[Webinar] Streamline your web hosting managementRegister Today

x
?
Solved

Help with trivial validation function

Posted on 2009-07-06
4
Medium Priority
?
270 Views
Last Modified: 2013-12-13
Hi everyone,
can anyone please help me do this right? I am trying to validate a form with category id, of which list I pull out of the DB to make sure the id is valid. But seem to get errors of invalid argument for "foreach". I am attaching the function below. I am a sort of beginner, so I am sure that everyone will have much better solution, but I am trying to do it myself (with this exception) and just need to find an error in my script.. not an entire new solution. that will be next :-)
$category_id_errors = check_cagegory_id(array('category')); 
 
 
function check_cagegory_id($check_category_id) {  //this is the funtion referred to in the code above
	$form_errors = array();
	
	$subject_set = get_all_subjects();
	if(mysql_num_rows($subject_set)>=1) {
		$subjects = array();
		while($subject = mysql_fetch_array($subject_set)) {
			$subjects[] = $subject['id'];
		}
		foreach($check_cagegory_id as $fieldname) {
			if((!isset($_POST[$fieldname]))||(!in_array($_POST[$fieldname], $subjects))) {
				$form_errors[] = $fieldname ."doesn't exist";
			}
		}
	} else {
		$form_errors[] = $fieldname." - no category to associate a new page with";
	}
	return $form_errors;

Open in new window

0
Comment
Question by:czechmate1976
  • 2
  • 2
4 Comments
 
LVL 12

Expert Comment

by:kevin_u
ID: 24786760
In the foreach,

$check_cagegory_id

is spelled wrong

It should be:

$check_category_id

0
 

Author Comment

by:czechmate1976
ID: 24787260
Thanks, I corrected it but still get "Warning: Invalid argument supplied for foreach() in C:\xampplite\htdocs\FGEnew\siteadmin\includes\functions.inc.php on line 218"

which is the following:
foreach($check_category_id as $fieldname) {
                  if((!isset($_POST[$fieldname]))||(!in_array($_POST[$fieldname], $subjects))) {
                        $form_errors[] = $fieldname ."doesn't exist";
                  }
0
 
LVL 12

Accepted Solution

by:
kevin_u earned 1000 total points
ID: 24787336
The underlying issue is that $check_category_id is not an array.  Thats the cause of the error message.   In your short example, it is surely an array.  But is it called the same way in the full script?
0
 

Author Comment

by:czechmate1976
ID: 24787404
resolved. Sorry, after days staring at the code I can hardly see.. Thanks for your help
0

Featured Post

The new generation of project management tools

With monday.com’s project management tool, you can see what everyone on your team is working in a single glance. Its intuitive dashboards are customizable, so you can create systems that work for you.

Question has a verified solution.

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

The title says it all. Writing any type of PHP Application or API code that provides high throughput, while under a heavy load, seems to be an arcane art form (Black Magic). This article aims to provide some general guidelines for producing this typ…
Strategic internal linking is often considered an SEO power technique, especially for content marketing. Do you need to hire an SEO agency to optimize you internal linking? No, this article will help you understand the basics of internal linking and…
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 create and use a small PHP class to apply a watermark to an image. This video shows the viewer the setup for the PHP watermark as well as important coding language. Continue to Part 2 to learn the core code used in creat…
Suggested Courses

590 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