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使用用户名在页面上显示用户信息

[英]Displaying user info on a page using their ID

Im trying to create a dynamic web page in php and mysql. 我试图在php和mysql中创建一个动态网页。 I have the below code on the profile.php page. 我在profile.php页面上有以下代码。 The issue im having is on the "while" line im not sure how i would go about getting the information from the DB. 我所遇到的问题是在“ while”行上,我不确定我将如何从数据库获取信息。 I want to display details like email, country First name, etc. Any help appreciated. 我想显示详细信息,例如电子邮件,国家/地区的名字等。任何帮助表示赞赏。 Thanks in advance! 提前致谢! CODE: 码:

$userid = (isset($_GET['id']) ? $_GET['id'] : NULL);

if ($userid) {
    $userinfo = $DB->query_first("SELECT * FROM `users` WHERE `id` = '$userid'");
    print_r($userinfo);
    }
        //{
            while($row = sql_fetch_assoc($DB)){

                echo'<div class="container">
            <div class="jumbotron" align="block">';
                echo $row['first_name'];
            echo $row['last_name'];
            echo $row['country'];
                echo $row['username'];
            echo'</div>';
            echo'</div>';
            }

Try by using this 尝试使用此

<?php
$userid = (isset($_GET['id']) ? $_GET['id'] : NULL);

if ($userid) {
   $userinfo = $DB->query_first("SELECT * FROM `users` WHERE `id` = '$userid'");
   print_r($userinfo);
?>
<table>
<tr>
    <td>First Name</td>
    <td><?=$userinfo['first_name']?></td>
</tr>
<tr>
    <td>Last Name</td>
    <td><?=$userinfo['last_name']?></td>
</tr>
............
............
</table>
<?php } ?>

do not use null use a empty string "" that is what you want sql well do nothing with an empty string it is just white space, enough about sql injection attacks wtf ever topic i enter. 不要使用null使用空字符串“”,这是您想要sql很好使用的空字符串,它只是空白,对于我所输入的主题进行的SQL注入攻击已经足够了。 actually make it equal to wither where userid='$userid' or "". 实际上使它等于在userid ='$ userid'或“”时枯萎。 there is only one row so the while statement is redundant just row = mysql_fetch_array. 只有一行,所以while语句是多余的,只是row = mysql_fetch_array。

You can try something like this: 您可以尝试如下操作:

<?php
    $userid = (isset($_GET['id']) ? $_GET['id'] : NULL);

    if ($userid) {
       $userinfo = $DB->query_first("SELECT * FROM `users` WHERE `id` = '$userid'");
       print_r($userinfo);
       echo'
           <div class="container">
               <div class="jumbotron" align="block">
                   '.$userinfo['first_name'].' <br>
                   '.$userinfo['last_name'].' <br>
                   '.$userinfo['country'].' <br>
                   '.$userinfo['username'].' <br>
               </div>
           </div>';
?>

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