Get database value based on another input field

Question

I've been searching for this kind of problem and I couldn't find one. I am using ajax to solve this problem but it didn't work out. I really want to have this scenario where after the user scanned his/her QR code, its data (account ID) will be in the input field. Then the other input field will automatically show its corresponding data from the database based on the data from QR code. This is so far my code.

This is from the main page:

<label id="acc">Account ID 
<input type="text" name="accId" id="accID" class="idnum" required="" value="">
</label>
<label id="label">QR ID
<input type="text" readonly=" " name="qrId" id="qrId" style="width: 108px;margin-right: 15px;" value="" > 
</label>

Ajax:

<script>
$("#accID").change(function() {
var accID = $(this).val(); 
$.ajax({
  url: 'loadata.php',
  type: 'POST',
  data: 'accID='+accID,
  success: function(html) {
    $("#qrId").html(html);
  }
});
});
</script>

this is the loadata.php:

<?php
session_start();
include ('connection.php');
if(isset($_POST['accID']))
{
$accID = $_POST['accID'];

$sel = "SELECT * FROM qrcode WHERE Cus_IDNum = '$accID'";
$sel_run = $conn->query($sel);
if($sel_run->num_rows>0)
{
    while($rows = $sel_run->fetch_assoc())
    {
        ?>
        <input type="text" readonly=" "id="qrId" name="qrId" style="width: 108px;margin-right: 15px;" value="" > 
        <?php
    }
}
}
?>

Thank you very much for your time! :)

@KenLee This is NOT correct. The documentation says, that data can be a string too. And yes, it will used as POST body, if POST is used: https://api.jquery.com/Jquery.ajax/ So, that part should be correct. — t.h3ads, Jan 25 '22 at 09:13
Thanks for correcting me. (actually I did double check the documentation and delete my comment before seeing your comment above) — Ken Lee, Jan 25 '22 at 09:38
**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, Jan 25 '22 at 11:33

score 0 · Answer 1 · answered Apr 25 '22 at 14:34

Are you getting any data to return? Have you tried changing your input field in loadata.php to a div with the same ID? Right now you're trying to place an input within an input.

Also, you don't need to wrap your inputs within the label. As long as the label and input share the same ID, they will always be together. Currently, you are not doing that.

score -1 · Answer 2 · answered Jan 25 '22 at 09:11

Setting $("#qrId").html(html); will do nothing, as #qrId is an input field. I think, what you want to do is to set the value of the input field. This should work like this: $("#qrId").val(html);

Then, there is a second problem as your PHP script returns HTML of an input field rather than just the value to set. Also, it may return multiple values as you loop through the database results. You could try to change your script to something like this to just return the value of the first selected database record. Replace qrCodeValue with the real column name to use:

<?php
session_start();
include ('connection.php');
if(isset($_POST['accID']))
{
    $accID = $_POST['accID'];

    $sel = "SELECT * FROM qrcode WHERE Cus_IDNum = '$accID'";
    $sel_run = $conn->query($sel);
    if($sel_run->num_rows>0)
    {
        $row = $sel_run->fetch_assoc();
        print $row['qrCodeValue'];
        exit;
    }
}
?>

Thanks for responding. But still it didn't work hehe – gellooo Feb 20 '22 at 07:15 — gellooo, Feb 20 '22 at 07:15

Get database value based on another input field

2 Answers2