CODEWITHSUNDEEP
javascriptphpmysqlarrays
Brandon Adkinson
Brandon Adkinson

Reputation: 1

Im trying to populate a dropdowm menu with a database column in php

I am trying to make a dropdown menu from a database. I am returning nothing right now. For the menu, I need to choose a object then run a SQL query on it. This query will populate a table and will be dynamic. Here is the code, please help.

<html>
<?php
include('header.php');
?>

  <h1>Chemicals Search</h1>

  <br/ >
  <br/ >

  </head>


  <h1>
    <center>Chemical Search</center>
  </h1>
  <form action="chemicals.php" method="post">
    <label>Search By Product:</label>
    <?php
    //making the chemical array
    $con = mysqli_connect("...");

    if (mysqli_connect_errno()) {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $p = mysqli_query($con, "SELECT product_name FROM product");

    $products = mysqli_fetch_array($p);

    echo '<select name="product">';
    foreach ($product as $key => $value) {
      echo "<option value=\"$key\"> $value</option>\n";
    }
    echo '</select>';
    echo "<br>";
    ?>
  </form>

Upvotes: 0

Views: 41

Answers (1)

Meenesh Jain
Meenesh Jain

Reputation: 2528

Your Query only selects product name make a little changes to it 

 $p = mysqli_query($con,"SELECT product_id,product_name FROM product");
 echo'<select name="product">';
 if(mysqli_num_rows($p) > 0)
 {
  while($products =mysqli_fetch_array($p))
  {
    echo"<option value='".$products['product_id']."'>".$products['product_name']."</option>\n";       
  }
 }
echo '</select>';

echo"<br>";

I am sure this will works, works for me everytime

Upvotes: 1

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