execute a php file on selection of dropdown value

Question

i have been searching a solution for days and i have tried ajax/jquery methods posted online but it just wouldnt work. I have a drop down list which gets its value from the database. On selecting any value apart from "Select", i want to display a value which is called upon by a php file:

here's my code for the form:

         <tr>  
        <fieldset id="Date">
          <td class="select"><label><span class="text_9">Date:</span></label></td>
          <td><select name="date" id="date">
          <option value="">Select</option>
          <?php include_once "selectdate.php"?></td>
          </select>
     </tr>
        </fieldset>
   </table>

and here's the php to run on selection of the drop down (called retrieve.php)

<?php
include_once "connect.php";
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password); 

$form=$_GET;
$trade=$form['tradetype'];
$metal=$form['metal'];
$amount=$form['amount'];
$date=$form['date'];

$stmt = $conn->query("SELECT Discount FROM Contracts WHERE Trade='$trade' AND Metal='$metal' AND Amount='$amount' AND ExpiryDate='$date'");
while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) { 
   echo ($row['Discount']); 
} 
?>

As you can see, the php to be run uses the value from multiple form elements...

I am very new to jquery/ajax... any help is appreciated as i want the result to be displayed on the same page as the form is. Thank you!

Where are you sending ajax? What is $form? You are open to SQL injections with this code. Use parameterized prepared statements. php.net/manual/en/pdo.prepared-statements.php — chris85
– chris85, Commented Nov 2, 2015 at 4:16
The tr and fieldset open and close tags do not match. Just FYI. — Sverri M. Olsen
– Sverri M. Olsen, Commented Nov 2, 2015 at 5:09

Niranjan N Raju · Accepted Answer · 2015-11-02 05:25:41Z

1

If you want to get data from another file to a select, you should add as option. Plain text inside select will not help you out. So wrap the vales with <option></option>

So change this line,

echo "<option>{$row['Discount']}</option>";

If you want to give values,

 echo "<option value='{$row['Discount']}'>{$row['Discount']}</option>";

EDIT

Now onchange of deopdown, date call ajax to do next stuff.

$(document).on("change","#date",function() {
     var tradetype = //capture the value here;
     var metal = //capture the value here;
     var amount = //capture the value here;
     $.ajax({
          url:"path/filename.php",
          data:{date:$(this).val(),"tradetype":tradetype,"metal":metal,"amount":amount},
          type: "POST",
          success: function(data){
               alert(data);//this will alert what you have echoed on php file
          }
     });

});

edited Nov 2, 2015 at 5:25

answered Nov 2, 2015 at 4:15

Niranjan N Raju

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17 Comments

chris85 Over a year ago

Need to give it a value as well.

Niranjan N Raju Over a year ago

He has included a file and echoed the values from db inside select. He is just echoing the value in php file. He shud wrap with <option> tag.

Niranjan N Raju Over a year ago

I was about to add that in answer. But you asked for clarification.

chris85 Over a year ago

Yea, answer was very vague to begin with. OP still has multiple issues with the current approach.

Niranjan N Raju Over a year ago

Yes, i will try changing my answer.

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