Output PHP mySQL query into dropdown

Question

Been at this for a bit and I cannot figure out what is wrong. I am trying to output a mySQL query into a drop down (just using PHP). The dropdown works, but there are no values in the list. However, it does seem to be aware of how many rows are in the table. Please advise. Thanks!

<html>
<head>
<body>

<table border="1">


<?php
require_once ('includes/database.php');         

$query = "SELECT User_ID FROM User_Account";
$result = mysqli_query ($my_dbhandle, $query);
echo "<select name=dropdown value=''>Dropdown</option>";
while($r = mysqli_fetch_array($result)){
echo "<option value=" . $r['User_ID'] . ">" . $r['User_ID'] . "</option>";
}
echo "</select>";



mysqli_close($my_dbhandle);
?>

</table>
</body>
</head>
</html>

For one thing, <select> does not have value, only <option>. Plus you should quote things, that's why it's not working. View HTML source. — Funk Forty Niner
– Funk Forty Niner, Commented Apr 11, 2015 at 16:32
Plus, your other question already contains your solution for this. Just do the same for the option echo. — Funk Forty Niner
– Funk Forty Niner, Commented Apr 11, 2015 at 16:38

tim · Accepted Answer · 2015-04-11 16:34:22Z

2

You forgot to add the actual text. So this:

"<option value=$r[User_ID]></option>";

should be this:

"<option value=" . $r['User_ID'] . ">" . $r['User_ID'] . "</option>";

If User_ID is in any way editable by users, you also should add protection against XSS.

answered Apr 11, 2015 at 16:34

tim

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1 Comment

Kripros Over a year ago

Thank you @tim. That was it. I will mark it correct ASAP.

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