sql查询无法在我的网页中工作

I wrote a MySQL query in my PHP code to join two tables, but I don't want all the columns in the result

My place details table:

my place details table pic

My registration table:

my registration table oic

My Code

         <?php
$con=  mysqli_connect("localhost", "root", "", "project");

        if(!$con)
       {
           die('not connected');
       }
            $result=  mysqli_query($con, "SELECT placedetails.name,placedetails.doj,placedetails.total,placedetails.stay,placedetails.food,placedetails.travel,registration.username 
          FROM placedetails inner join registration on placedetails.id=registration.id ;" );



?>
<div class="container">
<CENTER><h2>view Booking Details</h2> 
<table class="table table-bordered">

  <th>place</th>
  <th>Travelling Date</th>
  <th>Total Cost</th>
  <th>Stay cost</th>
  <th>Food cost</th>
  <th>Travelling cost</th>
  <th>User ID</th>


        <?php

             while($row =mysqli_fetch_array($result))

             {
                 ?>
            <tr>
                <td><?php echo $row['name']; ?></td>
                <td><?php echo $row['doj']; ?></td>
                <td><?php echo $row['total'] ;?></td>
                <td><?php echo $row['stay'] ;?></td>
                <td><?php echo $row['food'] ;?></td>
                <td><?php echo $row['travel'] ;?></td>
                <td><?php echo $row['username'] ;?></td>

                <td><img src='".$row['choose']."'/></td>";
            </tr>
        <?php
             }
             ?>
             </table>
            </div>

i wrote my code,its executing but the problem is the data is not executing,can someone please help me out with my problem.

MysqlAdmin

In your table place details create a new entry call ID_regist. After that change or give a number that correspond to your id in tabel Regist.

code php

For the connection create a connection.php.

<?php
     $con=  mysqli_connect("localhost", "root", "", "project");

        if(!$con)
       {
           die('not connected');
       }
?>

Makes your database more difficult to attack

Main code

<?php
include("connection.php");
?>
<div class="container">

<CENTER><h2>view Booking Details</h2> 
<table class="table table-bordered">

<th>place</th>
<th>Travelling Date</th>
<th>Total Cost</th>
<th>Stay cost</th>
<th>Food cost</th>
<th>Travelling cost</th>
<th>User ID</th>

<?php 
    $sql= 
    "SELECT ". 
    "pld.name, ".
    "pld.doj, ". 
    "pld.total, ".
    "pld.stay, ".
    "pld.food, ".
    "pld.travel, ".
    "pld.username ".
    "FROM placedetails pld ".
    "inner join registration reg ".
    "on pld.ID_regist=reg.id;";
    $result=  mysqli_query($con, $sql);
    if($result){
    while($row =mysqli_fetch_array($result))
         {
?>
            <tr>
                <td><?php echo $row['name']; ?></td>
                <td><?php echo $row['doj']; ?></td>
                <td><?php echo $row['total'] ;?></td>
                <td><?php echo $row['stay'] ;?></td>
                <td><?php echo $row['food'] ;?></td>
                <td><?php echo $row['travel'] ;?></td>
                <td><?php echo $row['username'] ;?></td>
                <td><img src='".$row['choose']."'/></td>"; // first there is no row, call chose
            </tr>
<?php
       }
     }
     else{
       die("Error"); //Something of the type mysqli_error(connection);
     }
?>
       </table>
 </div>

Attention to the row call chose, that you don't use in your sql query. If that does not solve your problems, tell me what happens when I try to help you.

You have to alter the place table and add registration_id. Then join with this field.

SELECT placedetails.name,placedetails.doj,placedetails.total,placedetails.stay,placedetails.food,placedetails.travel,registration.username 
FROM placedetails inner join registration on registration.id=place.registration_id