Still celebrating National IT Professionals Day with 3 months of free Premium Membership. Use Code ITDAY17

x
?
Solved

Help with PHP query

Posted on 2012-04-02
3
Medium Priority
?
250 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
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
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 111

Accepted Solution

by:
Ray Paseur earned 2000 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

Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

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

When table data gets too large to manage or queries take too long to execute the solution is often to buy bigger hardware or assign more CPUs and memory resources to the machine to solve the problem. However, the best, cheapest and most effective so…
Many old projects have bad code, but the budget doesn't exist to rewrite the codebase. You can update this code to be safer by introducing contemporary input validation, sanitation, and safer database queries.
Explain concepts important to validation of email addresses with regular expressions. Applies to most languages/tools that uses regular expressions. Consider email address RFCs: Look at HTML5 form input element (with type=email) regex pattern: T…
The viewer will learn how to look for a specific file type in a local or remote server directory using PHP.

704 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