试图显示两个表中的数据

我试图根据另一个表中找到的id从一个表中获取$_GET数据。我的第一张表叫做user_thoughts，这张表包含用户在我的“社交媒体网站”上发布的所有公共帖子的数据。我有另一个表users，它存储了该网站的注册用户的所有细节。试图显示两个表中的数据

我想在网站的主页上显示所有的user_thoughts，但我努力为“思想”的作者展示正确的数据。

的想法是让added_by（作者）的user_thought的然后使用变量保存的added_by的价值，并将其与表users发现username。

这里是我的表的，它的领域：

user_thoughts：

id 
message 
date_of_msg_post 
time_of_msg_post 
attachment 
added_by

users（只显示相关领域）

id 
first_name 
last_name 
username 
profile_pic

目的：获取ŧ他添加了user_thoughts的帖子，然后使用added_by从表users的用户名字段中获取added_by的详细信息。注意：added_by和username都会保持相同的值。

这是我曾尝试：

/* How it works: The id of each user_thought will be used to determine which user posted it. 
* Then display their details accourdingly. 
*/ 
$get_thoughts_from_db = mysqli_query($connect, "SELECT * FROM user_thoughts ORDER BY id DESC"); // newest posts first 

while ($row = mysqli_fetch_assoc($get_thoughts_from_db)) { 
    $thought_id  = $row['id']; 
    $msg_content  = $row['message']; 
    $date_of_msg  = $row['date_of_msg_post']; 
    $thoughts_by  = $row['added_by']; 
    $time_of_msg  = $row['time_of_msg_post']; 
    $attachent  = $row['attachment']; 
    $get_user  = $_GET['id']; 
} // while closed 

    // Get the details of the user based on the ID of the user thought. 
    $get_data = mysqli_query($connect, "SELECT * FROM users WHERE id = '$get_user'");         
    $get_user_data = mysqli_fetch_assoc($get_data); 
     $author_fname = $get_user_data['first_name']; 
     $user_profile_dp = $get_user_data['profile_pic']; 

// displaying all the posts in the database on the main page. 
// Will limit 15 posts per page, and will order them by the date and time posted (latest posts first) 
$get_all_posts_q = mysqli_query ($connect, "SELECT * FROM user_thoughts ORDER BY id DESC "); 
    $check_rows = mysqli_num_rows($get_all_posts_q); 
    while ($get_row = mysqli_fetch_array($get_all_posts_q)){ 
     $message   = $get_row['message']; 

     /**** Between the while loop is where I echo the div(s) 
        which display the above details. *******/ 

}

当前的行为： 目前，与上面的代码，它会显示所有user_thoughts，即有三个表中的行，并显示每一个，但是，它的细节不是基于这些职位的作者。例如，撰写作者的用户的profile_pic未显示。

来源

2016-03-01 Freddy

你很容易受到[SQL注入攻击（http://bobby-tables.com）。并且请注意，您正在循环中获取您的user_thoughts结果，但在循环完成之后才会使用这些结果，因此您只能使用您提取的LAST思想。 –

您是试图显示来自单个用户的想法列表，还是显示所有用户的想法列表？ – OscarJ

@OscarJ - user_thoughts表中的每一行都应显示在主页上。所以要回答你的问题，所有的用户。 – Freddy

这会链接用户和user_thoughts。记得清理你的输入。

$get_thoughts_from_db = mysqli_query($connect, "SELECT * FROM user_thoughts INNER JOIN users ON user_thoughts.added_by = users.username ON ORDER BY id DESC"); // newest posts first

来源

2016-03-01 15:05:40 Orion

$query = <<<EOD 
    SELECT t.id AS tid, u.id AS uid, * FROM user_thoughts AS t, users AS u 
    WHERE t.added_by=u.id ORDER BY t.id DESC 
EOD; 

$result = mysqli_query($connect, $query); 

while ($row = mysqli_fetch_assoc($result)) { 
    $thought_id  = $row['tid']; 
    $msg_content  = $row['message']; 
    $date_of_msg  = $row['date_of_msg_post']; 
    $thoughts_by  = $row['added_by']; 
    $time_of_msg  = $row['time_of_msg_post']; 
    $attachment  = $row['attachment']; 
    $author_fname = $row['first_name']; 
    $user_profile_dp = $row['profile_pic']; 

    // echo your div 
}

来源

2016-03-01 16:13:05 OscarJ

试图显示两个表中的数据

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