Populating a DropDown from SQL database in PHP - php

Question

I'm attempting to get data from a SQL database in order to populate a couple drop downs. This is an excerpt, but I can post more if you'd like. I didn't include it all because its more than a couple lines.
$queryData = mysql_query("SELECT DISTINCT DateTime AS DateTime FROM 'historicaldata' ORDER BY YEAR(DateTime), DAYOFYEAR(DateTime)");
$queryGroups = mysql_query("SELECT DISTINCT histgroupname AS GroupName FROM 'historicalgroups' WHERE `histgroupID` < 10 ORDER BY `histgroupname`");
$tracker = 0;
$dataArray = array();
$groupsArray = array();
$DateFormat1 = array();
$DateFormat2 = array();
$DayNumber = array();
$Month = array();
$Year = array();
while ($row = mysql_fetch_array($queryData)) {
$dataArray[$tracker] = $row['DateTime'];
$tracker++;
}
$tracker = 0;
while ($row = mysql_fetch_array($queryGroups)) {
$groupsArray[$tracker] = $row['GroupName'];
$tracker++;
}
$tracker = 0;
foreach ($dataArray as $l) {
$p = strtotime($l);
$x = getdate($p);
$DateFormat1[$tracker] = date("D M d, Y", $x);
$DateFormat2[$tracker] = date("M Y", $x);
$DayNumber[$tracker] = date("z", $x);
$Month[$tracker] = date("n", $x);
$Year[$tracker] = date("Y", $x);
$tracker++;
}
echo "<div id='Period1'> <span class='regblue'>Start</span><select name='startdate'><option value=''></option>";
foreach($DateFormat1 as $x)
echo "<option selected value='$x'>$x</option>";
echo "</select> </div>";
For some reason, the drop down remains empty no matter what I try.

Answer 1

Why are you using such a complex code. Use the power of php of integrating itself with HTML.
Try this Style.
And check if you have established a connection with the database or not.
<?php
require_once('connection.php'); //establish the connection with the database on this page.
$queryData = mysql_query("SELECT DISTINCT DateTime AS DateTime FROM 'historicaldata' ORDER BY YEAR(DateTime), DAYOFYEAR(DateTime)");
$queryGroups = mysql_query("SELECT DISTINCT histgroupname AS GroupName FROM 'historicalgroups' WHERE `histgroupID` < 10 ORDER BY `histgroupname`");
$result = mysql_fetch_array(mysql_query($queryData)); //$result now has database tables
$resultGroups = mysql_fetch_array(mysql_query($qrueryGroups)); //$resultGroups has now database tables
?>
<select name='Date'>
<?php
while($row = mysql_fetch_array($result))
{
?>
<option values=<?php echo($row['DateTime']); ?><?php echo($row['DateTime']); ?></option>
<?php
}
?>
</select>
<?php
?>

Answer 2

You may try like this
<?php
require_once('db_connect.php'); //connect with the database.
$queryData = mysql_query("SELECT DISTINCT DateTime AS DateTime FROM 'historicaldata' ORDER BY YEAR(DateTime), DAYOFYEAR(DateTime)");
$queryGroups = mysql_query("SELECT DISTINCT histgroupname AS GroupName FROM 'historicalgroups' WHERE `histgroupID` < 10 ORDER BY `histgroupname`");
$result = mysql_fetch_array(mysql_query($queryData)); //$result now has database tables
$resultGroups = mysql_fetch_array(mysql_query($qrueryGroups)); //$resultGroups has now database tables
echo '<select name="Date" id="Date">';
while($row = mysql_fetch_assoc($result))
{
echo '<option values=' . $row["DateTime"] . '>' . $row["DateTime"] . '</option>';
}
echo '</select>';
?>