分别从两个单独的MySQL表中删除信息


Delete Info From Two Separate MySQL Tables Individually?

我正在制作这个播客分享网站,用户可以上传音频播客,如果他们登录到管理页面可以编辑每个标题和每个URL,并删除任何播客。下面是我的代码,我将在后面解释我的错误:

    <li><form action="admin.php" method="POST"><input type="submit" name="1" value = "Insects and Plants" /><input type="submit" name="2" value = "Dr. Seuss" /></forM>
    <li><p><?php
    function disp($titid,$titol,$aid){
            if($_GET['del']){
            $delete_id=$_GET['del'];
            mysql_query("DELETE FROM `$titid` WHERE `$titid`.`inid` = $delete_id");
            header("location: admin.php");
            }
    echo "<a name='$aid'><h3>" . $titol . "</h3></a>";
    $result=mysql_query('SELECT * FROM `'.$titid.'` ORDER BY inid ASC');
            while($row=mysql_fetch_array($result))
            {
            $title = $row['title'];
            $url=$row['url'];
            $id = $row['inid'];
            echo '<div class="inneredit">';
            echo $title . '</br>';
            echo $url . '</br>'.$id.'</br>';
                    echo "<form action='admin.php' method='POST'><input type='text' name='nameedit".$id."' /><input type='submit' name='nameit$id' value='Edit Name' /></form>";
            echo "<form action='admin.php' method='POST'><input type='text' name='urledit".$id."' /><input type='submit' name='redit$id' value='Edit URL' /></form>";
            echo "<input type='button' id='delete' value='Delete Podcast' onclick='return Deleteqry($id)' />";
            echo "</div>";
            if(isset($_POST['urledit'.$id]));
                    if(isset($_POST['redit'.$id]))
                    {
                        $newd = $_POST['urledit'.$id];
                        mysql_query("UPDATE `$titid` SET url = '$newd' WHERE $titid.inid = $id ");
                        header("location: admin.php");
                    }
                                    if(isset($_POST['nameit'.$id]))
                    {
                        $newd = $_POST['nameedit'.$id];
                        mysql_query("UPDATE `$titid` SET title = '$newd' WHERE $titid.inid = $id ");
                        header("location: admin.php");
                    }
            }
            }
            if(isset($_POST['2'])){
            disp("DrSeuss","Dr. Seuss","Seussa");
            } else {
            disp("insects","Insects and Plants","Insectsa");
    }
            ?>
    <script>
            function Deleteqry(id)
            { 
              if(confirm("Are you sure you want to delete this audio file?")==true)
                       window.location="admin.php?del="+id;
                return false;
            }
    </script>

            ?>

所以现在当我选择'Dr. Seuss'后点击删除时,它删除了播客(具有相同的id),但在昆虫表中,而不是Dr. Seuss表中。谁能告诉我如何帮助PHP区分这两个表?

看一下你的js函数告诉我,你没有$_POST['2']设置,所以你对DrSeuss的if条件不会执行