显示两个表中的数据 - PHP mySQL

问：

我在数据库 1 中有两个表。post 和 2.操作系统

post:
+----+------------+-------+-------+-----+--------+------+-----+-------+
| id |    title   | metaD | metaK | img | author | date | cat |  text |
+----+------------+-------+-------+-----+--------+------+-----+-------+
| 1  | some title |  data | data  |data | data   | data | app |  data |
| 2  | title2     |  data | data  |data | data   | data | os  |  data |
| 3  | title3     |  data | data  |data | data   | data | oth |  data |
+----+------------+-------+-------+-----+--------+------+-----+-------+

os:
+------+----------+----------+--------+--------+--------+--------+-------+-------+------+------+--------+
| osId |  ostitle |   osimg  | oscpuM | oscpuR | osramM | osramR | oshdM | oshdR | osgM | osgR | osdown |
+----+------------+----------+-----------------+--------+--------+-------+-------+------+------+--------+
| 1  | title8     |  name    | here comes text | data   | data   | data  | data  | data | data | data   |
| 2  | title2     |  name    | here comes text | data   | data   | data  | data  | data | data | data   |
| 3  | title4     |  name    | here comes text | data   | data   | data  | data  | data | data | data   |
+----+------------+----------+-----------------+--------+--------+-------+-------+------+------+--------+

和代码：

//get data from "post" 
$post = "SELECT * FROM post WHERE id='{$_GET['id']}'";
    $post1 = mysql_query($post);
    $postV = mysql_fetch_array($post1);

    $postID = $postV['id'];
    $postTitle = $postV['title'];
    $postD = $postV['metaD'];
    $postK = $postV['metaK'];
    $postImg = $postV['img'];
    $postAuthor = $postV['author'];
    $postDate = $postV['date'];
    $postCat = $postV['cat'];
    $postText = $postV['text'];

//get data from "os"
    $os = "SELECT * FROM os WHERE ostitle='{$_GET['ostitle']}'";
    $os1 = mysql_query($os);
    $osV = mysql_fetch_array($os1);

    $osID = $osV['osId'];
    $osTitle = $osV['ostitle'];
    $osImg = $osV['osimg'];
    $osCpuM = $osV['oscpuM'];
    $osCpuR = $osV['oscpuR'];
    $osRamM = $osV['osramM'];
    $osRamR = $osV['osramR'];
    $osHdM = $osV['oshdM'];
    $osHdR = $osV['oshdR'];
    $osGM = $osV['osgM'];
    $osGR = $osV['osgR'];
    $osDown = $osV['osdown'];

    //dispaly post
    <?php
    echo "<img class='view_newsimg' src='$postImg'>
    <h3 class='lath'>$postTitle</h3>
    <ul class='det'>
    <li class='adc'>avtori: $postAuthor</li>
    <li class='adc'>TariRi: $postDate</li>
    <li class='adc'>kategoria: $postCat</li>
    </ul>
    <p class='news'>
    $postText";

//display from "os"
    $osif = $postV['cat'];

    if ($osif == 'os')
    {
    echo "<div class='os1'>
    <div>1</div>
    <div class='os1_1'>procesori</div>
    <div class='os1_1'>operatiuli mexsiereba</div>
    <div class='os1_1'>adgili myar diskze</div>
    <div class='os1_1'>grafikuli baraTi</div>
    </div>
    <div class='os1'>
    <div>minimaluri</div>
    <div>$osCpuM</div>
    <div>$osRamM</div>
    <div>$osHdM</div>
    <div>$osGM</div>
    </div>
    <div class='os1'>
    <div>rekomendebuli</div>
    <div>$osCpuR</div>
    <div>$osRamR</div>
    <div>$osHdR</div>
    <div>$osGR</div>
    </div>";
    }
    ?>

此页面的 URL 是：“.com/view.php？id=2ostitle=title2” 但我有这个错误：“注意：未定义的索引：第 54 行的 D：\XAMPP\htdocs\LinuxOid.com\blocks\db3.php 中的 ostitle”（这是：$os = “SELECT * FROM os WHERE ostitle='{$_GET['ostitle']}'”;

我该如何解决这个问题？