找不到朋友的ID將它存儲到數據庫中

我真的很困擾這一段時間，似乎無法讓它工作。在members.php中（我顯示所有註冊用戶），我有一個列表打印出一個鏈接「添加到朋友」旁邊的每個用戶。找不到朋友的ID將它存儲到數據庫中

我管理，出於測試的目的，以顯示每個成員身份證好（所以它得到的ID），但當我點擊鏈接，它指向的好友似乎是錯誤的地方。我不知道如何獲得朋友的ID我點擊了在friends.php文件中。請看看！

members.php

<?php 
     include 'connect.php'; 
     include 'header.php'; 

     if(isset($_SESSION['signed_in']) == false || isset($_SESSION['user_level']) != 1) 
      { 
       //the user is not an admin 
       echo '<br/>'; 
       echo 'Sorry! You have to be <a href="/signin.php"><b>logged in</b></a> to view all the <a href="signup_2.php" title="Become a registered user!"><b>registered</b></a> members.'; 
       echo '<br/><br/>'; 
      } 
      else 
      { 
       echo '<h2>Registered users:</h2>'; 

       $sql = "SELECT * FROM users ORDER BY user_name ASC"; 
       $result = mysql_query($sql); 

       $num=mysql_numrows($result); 

       $i=0; 
       while ($i < $num) 
       { 
        //$name = mysql_result($result,$i,"user_name"); 
        //$id = mysql_result($result,$i,"user_id"); 
        //$picture = mysql_result($result,$i,"pic_location"); 

        //?friend_id="'. $id .' 
        while($user = mysql_fetch_array($result)){ 
         echo $user['user_name'].'<br/><br/><a href="friends.php?user='.$user['user_id'].'">ADD TO FRIENDS</a><br/>'; 
         echo $user['user_id']; 
         echo '<br/><br/>'; 
        } 

        $i++; 
       } 

       /////////////////////////////// 
       /// adding users as friends /// 
       /////////////////////////////// 

       //while($user = mysql_fetch_array($result)) 
       //echo $user['user_name'].' 
        //<a href="addfriend.php?user='.$user['id'].'">ADD TO FRIENDS</a><br/>'; 


       //NOW I WANT TO MAKE A SPECIFIC "ADD AS FRIEND" LINK NEXT TO EACH USER 


      } 
      include 'footer.php'; 
?>

正如我說我不知道如何得到這個，所以請看看！謝謝！

friends.php

<?php 
    include "connect.php"; 
    include "header.php"; 

    if(isset($_SESSION['signed_in']) == false || isset($_SESSION['user_level']) != 1) 
    { 
     //the user is not an admin 
    echo '<br/>'; 
    echo 'Sorry! You have to be <a href="/signin.php"><b>logged in</b></a> if you want to add the person as a friend!'; 
    echo '<br/><br/>'; 
    } 
    else 
    { 
     $sql = "SELECT * FROM users"; 
     $result = mysql_query($sql); 

       //friend_id is the ID of the friend that is clicked on... 
       //HOW DO I GET THAT ID THAT IS CLICKED ON IN THE WHILE "loop" in members.php? 

       $friends = ("INSERT INTO friends SET user_id='" . $_SESSION['user_id'] . "', friend_id='".$id."', status='0'"); 
       $result_friends = mysql_query($friends); 


       if(!$friends) 
       { 
        //When you can't add this person as a friend this error will show! 
        echo 'You cannot add this user at this time. Please try again later!'; 
       } 
       else 
       { 
        //When the friend is now added to the system! 
        echo 'Great! Now the user needs to approve before you can be friends!'; 
       } 
    } 
?>

來源

2011-10-20 Joe_B

在您friends.php使用

$_GET['user_id']

相反的$ id的，$ id爲未定義，讓id的值從查詢字符串中使用$ _GET變量調用它，如

$ _GET ['name_of_query_string_value']

來源

2011-10-20 11:12:11 NVTOnline

我猜'$ _SESSION ['user_id']'保存訪問該頁面的用戶的ID，並且它被設置在別的地方。 '$ _GET ['user']'持有這個用戶想要添加的朋友的ID。 – AJJ

好酷！我理解得更好，但現在它說第20行中的C：\ xampp \ htdocs \ friends.php中的未定義索引：user_id。那麼，如何「聲明」該特定的user_id？ –

@AJweb：你真棒！但我似乎無法得到它的工作！ :( –

找不到朋友的ID將它存儲到數據庫中

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