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Solved

My drop-down menu isn't getting populated

Posted on 2012-04-03
4
Medium Priority
?
246 Views
Last Modified: 2012-04-03
Here is the code:

When looking at it in a browser, I only get an empty drop-down menu.

<?php session_start(); ?>

<HTML>

<HEAD><TITLE>Florida Pet Clinics Inc.</TITLE>

</HEAD>

<BODY>

<H1 ALIGN="CENTER">Florida Pet Clinics Inc. </H1>



<?php>

define("MYSQLUSER", "???");
define("MYSQLPASS", "???");
define("HOSTNAME", "d???");
define("MYSQLDB", "???");

$customer = $_SESSION['curr_customer'];

echo "You are viewing the information for customer number " . $customer . "<br />";

$connection = new mysqli(HOSTNAME, MYSQLUSER, MYSQLPASS, MYSQLDB);
if ($connection->connect_error) {
	die('Connect Error: ' . $connection->connect_error);
} 

else {

	$query = "SELECT PetNumber, PetName FROM Pet WHERE CustomerNumber = $customer";
	

	$result = $connection->query($query);

	if ($result->num_rows < 1) {
		echo 'No pets have been added for this customer';
	}
	else {

		echo "<select id='pet' name='pet'>";
		echo "<option value=' '></option>";

		while ($row = $result->fetch_array(MYSQLI_ASSOC)) {
			echo "<option value='{$row->PetNumber}'>{$row->PetName}</option>";	
		}

		echo "</select>";
	}
}

?>

<P><A HREF="addpetinfo.php">Add New Pet</A>

</BODY>

</HTML>

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0
Comment
Question by:InquisitiveProgrammer
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4 Comments
 
LVL 35

Expert Comment

by:gr8gonzo
ID: 37802764
What does the HTML source look like after the page finishes loading?
0
 
LVL 1

Author Comment

by:InquisitiveProgrammer
ID: 37802790
My syntax is wrong for how I'm providing my database row results, but I'm not sure what I'm doing wrong:

<HTML>

<HEAD><TITLE>Florida Pet Clinics Inc.</TITLE>

</HEAD>

<BODY>

<H1 ALIGN="CENTER">Florida Pet Clinics Inc. </H1>



You are viewing the information for customer number 16<br /><select id='pet' name='pet'><option value=''></option><option value=''></option></select>
<P><A HREF="addpetinfo.php">Add New Pet</A>

</BODY>

</HTML>

Open in new window

0
 
LVL 35

Accepted Solution

by:
gr8gonzo earned 2000 total points
ID: 37802816
Try changing:

$row->PetNumber

to:

$row["PetNumber"]
0
 
LVL 1

Author Comment

by:InquisitiveProgrammer
ID: 37802870
I have this code now:

else {

	$query = "SELECT PetNumber, PetName FROM Pet WHERE CustomerNumber = $customer";
	

	$result = $connection->query($query);

	if ($result->num_rows < 1) {
		echo 'No pets have been added for this customer';
	}
	else {

		echo "<select id='pet' name='pet'>";
		
		while ($row = $result->fetch_array()) {
			$rows[]=$row;	
		}

		foreach($rows as $row) {
			echo "<option value='$row[\"PetNumber\"]'>'$row[\"PetName\"]'</option>";
		}

		echo "</select>";
	}
}

Open in new window


and I get the following errors:

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /homepages/35/d387195482/htdocs/lodwig/customerinfo.php on line 50
0

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