php - 如何在未登录的情况下根据url id放置在浏览器地址上来获取特定数据表

Question

我是一个新人，刚刚学习，我对从表到 url id 获取数据表感到困惑，我想在写入 url 时从表中获取数据，例如 localhost/index.php?=user='username' 并且表将根据 url 用户名显示

这里是表格示例

内容表

code_content    |  judul_content  |  deskripsi  |  nama_lengkap
   2                  abc               tes1        john
   3                  efg               tes2        gerald
   4                  hij               tes3        john 

用户表

username   |  password  | nama_lengkap
 user1          123         john 
 user2          234        gerald

当我写这个 url localhost/index.php?=user=user2 时，我希望显示此输出

**

judul_content = "efg" 
nama_lengkap  = "gerald"

**

<?php 
    $koneksi = mysql_connect("localhost","root","1234");
    mysql_select_db("my_db");
    $hal = 0;
    if(isset($_GET['page'])){
        $page = $_GET['page'];
    }else{
        $page = 1;
    }

    $limit = 2;
    $offset = ($page - 1)*$limit;   
    $hasil = mysql_query("select * from content order by code_content desc limit $offset,$limit;");
    $get_total = mysql_query("select count(*) as total from content");
    $total_artikel = 0;
    while($h_total = mysql_fetch_array($get_total)){
        $total_artikel = $h_total['total'];
    }   
    $total_page = ceil($total_artikel/$limit);
?>

        <?php while($data = mysql_fetch_array($hasil)){ ?>

           <h2><a href="index.php?page=full&kode=<?php echo $data["code_content"]; ?>" style="text-decoration:none;">
           <?php echo $data["judul_content"]; ?></a></h2>            
            <?php echo $data["tanggal"]; ?> by <a href="#" target="_blank"><?php echo $data["nama_lengkap"]; ?></a>            
      <div id="posts-list" class="cf">
                <div class="right" style="float:left; text-align:justify; width:100%;">
                    <?php 
                        echo implode(array_slice(explode(" ",$data["content"]),0,75)," ").".....";
                    ?><br /><br />
                    <a href="index.php?page=full&kode=<?php echo $data["code_content"]; ?>">Continue reading...</a> <!--| <a href="#">Comments (60)</a>-->
                </div>                
                <div class="cleaner"></div>
            </div>            
                <?php } ?>  
        <div>
            <?php if($page > 1){ ?>

如何在没有登录的情况下根据url用户显示表格谢谢

Answer 1

由于您尝试从 2 个表中检索数据，因此您必须在两个表之间建立关系(将内容中的 user_id 作为外键传递)，然后您可以借助 URL 传递的值轻松检索数据。

用户表

id | username   |  password  | nama_lengkap
1    user1          123         john 
2    user2          234        gerald

内容表

user_id | kode_content    |  judul_content  |  deskripsi  |  nama_lengkap
  1         2                  abc               tes1        john
  2         3                  efg               tes2        gerald

localhost/index.php?=user='用户名'

if(isset($_GET['user'])) {
   $user_name = $_GET['user'];

 $check_user = "SELECT username FROM user WHERE username = '$user_name' LIMIT 1";
 if(!empty($check_user))
 {
   $user_data = "SELECT contents.judul_content, users.nama_lengkap
   FROM users
   INNER JOIN contents
   ON users.id=contents.user_id 
   WHERE users.username = '$check_user'";
   print_r ($user_data);
  }else{
   echo "No User Found."
  }

}