Question

我正在制作这样的播客共享网站，用户可以上传音频播客，如果他们登录管理页面，则可以编辑每个标题和每个网址，并删除任何播客。这是我的代码，之后我会解释我的错误：

    <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'博士后点击删除Seuss'它删除播客（具有相同的id）但在昆虫表中，而不是Dr. Seuss表。任何人都可以帮助如何帮助PHP区分这两个表吗？

Answer 1

查看你的js函数告诉我你没有设置$ _POST ['2']，所以你在DrSeuss上的if条件不会执行。

从两个单独的MySQL表中单独删除信息？

1 个答案: