Solved

Help with PHP query

Posted on 2012-04-02
3
244 Views
Last Modified: 2012-06-27
I'm struggling to get this query working.

I think my syntax is wrong, but I don't know how I need to modify it.

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

else {

	$query = "SELECT * FROM `Pet` WHERE 'CustomerNumber' = '$customer'";
	

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

	while($result = $result_obj->fetch_array(MYSQLI_ASSOC)) {
		print_r($result);
		echo '<br />';
	}

	

}

Open in new window

0
Comment
Question by:InquisitiveProgrammer
3 Comments
 
LVL 24

Expert Comment

by:johanntagle
ID: 37798738
CustomerNumber should be enclosed by backticks, not single quotes.  Just like what you did with Pets.
0
 
LVL 1

Author Comment

by:InquisitiveProgrammer
ID: 37798795
Unfortunately that isn't the problem I'm having. I have added the complete code below:

<?php session_start(); ?>

<HTML>

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

</HEAD>

<BODY>

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

<?php

$customer = $_SESSION['curr_customer'];

echo "You are viewing the information for customer number " . $customer;

?>

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

<?php>

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

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

else {

	$query = "SELECT * FROM `Pet` WHERE `CustomerNumber` = '$customer'";
	

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

	while($result = $result_obj->fetch_array(MYSQLI_ASSOC)) {
		print_r($result);
		echo '<br />';
	}

	

}

?>

</BODY>

</HTML>

Open in new window


And this is the error I get:

Fatal error: Call to a member function fetch_array() on a non-object in /homepages/35/d387195482/htdocs/herename/customerinfo.php on line 43
0
 
LVL 108

Accepted Solution

by:
Ray Paseur earned 500 total points
ID: 37798964
Lines 43-46
	while($row = $result->fetch_array(MYSQLI_ASSOC)) {
		print_r($row);
		echo '<br />';
	}

Open in new window

I am not sure this is all that is wrong, and I cannot test because I do not have your data base, but it seems like this is something worth trying.  Best, ~Ray
0

Featured Post

Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Does the idea of dealing with bits scare or confuse you? Does it seem like a waste of time in an age where we all have terabytes of storage? If so, you're missing out on one of the core tools in every professional programmer's toolbox. Learn how to …
Creating and Managing Databases with phpMyAdmin in cPanel.
Learn how to match and substitute tagged data using PHP regular expressions. Demonstrated on Windows 7, but also applies to other operating systems. Demonstrated technique applies to PHP (all versions) and Firefox, but very similar techniques will w…
The viewer will learn how to count occurrences of each item in an array.

910 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

24 Experts available now in Live!

Get 1:1 Help Now