php - 在页面上获取多个mysql结果

您好，所以我在这里遇到了一些麻烦，我想要做的是使用下面的代码在同一页面上显示所有用户朋友的请求，但是正在发生的情况是该页面一次仅显示一个请求，然后必须刷新或重新加载页面，以使剩余的请求一个接一个地显示，而不是关闭页面下方列出的所有请求，因为我正在尝试学习谢谢，所以任何帮助都将非常有用

<?php
include ('views/header.php');
require_once ('config/config.php');
include ('config/connection.php');


{

    global $user_name,$page_owner,$username;

    $user_name = trim(strip_tags($_SESSION["user_name"])); //This is the user who logged into the system or logged in session

    $page_owner = trim(strip_tags($_SESSION["user_name"])); // This is the owner of the page viewed
            $username = mysql_query("select * from request where friend ='".$user_name."'");
    $user_id = mysql_query("select user_id from users where user_id = 'user_id'");


    //This is the page that checks for Friend Request


 $check_request = mysql_query("select * from request where friend = '".$user_name."'"); //First Request receive, first to respond to


                    if(intval(mysql_num_rows($check_request))==0); //If there is a friend request for the logged in user then show it to the user otherwise do nothing




                            $get_request_details = mysql_fetch_array($check_request);


                            //Check friend who sent the request full info from the users table

                            $check_request_info = mysql_query("select * from `users` where `user_name` = '".mysql_real_escape_string($get_request_details["username"])."'");

                            //Get friend who sent the request full info from the users table

                            $get_request_info = mysql_fetch_array($check_request_info);


                            //Check logged in user full info from the users table

                            $check_logged_in_user_info = mysql_query("select * from `users` where `user_name` = '".$_SESSION['user_name']."'");

                            //Get logged in user full info from the users table

                            $get_logged_in_user_info = mysql_fetch_array($check_logged_in_user_info);

?>


new requests(<?php echo intval(mysql_num_rows($check_request)); ?>)







                           <div>Hello <?php echo strip_tags($get_logged_in_user_info["user_name"]);?><div>

                            <div style="font-family:Verdana, Geneva, sans-serif; font-size:11px; line-height:18px;" align="left">Here are your friend requests.</div>

                            <a href="userpro.php?id=<?php echo $get_request_info["user_id"]; ?>"><font style="color:blue;font-family:Verdana, Geneva, sans-serif; font-size:14px;"><?php echo strip_tags($get_request_info["user_name"]); ?> wants to be friends</font></a><div>


                            <div>



<div>

<a href="af.php?username=<?php echo  $get_request_info["user_name"];
?>"class="square">Accept</a>


                            <a href="df.php?username=<?php echo $get_request_info["user_name"]; ?>"class="square">Decline</a>


 <?php




                    }













    {

            //Unknown page realized

    }



?>

最佳答案

这是因为您只打印了一张。您永远不会遍历包含所有好友请求的数组-您回显数组中的第一个请求。如果您拒绝或接受其中一个，则会显示下一个，依此类推。

见Populate PHP Array from While Loop