PHP查詢檢索朋友信息和列表縮略圖鏈接到朋友信息

Question

我正在一個社交網站，我想要一個用戶的朋友顯示在PHP生成的表。我想在這些縮略圖下方顯示朋友縮略圖和其他信息，這樣如果您單擊縮略圖，它會將您帶到該用戶的個人資料。

我的用戶表包含字段，例如

user_id, username, user_first_name, user_last_name...etc 

我有一個好朋友表：（user1_id，user2_id是PK USER_ID的FK在用戶表）

pal_id user1_id  user2_id  status  timestamp 
6    92   95   1  2011-02-02 21:44:58 
7    98   97   0  2011-02-02 21:44:28 
8    92   98   0  2011-02-02 21:44:28 

圖片其中化身供顯示檢索表。如果化身=「1」，即用戶化身：

picture_id picture_url    picture_thumb_url     user_id avatar timestamp 
73 ../User_Images/66983.jpg ../User_Images/Thumbs/66983.jpg   92   1  2011-02-01 21:41:59 
74 ../User_Images/56012.jpg ../User_Images/Thumbs/56012.jpg   94   0  2011-01-25 12:09:58 

從上面，用戶92與用戶95，因爲他們已經證實友誼帕爾斯（示出爲狀態=「1」）

user1_id在好友獲得友誼的發起者的用戶ID。對於pal_id = 6，用戶92請求友誼，95確認它。如果用戶95請求友誼，並已確認，user2_id將讀取92 ...

現在，我將如何顯示我的PHP查詢以顯示朋友縮略圖和信息？

我的PHP代碼：

<?php require_once('Connections/connections.php'); ?> 
<?php 
//query username 
$user_id = $_SESSION['UserSession']; 
$user_id = mysql_real_escape_string($user_id); 
$query_user_info = "SELECT username FROM users WHERE user_id='$user_id'"; 
$user_info = mysql_query($query_user_info, $connections) or die(mysql_error()); 
$row_user_info = mysql_fetch_assoc($user_info); 

//code for displaying all your pals 
$query_pal_no = "SELECT * FROM pals WHERE (user1_id='$user_id' AND status='1') OR (user2_id='$user_id' AND status='1')"; 
$pal_no_result = mysql_query($query_pal_no, $connections) or die(mysql_error()); 
$row_pal_no = mysql_num_rows($pal_no_result); 

while ($pal_no = mysql_fetch_assoc($pal_no_result)) 
{ 
    //get pal info 
    $query_pal_info = "SELECT users.user_id, user_first_name, user_last_name, username, picture_thumb_url, avatar FROM users LEFT JOIN picture ON 
    users.user_id = picture.user_id 
    AND picture.avatar=1 WHERE users.user_id = {$pal_no['user2_id']}"; 
    $pal_info = mysql_query($query_pal_info , $connections) or die(mysql_error()); 
    $totalRows_pal_info = mysql_num_rows($pal_info); 

    //echo table with pal information 
    echo "\n<table>"; 
    $j = 5; 
    while ($row_pal_info = mysql_fetch_assoc($pal_info)) 
    { 
     if($j==5) echo "\n\t<tr>"; 
     $thumbnail_user = $row_pal_info['picture_thumb_url'] != '' ? $row_pal_info['picture_thumb_url'] : '../Style/Images/default_avatar.png'; 
     echo "<td width='100' height='100' align='center' valign='middle'><a href = 'user_view.php?user_id2={$row_pal_info['user_id']}'> 
     <img src='/NNL/User_Images/$thumbnail_user' border='0'/></a></td>\n"; 
     $j--; 
     if($j==0) { 
     echo "\n\t</tr>\n\t<tr>"; 
     $j = 5; 
     } 
    } 
    if($j!=5) echo "\n\t\t<td colspan=\"$j\"></td>\n\t</tr>"; 
    echo "\n</table>"; 
} 
?> 
<table width="500" border="0"> 
    <tr> 
    <td height="20"><div class="heading_text_18"><?php echo $row_user_info ['username']; ?>'s&nbsp;pals <?php echo $row_pal_no ?></div> </td> 
    </tr> 
    <tr> 
     <td class="interactionLinksDiv" align="right" style="border:none;"><a href="#" onclick="return false" 
     onmousedown="javascript: toggleInteractContainers('pal_requests');">Pal Requests</a></td> 
    </tr> 
    <tr> 
     <td height="5"></td> 
    </tr> 
</table> 

我知道問題出在查詢中的WHERE後：

$query_pal_info = "SELECT users.user_id, user_first_name, user_last_name, username, picture_thumb_url, avatar FROM users LEFT JOIN picture ON 
    users.user_id = picture.user_id 
    AND picture.avatar=1 WHERE users.user_id = {$pal_no['user2_id']}"; 

正如我現在有它，它顯示了用戶92的唯一的朋友是誰的用戶95這僅僅是因爲92是發起者和95接受者。如果95是發起者，查詢將不起作用，因爲查詢將返回用戶92（他自己）作爲他的朋友。

什麼是正確的查詢？感謝您的幫助提前

Answer 1

SELECT * FROM users u 
INNER JOIN pal p 
ON p.user1_id = u.user_id 
INNER JOIN pictures pi 
ON pi.user_id = p.user2_id 
WHERE u.user_id = 92 

這應該給你所有的用戶ID 92.