Displaying mysqli_fetch array results in dropdown php

Question

I have a drop down that is being populated from a static select. Then when a choice is made in the first drop down a query runs and the second drop down is populated from the database depending on selection in first select box. Here is the code. I'm having a problem displaying the second drop down with the data.

$selectvalue = mysqli_real_escape_string($mysqli, $_GET['selectvalue']);

$result = mysqli_query($mysqli, "SELECT DISTINCT '$selectvalue' FROM accounts ");

echo '<option value="">Please select...</option>';
while($row = mysqli_fetch_array($result))
{
echo '<option value="'.$row['$selectvalue'].'">' . $row['$selectvalue'] . "</option>";
//echo $row['drink_type'] ."<br/>";
 }

 mysqli_free_result($result);
 mysqli_close($connection);

 ?>

select distict 'foo' is selecting the string foo from your table, you're not fetching a field. field/table names can NOT be quoted with '. you have to use backticks: select `foo` from — Marc B
– Marc B, Commented Dec 16, 2014 at 16:32
is this the second select box? You are changing depending on first select box...don't you using ajax? — Riad
– Riad, Commented Dec 16, 2014 at 16:56
Yes, this is the second select box and I'm using ajax to send variable from first drop down. — California Steve
– California Steve, Commented Dec 16, 2014 at 17:12

jay.jivani · Accepted Answer · 2014-12-16 17:16:41Z

1

<?php
$selectvalue = mysqli_real_escape_string($mysqli, $_GET['selectvalue']);
$result = mysqli_query($mysqli, "SELECT DISTINCT * FROM accounts WHERE col_name = '".$selectvalue."' ");
echo '
<select name="some_name">
<option value="">Please select...</option>';
while($row = mysqli_fetch_array($result)) {
    echo '<option value="'.$row['col_name'].'">'.$row['col_name']."</option>";
}
mysqli_free_result($result);
mysqli_close($connection);
?>

answered Dec 16, 2014 at 17:16

jay.jivani

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