php - 从 mysqli select 返回错误的数值

Question

我正在尝试将用户信息存储在 mysql 数据库中并将其返回到浏览器窗口，我得到的是数据父级的行号，而不是应存储在其中的数据。自从我使用 mysql 或 php 以来已经有一段时间了(请原谅这个双关语)。

<?php
//connect to database
include('db-connect.php');
echo "<br/> echo user email: ". $user_profile['email'];

$fb_id = $user_profile['id'];
$fb_id = mysqli_real_escape_string($dbc,$fb_id);
//mysqli_real

//check if user exists
$check_user = mysqli_query($dbc , "SELECT * FROM users WHERE fb_id = '$fb_id';");

$check_user_result = mysqli_fetch_row($check_user);

if ($check_user_result[0] == 0) {
    echo ' <br/> no such user exists';
    $add_user = mysqli_query($dbc , "INSERT INTO users (user_id, email, first_name, last_name, fb_id) VALUES('','".$user_profile['email']."','".$user_profile['first_name']."','".$user_profile['last_name']."','".$user_profile['id']."')");

}
else{
echo " <br/> already registered  <br/>  should be an facebook ID before this". $fb_id;
$get_user_db_info = mysqli_query($dbc , "SELECT * FROM users");

while($row = mysqli_fetch_array($get_user_db_info)){
   // then your for loop
   echo '<br/>user_id: '.$row['user_id'].'<br/>email: '.$row['user_id'].'<br/>First Name: '.$row['user_id'];
}
}

?>

任何帮助都会很棒，并且可以根据需要提供更多信息。

Answer 1

您一遍又一遍地输出相同的值:

echo '<br/>user_id: '.$row['user_id'].'<br/>email: '.$row['user_id'].'<br/>First Name: '.$row['user_id'];
                            ^^^^^^^                        ^^^^^^^                             ^^^^^^^