While循環上的特定ID只

Question

<?php 

class Friends { 
    public function __construct() { 
    global $DB, $name, $f_name; 
    $name = $_SESSION['name']; 
    if (isset($_POST['submitted'])) { 
     $f_name = mysql_real_escape_string($_POST['f_name']); 
    } 
    } 

    public function AddList() { 
    global $DB,$name; 
    echo "<center>"; 
    $result = $DB->query(
     "SELECT * FROM friends WHERE user_name!='{$name}' AND friend_name!='{$name}' " 
    ); 

    while($row = mysql_fetch_array($result)) { 
     echo "<form method='post'>"; 
     echo '<input type="hidden" value=' . $row['user_name'] . ' name="f_name"</th> '; 
     echo '<tr>'; 

     echo '<th><b><font color="#663300">' . $row['user_name'] . '</font></b></th>'; 

     $this->auth(); 
     echo "</form>"; 
     echo "</tr>"; 
     echo "</tr>"; 
    } 

    echo "</table>"; 
    echo "</center>"; 
    } 


    public function auth() { 
    global $name,$f_name,$DB; 
    $authprove = $DB->query("SELECT * FROM approval WHERE sender_id='{$name}'"); 

    if (mysql_num_rows($authprove) > 0) { 
     echo '<th><b>Approving...</th>'; 
    } else { 
     echo '<th><b><input type="submit" name="submitted" value="Add"></th>'; 
    } 
    } 
} 

$F= new Friends(); 

?> 

有點交代：

這有可能使只有特定的ID輸出認同與另一個輸出添加。

像這樣

當表（審批）：

---- SENDER_ID || receive_id

---- Admin ||客戶

輸出Friends.php：

----添加好友：

----嘉賓 - >等待審批

---- Guest2 - >添加爲好友

---- Guest3 - >添加爲好友

Answer 1

你很unefficiently執行你的任務：你會運行N + 1個查詢，如果你有N個用戶。您可以使用JOIN使用單個查詢來完成此操作。在這種情況下，你只需要一個循環就可以完成你所需要的任何事情。

喜歡的東西：

$query = " 
    SELECT 
     f.id, 
     f.user_name, 
     f.fullname, 
     a.status 
    FROM 
     friends f 
     LEFT OUTER JOIN approval a ON (f.id=a.sender_id) 
    WHERE 
     user_name!='{$name}' 
     AND friend_name!='{$name}' 
"; 


$result = $DB->query($query); 
while($row = mysql_fetch_array($result)) { 
    echo $row['f.fullname']; 
    if ($row['a.status'] == 'approved') echo "Add as Friend"; 
    else echo "Add as Friend"; 
}