone_set('America/New_York');
if(isset($_POST['Find'])) 
{
            $connection = mysql_connect("ocelot.aul.fiu.edu","userName","password");

            // Check connection
            if (!$connection)
        {
            echo "Connection failed: " . mysql_connect_error();
        }
            else
            {  
                //select a database
                $dbName="spr15_xgotz001";
                $db_selected = mysql_select_db($dbName, $connection); 

                //confirm connection to database
                if (!$db_selected)
                {
                    die ('Can\'t use $dbName : ' . mysql_error());
                }
                else
                {
                    $result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID =<?php$_POST[KittenID]?>;)

                    while($row = mysql_fetch_array($result))
                    {
                    $Name = $row['Name'];
                    $KittenID = $row['KittenID'];
                    $KittenAge = $row['KittenAge'];
                    $Email = $row['Email'];
                    $Comments = $row['Comments'];
                    $Gender = $row['Gender'];
                    $Personality = $row['Personality'];
                    $Activity = $row['Activity']; 
                    echo $row['Comments'];

                    }
                }
            }

            mysql_close($connection);
}


?>

使用

而不是

$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID =<?php$_POST[KittenID]?>;)

$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID =<?php$_POST[KittenID]?>;)

$result=mysql\u查询（$connection，“从Kittenz中选择*，其中KittenID=；）

注意：请在您未来的项目中使用mysqli，您需要提供更多的上下文。您如何设置$\u GET['id']…它实际上是否存储为$\u GET['KittenID']（例如）。如果是这样

您可以设置一个变量并声明“KittenID”

$kittenid = $_POST['KittenID'];

$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID = $kittenid");

我建议提供更多的上下文。您遇到了什么错误？您的参数是什么样子的？
使用

$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID = " .$_SERVER['KittenID']);

而不是

$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID =<?php$_POST[KittenID]?>;)

$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID =<?php$_POST[KittenID]?>;)

$result=mysql\u查询（$connection，“从Kittenz中选择*，其中KittenID=；）
“它不工作”不是一个问题描述。描述代码应该做什么，它通常做什么，并发布任何错误消息。我有点不知所措，这是由于在8小时内抱歉。