PHP - Moved: Can't Get Z-index To Stack Things Properly

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Hi

I am buidling a Blog System using OO techniques (or at least attempting to)..

I keep getting an error :

Notice: Undefined variable: blog_categoryName in C:\wamp\www\atkinsonsCMS\admin170976\blog_categories.php on line 13
...And then underneath all that is is displaying 'Call Stack'. I've never come across this notification before so I guess it is an OO term, and the a list of unexpected outputs: #, time, memory, function and location....each with an associated value.
I attempted to resolve the variable issue in the usual way (as you can see in my code), but still the errors/warnings persist.

The purpose of this part of the code is to simply retrieve SQL data (which is a list of blog categories), allowing the user to select one and then click through to be directed to the appropriate category page (although this last requirement of the code has not yet been programmed).

Could those who have experience of this please take a look at my code and let me know where I am going wrong?

Thanks

blog_categories.php
This file calls on the the blogFunctions.php Category Class, and the displays the results.


session_start();
$pageTitle = "Blog Categories";
include("../includes/admin_header.php");
include("../classes/blogFunctions.php");
echo '<div class="admin_main_body"><br /><br />';
echo "<h2>Category List</h2><br />";
$content = new Category;
$content -> process();
//print_r(debug_backtrace());
echo "<a href=blog.php>$blog_categoryName</a>";
echo '</div>';
echo '</div>';
include("../includes/admin_footer.php");




blogFunctions.php
This class deals with the retreaval of the SQL data, and then passes it to blog_categories.php


class Category {
               public $errors;
               public $blog_categoryName = NULL;
public function __construct() {
                $this->errors = array();
                $blog_categoryName = $this->blog_categoryName;
                }                       
public function getCategories(){
                if(!isset($blog_categoryName))
                {
                $blog_categoryName = 'value';
                }
                ($GLOBALS["___mysqli_ston"] = mysqli_connect("localhost", "atkinson", "XYZ111WA")) or die(((is_object($GLOBALS["___mysqli_ston"])) ? mysqli_error($GLOBALS["___mysqli_ston"]) : (($___mysqli_res = mysqli_connect_error()) ? $___mysqli_res : false)));
                ((bool)mysqli_query($GLOBALS["___mysqli_ston"], "USE atkinsonscms")) or die (((is_object($GLOBALS["___mysqli_ston"])) ? mysqli_error($GLOBALS["___mysqli_ston"]) : (($___mysqli_res = mysqli_connect_error()) ? $___mysqli_res : false)));
                mysqli_query($GLOBALS["___mysqli_ston"], "SELECT * FROM blog_category ORDER BY ASC");
                $result = @mysqli_query($GLOBALS["___mysqli_ston"], $query);
                if ($result) {
                while ($row = $result->fetch_object()) {
                $blog_categoryName = $row->blog_categoryName;
                 }
                //need to use mysqli query to retrieve results. 
                if(mysqli_affected_rows($GLOBALS["___mysqli_ston"])< 1)
                $this->errors[] = 'Could not retrieve data';
                }
                                }
public function show_errors(){
                echo "<h3>Errors</h3>";
                foreach($this->errors as $key=>$value)
                echo $value."<br>";
                                    }
public function process() {
                $this->getCategories();
                return count($this->errors)? 0 : 1;
                                    }
} //Closes Category Class.            

hello dear php-experts and freaks

how to view photos and to rapidly change the major categories

i have several hundred images (photographs) taht i want to view on opensuse linux version 13.1


i want to view the images - and of while doing so i want to turn the colored images into b/w

Question; which is the best - ie. the quickest way to do so.

note; i have installed the following thigns on my opensuse 13.1:

- gimp the great grahical tool
- digicam - the great tool to view images and pictures
- Gwenview Version 4.11.4 on KDE 4.11.5


so again here the question: which tool allowes to view the color(ed) image and to turn it - on the fly - in to black and white?

BTW: can i tell linux to swith the colors - in general - to black and white? is this doable...

looks like we could write a small shell script, utilizing imagemagick.

maybe like this:
Code:

for i in *jpg ; do mogrify -colorspace Gray "$i" ; done

this is just a quick hack. i have tried. See more bleow: Also, it's not interactive and simply changes all jpg's to grayscale, even if they already are grayscale.

please note that we need to have some more things about imagemagick.
  well see the results:

martin@linux-70ce:~/Bilder>
martin@linux-70ce:~/Bilder> for i in *jpg ; do mogrify -colorspace Gray "$i" ; done
mogrify: unable to open image `*jpg': Datei oder Verzeichnis nicht gefunden @ error/blob.c/OpenBlob/2643.
mogrify: no decode delegate for this image format `*jpg' @ error/constitute.c/ReadImage/552.
martin@linux-70ce:~/Bilder>

 

I really dont no what the problem is and im new here so i dont no if this is the right place to put this but im trying to make a registration page for my server and when i bring it up it says Undefined index: action in C:\xampp\htdocs\index.php on line 56   and idk what to do >.<
this is the line that has the problem

if($_POST['action']!="signup")

so hopefully u guys could help me :/ Im using xampp btw

Hi there,

I was wondering if there is a way to display content on only my index.php page? I believe there is a way and I have done it before, but some of my pages use index.php?category etc.

Is there a way to show content just on index.php and not on any dynamic pages?

Thanks

Hello everyone 
I am working on a simple php mysql project (TodoList with crud funtions with fields- head, content ,date and time ) and i am stuck on this error Fatal error: Uncaught Error: Call to a member function close() on bool Stack trace: #0 {main} 
How can i fix it 
Here is link from the project https://easyupload.io/gxxv1w
 

You: As a PHP expert, you find the prospect of building yet another brochure-ware website distressing; you want to get your hands dirty building an advanced, MVC based application in a fun and supportive environment where you will be constantly challenged, always learning and delivering a product used by millions of people every day. Our company was founded and is managed by developers who wrote the original software the company was built upon; the technical team are the core of our business. As a member of that team, it will be you who decides upon the on-going development of our technical stack as well as the product itself. You will of course have complete control over your workstation.   Our stack:   Qualifications: A passion for programming, solving problems and building software customers love. Experience building web-based applications as distinct from brochure-ware websites. An expert in OOP/MVC/ORM programming techniques and PHP (v5.3+) frameworks. We happen to use Symfony2 but any relevant experience is fine. You can bend MySQL to your will. Solid front-end development experience with XHTML, CSS, and Javascript. Our stack includes AngularJS, CoffeeScript, jQuery and LESS. An appreciation for a beautiful GUI and excellent user experience. Expectation of a fast moving, agile environment. We ship code to customers daily – we’d expect you contribute from day 1. You can validate your skills for example with source code samples, a github profile, contributions to an open source project, developer blog, stackoverflow answers etc. What you will do: Design (with help) and develop new channels for our software. In particular we are looking at apps using twilio (for voice/sms), facebook and twitter as well as a web based screensharing tool. Integrate our software into other 3rd party applications. This could both be via APIs (salesforce, jira, 37signals) or through digging directly into the source code (sugarCRM, xenforo etc). Continue improve and refine our existing channels and product. We are never satisfied and always looking for that small improvement, refactor or redesign that can increase usability, speed and functionality for our customers. Work on our backend systems including internal developer tools and deployment processes, our SaaS application stack or the tools we have to manage our business. Work on our mobile apps, built using our API / HTML5 and PhoneGap About us: Our principle product is DeskPRO, the helpdesk software platform. We make it easy for organisations (companies – large and small, charities, public sector organisations) to communicate with their users via email, twitter, facebook, SMS, web forms, live chat (text and voice) as well as providing publishing self-help tools, sales management, co-working tools and community based question and answers. DeskPRO is a platform where customers install the helpdesk applications they need to manage their business. We sell this software to a range of organisations including large companies (e.g. Tumblr, Xerox, T-mobile, Fujitsu, Valve Software and AT&T) universities and the public sector all the way through to small family businesses. We have a dual license model; selling licenses to software installed on our clients’ servers (we don’t encrypt the code – so your code will get read!) and offering a fully managed SaaS solution. Millions of people use our software every day and a lot of agents spend their whole working day using it constantly.   What we offer: Competitive salary based upon experience. Friendly work environment at the Innovation Warehouse in Farringdon: http://theiw.org/. 30" monitor and control over your workstation setup. No corporate bureaucracy here. A mixture of autonomy over your role and real responsibilities to the team and business. Varied work. The DeskPRO product is large with lots of modules and technologies. Review our source code before even applying; just ask and we will add you to our github account. Never accept a job until you have seen the code you will be working with. We plan to have a lot of fun on this journey – please bring a sense of humour. Apply: Please send your CV and either a link to your online profile, github account or some sample source code to: jobs@deskpro.com

I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.   my form page test.php

<?PHP
require_once("./include/membersite_config.php");

if(!$fgmembersite->CheckLogin())
{
    $fgmembersite->RedirectToURL("login.php");
    exit;
}

?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?>  "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?>  "><br><br>

<button>Submit</button>

my editform.php

<?php

$con = mysqli_connect("localhost","root","user","pass");

if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");



header("Location: test.php");

?>

How can i make code which can do something when time is 00:00 ?
Is that possible in php, and if not any suggestions?

By not using the order by function in SQL.

Like I have $match_1 and $segment_1. They are two seperate tables so how would I order them like the ORDER BY in sql. Is there a way to do that?

ok so i have my login page all set up and working good , but i want to add a special thing to where after 3 login attempts it locks the account for like 30 mins , and notifies the use by email that someone is trying to login and failed... what would be the best way to do this?

Ok If anyone read my recent topics they will know I am making a forum system This system for users is currently about 1/3 complete and I now have a couple of issues.

Firstly:
When a user logs in I am using $_SESSION['RAYTH_MEMBER_ID'] to save their member ID for when they return. However every time the browser/page is closed it doesn't save and then they are logged out. How would I set it so it keeps them logged in until they personally logout?

Secondly:
Since It's a forum you expect there to be new lines and stuff when you Read posts in threads (like we do here). How would I set it when a user posts it replaces every return character from the input (enter/newline etc) and replace it with <br> in the mysql database?

Thanks for your help

ok , here is my mysql code to get all posts from the posts table .
Code: [Select]
$query = mysql_query("SELECT id,to_id,from_id,post,type,state,date FROM posts WHERE state='0' ORDER BY id DESC LIMIT 50"); and here is the code to display the users friends...
Code: [Select]
$sqlArray = mysql_query("SELECT friend_array FROM myMembers WHERE id='" . $logOptions_id ."' LIMIT 1");
while($row=mysql_fetch_array($sqlArray)) { $iFriend_array = $row["friend_array"]; }
$iFriend_array = explode(",", $iFriend_array);
if (in_array($id, $iFriend_array))see now i got as far as , if(in_array($id, $iFriend_array))
How would i put these togeather to where it would get the posts from the posts table that there friends posted?