How to remove duplicate of dropdown when selecting dropdown value

Question

I am quite new to the whole javascript and PHP. I have two issues with my code that i cannot seem to fix.

When i choose a category, it should show a table with the corresponding category. But it is including the dropdown as a duplicate, also there is an error that i can't see why is returned as an "Undefined variable: output in /var/www/html/phpDD.php on line 37" which is: "$output .= "tbody".

i hope you can help me to continue develop! :)

phpDD.php

<?php

    // include database connection file
    include_once "testSide.php";

    if(isset($_POST['InvestmentName']))
{
    $uid = $_POST['InvestmentName'];
}
    $qu = "select Prod.name, one.yield as yield_one, five.yield as yield_two, ten.yield as yield_three, twenty.yield as yield_four, one.stdev as st_one, five.stdev as st_two, ten.stdev as st_three, twenty.stdev as st_four, one.top1dd as top_one, five.top1dd as top_two, ten.top1dd as top_three, twenty.top1dd as top_four Select *
        from products
        where category_id = $uid) as Prod
    left join;";
        
        
            $result = $conn->query($qu);
            if (!$result) {
                trigger_error('Invalid query:' . $conn->error);
            }
        
            if ($result->rowCount() > 0) {
                $output .= "<table class='table table-hover table-border'>
                                <thead>
                                  <tr>
                                    <th>Name</th>
                                    <th>Afkast sidste år</th>
                                    <th>Genst sidste 5 år</th>
                                    <th>Genst stidste 10 år</th>
                                    <th>Genst sidste 20 år</th>
                                    <th>SR sidste år</th>
                                    <th>SR 5 år</th>
                                    <th>SR 10 år</th>
                                    <th>SR 20 år</th>
                                    <th>DD sidste år</th>
                                    <th>DD 5 år</th>
                                    <th>DD 10 år</th>
                                    <th>DD 20 år</th>
                                  </tr>
                                </thead>";
            while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
                $output .= "<tbody>
                <tr>
                  <td>{$row["name"]}</td>
                  <td>{$row["yield_one"]}</td>
                  <td>{$row["yield_two"]}</td>
                  <td>{$row["yield_three"]}</td>
                  <td>{$row["yield_four"]}</td>
                  <td>{$row["st_one"]}</td>
                  <td>{$row["st_two"]}</td>
                  <td>{$row["st_three"]}</td>
                  <td>{$row["st_four"]}</td>
                  <td>{$row["top_one"]}</td>
                  <td>{$row["top_two"]}</td>
                  <td>{$row["top_three"]}</td>
                  <td>{$row["top_four"]}</td>
                </tr>";
              
                }                       
                "</tbody>";               
                $output .= "</tbody></table>";
                echo $output;
            }else{
                echo "No records found";
            }
        ?>

Index.php

<!Doctype html>
<html lang="en">
<head>
  href="https://maxcdn.bootstrapcdn.com/bootstrap/4.5.0/css/bootstrap.min.css">
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
  <script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.16.0/umd/popper.min.js"></script>
  <script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.5.0/js/bootstrap.min.js"></script>
</head>
  <body>
    <div class="container" style="margin-top: 50px;">
      <h2 class="text-center">Væg en kategori </h2>
      <div class="row">
        <div class="col-md-4"></div>  
          <div class="col-md-4" style="margin-top:20px; margin-bottom:20px;">
            <form id="submitForm">
              <div class="form-group">
                <select class="form-control Investment" name="Investment" id="Investment">
                  <option value="">Vælg Kategori</option>
                      <?php foreach($users as $user): ?>
                        <option value="<?= $user['id']; ?>"><?= $user['name']; ?></option>
                      <?php endforeach; ?>
                </select>
              </div>
            </form>
          </div>
        </div>
        <div class="col-md-12">
          <div id="show-invest">
          </div>
        </div>    
      </div>
  </body>
</html>

<!---jQuery ajax load rcords using select box --->
<script type="text/javascript">
  $(document).ready(function(){
      $(".Investment").on("change", function(){
        var InvestmentName = $(this).val();
        if (InvestmentName !== "") {
          $.ajax({
            url : "phpDD.php",
            type:"POST",
            cache:false,
            data:{InvestmentName:InvestmentName},
            success:function(data){
              $("#show-invest").html(data);
            }
          });
        }else{
          $("#show-invest").html(" ");
        }
      })
  });
</script>

Website

DB

Can you show your database? Is that the full phpDD.php? (It doesn't have the " — Seth B, Dec 19 '20 at 19:34
**Warning:** You are wide open to [SQL Injections](https://php.net/manual/en/security.database.sql-injection.php) and should use parameterized **prepared statements** instead of manually building your queries. They are provided by [PDO](https://php.net/manual/pdo.prepared-statements.php) or by [MySQLi](https://php.net/manual/mysqli.quickstart.prepared-statements.php). Never trust any kind of input! Even when your queries are executed only by trusted users, [you are still in risk of corrupting your data](http://bobby-tables.com/). [Escaping is not enough!](https://stackoverflow.com/q/5741187) — Dharman, Dec 19 '20 at 23:17

score 1 · Answer 1 · answered Dec 19 '20 at 19:45

"Undefined variable: output in /var/www/html/phpDD.php on line 37"

If you look for the first instance of $output, it is:

$output .= "<table class='table table-hover table-border'>

It is performing two operations in one line: 1) concatenation, then 2) assignment.

It is understood that, to use a variable in an operation (in this case concatenation), it must first be defined. Since that is not the case, we get the error "Undefined variable: output ...".

Since the $output variable is only used in the if block, we can replace the concatenate-then-assign line with just an assignment:

if ($result->rowCount() > 0) {
    $output = "<table class='table table-hover table-border'>
    ...
}

However, if the $output is echo()'d outside the if block, then we must declare it outside/before the if block:

$output = '';

if ($result->rowCount() > 0) {
    $output .= "<table class='table table-hover table-border'>
    ...
}

echo $output;

Seth B · Accepted Answer · 2020-12-19T20:22:06.540

The problem is that you have not defined $output in your script. You can't have the first instance of $output being $output .= because the .= implies that the variable already exists.

To fix this, add $output = ''; somewhere before the $output .=

For example:

<?php

    // include database connection file
    include_once "testSide.php";

    if(isset($_POST['InvestmentName']))
{
    $uid = $_POST['InvestmentName'];
}

    $output = ''; //This is required

    $qu = "select Prod.name, one.yield as yield_one, five.yield as yield_two, ten.yield as yield_three, twenty.yield as yield_four, one.stdev as st_one, five.stdev as st_two, ten.stdev as st_three, twenty.stdev as st_four, one.top1dd as top_one, five.top1dd as top_two, ten.top1dd as top_three, twenty.top1dd as top_four Select *
        from products
        where category_id = $uid) as Prod
    left join;";
        
        
            $result = $conn->query($qu);
            if (!$result) {
                trigger_error('Invalid query:' . $conn->error);
            }
        
            if ($result->rowCount() > 0) {
                $output .= "<table class='table table-hover table-border'>
                                <thead>
                                  <tr>
                                    <th>Name</th>
                                    <th>Afkast sidste år</th>
                                    <th>Genst sidste 5 år</th>
                                    <th>Genst stidste 10 år</th>
                                    <th>Genst sidste 20 år</th>
                                    <th>SR sidste år</th>
                                    <th>SR 5 år</th>
                                    <th>SR 10 år</th>
                                    <th>SR 20 år</th>
                                    <th>DD sidste år</th>
                                    <th>DD 5 år</th>
                                    <th>DD 10 år</th>
                                    <th>DD 20 år</th>
                                  </tr>
                                </thead>";
            while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
                $output .= "<tbody>
                <tr>
                  <td>{$row["name"]}</td>
                  <td>{$row["yield_one"]}</td>
                  <td>{$row["yield_two"]}</td>
                  <td>{$row["yield_three"]}</td>
                  <td>{$row["yield_four"]}</td>
                  <td>{$row["st_one"]}</td>
                  <td>{$row["st_two"]}</td>
                  <td>{$row["st_three"]}</td>
                  <td>{$row["st_four"]}</td>
                  <td>{$row["top_one"]}</td>
                  <td>{$row["top_two"]}</td>
                  <td>{$row["top_three"]}</td>
                  <td>{$row["top_four"]}</td>
                </tr>";
              
                }                       
                "</tbody>";               
                $output .= "</tbody></table>";
                echo $output;
            }else{
                echo "No records found";
            }
        ?>
Index

As for the duplicate, you have the main file testSide.php being included when you use AJAX to load content.

Thanks man! It worked now! can you see why it prins two of the dropdowns as well? — AFT2233, Dec 19 '20 at 19:56
It should be in the picture above the database :) tou can see "vælg en katego" and the dropdown are there two times — AFT2233, Dec 19 '20 at 20:00
Can you give me the view source "view-source:URL" in chrome. It is showing only one for me. — Seth B, Dec 19 '20 at 20:04
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/226179/discussion-between-aft2233-and-seth-b). — AFT2233, Dec 19 '20 at 20:10

How to remove duplicate of dropdown when selecting dropdown value

2 Answers2