我想從我的數據庫中拉出一些信息,並把它放在一個模式。我去了基金會的網站,並試圖從他們的文檔部分弄清楚。我不完全明白它。因此,我有一部分網站允許用戶請求刪除他們上傳的歌曲。現在,如果他們點擊X模式應該彈出並要求確認。PHP在基金會模式
<div class="row">
<div class="large-8 column musicup">
<p> <?php echo "No music uploaded..."; ?> </p>
</div>
</div>
<?php
}else{
?>
<h2 style="margin-top:1em;">Music uploaded</h2>
<hr style="opacity:.4;">
<?php
while($row_a = mysql_fetch_array($res))
{
?>
<div class="row">
<div class="large-4 column musicup">
<p><?php echo $row_a['title']; ?></p>
</div>
<div class="large-3 column musicup"><span data-tooltip class="has-tip tip-top" title="<?php echo $row_a['reason']; ?>">
<div class="button <?php echo $row_a['status'];?>"><?php echo $row_a['status'];?></div>
</span></div>
<div class="large-3 column musicup_date">
<p><?php echo date('F j Y',strtotime($row_a['uploaded'])); ?> </p>
</div>
<div class="large-2 column musicup">
<p><a href="song_delete.php?id=<?php echo $row_a['song_id']; ?>" data-reveal-id="deleteMusic" data-reveal-ajax="true" style="font-weight:bold">X</a></p>
</div>
</div>
<?php
}
}
}
?>
</div>
因此,現在我有一個名爲song_delete.php的新頁面上的模態和所有數據庫查詢。
下面是該代碼:
<?php
include_once "functions.php";
$query = sprintf("SELECT * FROM songs WHERE user_id = %d AND song_id = %d",$_SESSION['user_id'], $_GET['id']);
$res = mysql_query($query) or die('Error: '.mysql_error());
$row_a = mysql_fetch_assoc($res);
$totalRows_a = mysql_num_rows($res);
?>
<div id="deleteMusic" class="reveal-modal medium">
<h2>Request to delete<span style="color:#F7D745;"> <?php echo $row_a['title']; ?></h2>
<p class="lead">Are you sure you want to delete this song? Please allow 2 full business weeks for deletion.</p>
<span style="float:right;"><a href="#" class="button close-reveal-modal cancelbtn">Cancel</a>
<a href="#" class="button submitbtn">Submit</a> </span>
<a class="close-reveal-modal">×</a>
</div>
感謝提前任何幫助。我很感激。
請不要告訴我的mysql_query中我應該怎麼使用PDO或庫MySQLi和OOP我知道這一點,但是這個網站目前並未與所有編碼..