根据第一个下拉框从数据库检索第二个下拉框的选项

Question

I am currently working on a small project. I have to retrieve some data from a DB (MySQL) and insert it into a webpage as a select (Drop Down box). The code I have written in PHP is:

<?php
// Connect to the db.
require ('mysqli_connect.php');

// Make the query:
$q = "SELECT employee_name from employee where dept_id=3 ORDER BY employee_id ASC"; 

// Run the query.
$r = mysqli_query ($dbc, $q);

if ($r) // If it ran OK, display the records. 
{ 
     echo '<select name="employee_name">';

     // Fetch and print all the records:
     while ($row = mysqli_fetch_array($r)) 
     {
          echo '<option value="'.$row['employee_name'] . '>"'.$row['employee_name'] .'</option>';

     }
         echo "</select>";    

}

mysqli_free_result ($r); // Free up the resources.
mysqli_close($dbc); // Close the database connection.
?>

When I execute the query in MySQL console, it returns the correct output. [It is a list of five names].

Can you help me find the error?

Answer 1

Use this:

if ($r = @mysqli_query ($dbc, $q))
{
    echo 'select ....

As $r is different to true on SELECT queries.

EDIT

You are closing the select tag on every iteration of the while . Try it like this:

if ($r = @mysqli_query ($dbc, $q))
{ 
     echo '<select name="employee_name">';
     // Fetch and print all the records:
     while ($row = mysqli_fetch_array($r)) 
     {
          echo '<option value="'.$row['employee_name'] . '>"'.$row['employee_name'] .'</option>';
     }
     echo "</select>";                         
//   ^
//   |__  Now the </select> is out of the loop
}

Answer 2

Echo your end select out of the loop.

 while ($row = mysqli_fetch_array($r)) 
 {
      echo '<option value="'.$row['employee_name'] . '>"'.$row['employee_name'] .'</option>';

 }
 echo "</select>";