如果数据库没有记录，则显示回显？

Question

所以我做了配置文件类型的事情，与状态更新，但是，当没有状态的数据库中，当然显示空白，我试图使它如此，如果数据库是empty，然后echo "You have no Feed"这里是我的代码：

<?php 
$status = mysql_query("SELECT * FROM status WHERE user='$username' ORDER BY date DESC"); 
while($row = mysql_fetch_array($status)) 

if($row['user'] == '') { echo "You have no Feed"; } else { 

{ 

echo "<table width='520'>"; 
echo "<tr>"; 
echo "<td width='53' height='57' valign='top' rowspan='3'><img src='../includes/images/profile/$image' height='50px' width='50px' style='border: 1px solid #000'></td>"; 
echo "<td width='411' valign='top'><font color='#CCCCCC'>". $row['sentby'] . "</font><br><font size='1px' color='#CCCCCC'>". $row['date'] . "</font></td>"; 
echo "</tr>"; 
echo "<tr>"; 
echo "<td></td>"; 
echo "</tr>"; 
echo "<tr>"; 
echo "<td>"; 
echo "</td>"; 
echo "</tr>"; 
echo "<tr>"; 
echo "<td colspan='2' valign='top'><font color='#CCCCCC'>". $row['status'] . "</font></td>"; 
echo "</tr>"; 
echo "</table>"; 
echo "</p>"; 

} 
} 

?> 

的方式我的个人资料的作品，是每个人都有自己的空间，如果你去别人的个人资料，并在他们的“墙”评论ofcourse，只会留在他们的墙上，我的数据库创建为'id' 'status' 'user' 'sentby''user' (being the username of who's wall it was posted on)和'sentby' (to display who sent it)

基本上，我需要'user'列匹配if无法找到用户名，如果找不到用户名，请回复邮件。

Answer 1

mysql_num_rows函数将返回查询获取的行数。这应该工作。

 $status = mysql_query("SELECT * FROM status WHERE user='$username' ORDER BY date DESC"); 


    if(mysql_num_rows($status) > 0){ 
     while($row = mysql_fetch_array($status)) 

     echo "<table width='520'>"; 
     echo "<tr>"; 
     echo "<td width='53' height='57' valign='top' rowspan='3'><img src='../includes/images/profile/$image' height='50px' width='50px' style='border: 1px solid #000'></td>"; 
     echo "<td width='411' valign='top'><font color='#CCCCCC'>". $row['sentby'] . "</font><br><font size='1px' color='#CCCCCC'>". $row['date'] . "</font></td>"; 
     echo "</tr>"; 
     echo "<tr>"; 
     echo "<td></td>"; 
     echo "</tr>"; 
     echo "<tr>"; 
     echo "<td>"; 
     echo "</td>"; 
     echo "</tr>"; 
     echo "<tr>"; 
     echo "<td colspan='2' valign='top'><font color='#CCCCCC'>". $row['status'] . "</font></td>"; 
     echo "</tr>"; 
     echo "</table>"; 
     echo "</p>"; 
     } 
     } else { 
     echo "<p>This user has not yet updated their status.</p>"; 
     }