MySQL Forums
Forum List  »  PHP

How do I get this drop down to work?
Posted by: Brian Powell
Date: September 02, 2014 11:46AM

Hi There,

I have the following code that I've pieced together from various sources. Neither PHP nor SQL are my main languages, so I'm still learning how this works. I'd like some advice on why this isn't working:

<?php
$con=mysqli_connect("localhost","root","","dwinvoice");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$getPM = "Select Distinct PMName from report";
$results = mysqli_query($con,$getPM)
?>
<select name="select1">
<?php	
	while ($line = mysql_fetch_array($getPM, MYSQL_ASSOC))
?>
	<option value="<?php echo $line['field'];?>"> <?php echo $line['field'];?> </option> 	
</select>

<?php
$result = mysqli_query($con,"SELECT * FROM report WHERE PMName = $results");

echo "<table border='1'>
<tr>
<th>Track Number</th>
<th>PM Name</th>
</tr>";

while($row = mysqli_fetch_array($result)) {
  echo "<tr>";
  echo "<td>" . $row['TrackNumber'] . "</td>";
  echo "<td>" . $row['PMName'] . "</td>";
  echo "</tr>";
}

echo "</table>";

mysqli_close($con);
?>

I am trying to build a dropdown menu that will pull all distinct names from the PMName column in the table 'report', then take that selection and utilize it in the HTML table down below. I know the syntax here:
$result = mysqli_query($con,"SELECT * FROM report WHERE PMName = $results");
is wrong, but I'm not sure how to make it right. The drop-down list does not actually pull any values, and I'm not sure what I'm doing wrong....

Options: ReplyQuote


Subject
Written By
Posted
How do I get this drop down to work?
September 02, 2014 11:46AM


Sorry, you can't reply to this topic. It has been closed.

Content reproduced on this site is the property of the respective copyright holders. It is not reviewed in advance by Oracle and does not necessarily represent the opinion of Oracle or any other party.