Data Not Displaying in Drop Down by mysqli - php

Question

I have a small data entry form which works well, but someone asked me if I can have the 'name' field as a drop down box of users as opposed to having to type in a name and risk a spelling mistake - yep makes sense.
This is pretty new to me and following some information on here and other sites I have tried to accomplish the first part.. populating the drop down box.. nope. No errors, just nothing in the box.
To power this I have tblStaffNames (userID, txtName)
The code I am using looks like this;
<?php
include("connect-db.php");
$queryNames = "SELECT txtName FROM tblStaffName";
$resultNames = $conn->query($queryNames);
?>
<select name="personname">
<?php
while ($rowNames = $resultNames->fetch_assoc()) {
echo "<option value=\"{$rowNames['txtName']}\">";
echo $rowNames['txtName'];
echo "</option>";
}
?>
</select>
The $conn is all good as on another page I can display data in a table from the database, including tblStaffNames - so I can rule out any sort of connection issues.
When I run the page, the little drop down box appears, very simple like but it's there, just no values.
I will end up using the value like this as part of the data entry form;
<td><select name="personname" style="width:100px" ><?php echo $RowNames; ?></select></td>
But I can't actually get to the point of displaying data.
Can anyone help me out with what I am doing wrong here?

Answer 1

I tested your code with slight changes. It works for me. Please check your DB connection is OK as I have done in my code.
<?php
$servername = "localhost";
$username = "user";
$password = "password";
$dbname = "staff_db";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$queryNames = "SELECT txtName FROM tblStaffName";
$resultNames = $conn->query($queryNames);
//Generating the Dropdown
echo "<select name=\"personname\">";
while ($rowNames = $resultNames->fetch_assoc()) {
echo "<option value=\"{$rowNames['txtName']}\">";
echo $rowNames['txtName'];
echo "</option>";
}
echo "</select>";
?>

Answer 2

I dont know what your error but for better understanding i change your code like this
$queryNames = "SELECT userID,txtName FROM tblStaffName";
$resultNames = $conn->query($queryNames);
?>
<select name="personname">
<?php
while ($rowNames = $resultNames->fetch_assoc()) {
?>
<option value="<?php echo $rowNames['userID']; ?>">
<?php echo $rowNames['txtName']; ?>
</option>
<?php
}
?>
</select>

Answer 3

Following code snippets works for me.
<?php
$servername = "localhost";
$username = "user";
$password = "password";
//DB Server Connection
$conn = mysql_connect($servername, $username, $password) or die("Connection establishment failed");
//DB Selection
$selected = mysql_select_db("staff_db", $conn) or die("Could not select DB");
//Query String
$queryNames = "SELECT txtName FROM tblStaffName";
//Query the DB
$resultNames = mysql_query($queryNames);
//Generating the Dropdown
echo "<select name=\"personname\">";
while ($rowNames = mysql_fetch_array($resultNames)) {
echo "<option value=\"{$rowNames['txtName']}\">";
echo $rowNames['txtName'];
echo "</option>";
}
echo "</select>";
?>