有人可以幫我解決我的配置文件中有用戶消息,它只顯示最後一條消息,我將如何顯示該用戶的所有消息。爲什麼只輸入最後一條消息
下面是代碼我已經:
<?php
$message_list = mysql_query("SELECT * FROM `messages`");
while($get_message = mysql_fetch_array($message_list)) {
$message_subject = $get_message['subject'];
$message_body = $get_message['body'];
$message_from = $get_message['from_user'];
$message_date = $get_message['date'];
$message_to = $get_message['to_user'];
}
if($message_to === $user_data['username']) {
echo
'
<center>
<table border="2" bordercolor="" style="background-color:" width="85%" cellpadding="3" cellspacing="3">
<tr>
<td width="15%"><center>From</center></td>
<td width="20%"><center>Date/Time</center></td>
<td><center>Message</center></td>
<td width="15%"><center>Reply</center></td>
<td width="7%"><center>Delete</center></td>
</tr>
</table>
</center>';
echo
'
<center>
<table border="1" bordercolor="" style="background-color:" width="85%" height="35px" cellpadding="3" cellspacing="3">
<tr>
<td width="15%"><center>'. $message_from .'</center></td>
<td width="20%"><center>'. $message_date .'</center></td>
<td><center><br>'. $message_body .'<br></center><br></td>
<td width="15%"><center>Reply</center></td>
<td width="7%"><center><form><input type="checkbox" name="delete"></form></center></td>
</tr>
</table>
</center>';
} else {
echo '<center>you have no messages</center>';
}
?>
其所有顯示的最後一條消息進入不是全部。
你需要把內環路您的HTML代碼中的其外部的循環,使最後的結果將顯示 –
感謝您@MKhalidJunaid什麼一個數字我是我錯過了}在結束 – HelpingHand