I'm using this a particular code to dynamically load content on a webpage.
Problem is, I want to use it to load multiple things on the same page. How do I go about making this possible? I've already isolated the problem to the fact that the second instance of the code that runs replaces the entire URL instead of appending an additional ?="pageurl" but I'm stuck right there.
I've got a second script that uses the variable p instead of b so the browser knows to load p's content in a different location.
This is what the code does:
http://www.youtube.com/watch?v=Qg-EBdNaUbo
[edit]
Not sure why you guys can't see the link, it explains everything. I'm calling the script via a link Here is the link again youtube.com/watch?v=Qg-EBdNaUbo
CODE:
<?php
$pages_dir = 'pages';
if(!empty($_GET['b']))
{
$pages = scandir($pages_dir, 0);
unset($pages [0], $pages [1]);
$b = $_GET['b'];
if(in_array($b, $pages))
{
include($pages_dir.'/'.$b);
}
else
{
echo 'sorry page not found';
}
}
else
{
}
?>
Related
I am Working on making the menu for our content management software using php and we are having this small issue. Since we want everything to eventually be called in chunks, were breaking certain page items into chunks and loading them via functions through an included file. Since this is hard to explain, I will post some example code of what i mean below.
This is the file page.php (removed needless html code).
This is the page the user is on:
<?php
define("CURRENT_PAGE", "page.php");
include_once("data/main.inc.php");
?><html>
Content loads here.
<? desktopMenu(); ?>
</html>
Okay and here's the function for desktopMenu() from main.inc.php:
function desktopMenu() {
// Query to get the top level navigation links with no parents
$query = mysql_query("SELECT * FROM menu WHERE p_id = '0'");
if(mysql_num_rows($query) > 0) {
while($result = mysql_fetch_array($query)) {
extract($result);
if($isparent == "1") {
// Just check if they have children items
$sub_menu_query = mysql_query("SELECT * FROM menu WHERE p_id = '$id'");
if(mysql_num_rows($sub_menu_query) > 0) {
// CODE TO SHOW THE MENU ITEM AND ITS SUBS
}
} else {
// CODE TO SHOW REGULAR MENU ITEMS
// WANT TO INSERT class='active' if the CURRENT_PAGE is this value..
echo "<li><a href='#'>link</a></li>";
}
} else {
echo "<li><a href='javascript:void(0);'>Error Loading Menu</a></li>";
}
}
I am wondering how I can get the CURRENT_PAGE on the included script so I can load the class="active" onto the correct page. I am already using the following:
$config = include('config.inc.php');
$GLOBALS = $config;
on the top of main.inc.php, above this menu function so I could set global variables and include my $config['database'] variables for calling the SQL database within a function (doesn't work otherwise).
How can I check the current_page variable so I can set it active in the menu? I have tried a few different things but nothing is showing the way we expect it to. Thanks guy.
First of all I would recommend looking at MVC architecture when building your apps. I believe the use of GLOBALS is frowned upon.
To answer your question:
Since you are defining a constant define("CURRENT_PAGE", "page.php"); then this will be globally available within the scope of the function desktopMenu()
so you may use something like:
$className = (isset(CURRENT_PAGE) && CURRENT_PAGE=='xxxxx')?'class="active"':'';
echo "<li>link</li>";
xxxx string is most likely a field output from you database as the page name which will match the defined constant.
$className = (isset(CURRENT_PAGE) && CURRENT_PAGE==$result['page_name'])?'class="active"':'';
This is the basic form and you will most likely need additional conditions for the 'active' menu switch mapping to different pages.
I've tried to answer your question with an example although the structure you have used run the app is not the recommended way to develop.
I would look at the way modern frameworks are structured (Laravel, Zend, Symphony...) and utilise these.
I would also try and automate the page mapping (e.g. look at the URL and pull out the page from a rewrite which matches to the menu in your database)
best of luck
There are multiple options. Including static functions, global variables and passing the variable or object into the function.
The consensus for various reasons is to pass the variable into the function
$myVar = new Object\Or\Data();
function myFunction($myVar) {
//do stuff with $myVar
}
//then call the function
myFunction($myVar);
There are lots of answers to this question on stackOverflow, so have a deeper search. Here is an example
I found the solution to my problem and thought I would share here. I first set the call on the page.php to use desktopMenu(CURRENT_PAGE); and then on my main.inc.php I added this line
$thispage = CURRENT_PAGE;
function desktopMenu($thispage) {
//REST OF FUNCTION
}
And I set a table variable on each menu item called menu-group, so I can define the current menu group for a menu item and have the appropriate menu item highlighted when you're on that page or one of it's sub pages.
Thanks so much for the answers guys!
In cakephp, how do we specify which theme to use for an Element. I am initializing View object in Controller. I need to pass the element content as ajax response.
Controller :
$view = new View($this);
$view->layout = 'theme2';
$view->theme = 'newNav';
foreach($ctps as $ctpName)
{
$ctp[] = $view->element($ctpName);
}
At first I thought of accessing it as
$ctp[] = $view->element('../Themed/new/Elements/'.$ctpName);
But it obviously does not take care about element's directory. As some of the elements are in app/View/Elements and some are in app/ViewThemed/new/Elements/ directory.
Please suggest.
As per Arilia's suggestion, I am now trying
$this->viewClass = "Theme";
$this->viewPath = 'Elements';
$view = new View($this);
$view->theme = "new";
$view->layout = "theme";
$ctp = $view->element('userprofile');
echo json_encode($ctp);
die;
I am making an ajax call to this code. and it returns
"Element Not Found: Elements/userprofile.ctp"
what I was trying to develop was to render the basic page at first so that it can be cached. and then with the ajax call fetch the parts of UI that are session dependent and then replace the respective elements on browser.
To implement this, I tried fetching cakephp elements in controller, json encoding them and echoing them back. The code above though it worked for a normal dynamic page, it did not work for ajax calls. Cakephp somehow misbehaved here.
Now, as a better alternative, I have moved the code to View itself and I am echoing the json encoded from there. and as expected it is working well. Code is as
app/View/Themed/new/Elements/get_ctps.ctp :
<?php
foreach($ctps as $ctpName)
{
$ctp[] = $view->element($ctpName);
}
echo json_encode($ctp);
die;
?>
Its not a usual case. However, it might be helpful for someone.
URL : http://www.sayuri.co.jp/used-cars
Example : http://www.sayuri.co.jp/used-cars/B37753-Toyota-Wish-japanese-used-cars
Hey guys , need some help with one of my personal projects , I've already wrote the code to fetch data from each single car url (example) and post on my site
Now i need to go through the main url : sayuri.co.jp/used-cars , and :
1) Make an array / list / nodes of all the urls for all the single cars in it , then run my internal code for each one to fetch data , then move on to the next one
I already have the code to save each url into a log file when completed (don't think it will be necessary if it goes link by link without starting from the top but will ensure no repetition.
2) When all links are done for the page , it should move to the next page and do the same thing until the end ( there are 5-6 pages max )
I've been stuck on this part since last night and would really appreciate any help . Thanks
My code to get data from the main url :
$content = file_get_contents('http://www.sayuri.co.jp/used-cars/');
// echo $content;
and
$dom = new DOMDocument;
$dom->loadHTML($content);
//echo $dom;
I'm guessing you already know this since you say you've gotten data from the car entries themselves, but a good point to start is by dissecting the page's DOM and seeing if there are any elements you can use to jump around quickly. Most browsers have page inspection tools to help with this.
In this case, <div id="content"> serves nicely. You'll note it contains a collection of tables with the required links and a <div> that contains the text telling us how many pages there are.
Disclaimer, but it's been years since I've done PHP and I have not tested this, so it is probably neither correct or optimal, but it should get you started. You'll need to tie the functions together (what's the fun in me doing it?) to achieve what you want, but these should grab the data required.
You'll be working with the DOM on each page, so a convenience to grab the DOMDocument:
function get_page_document($index) {
$content = file_get_contents("http://www.sayuri.co.jp/used-cars/page:{$index}");
$document = new DOMDocument;
$document->loadHTML($content);
return $document;
}
You need to know how many pages there are in total in order to iterate over them, so grab it:
function get_page_count($document) {
$content = $document->getElementById('content');
$count_div = $content->childNodes->item($content->childNodes->length - 4);
$count_text = $count_div->firstChild->textContent;
if (preg_match('/Page \d+ of (\d+)/', $count_text, $matches) === 1) {
return $matches[1];
}
return -1;
}
It's a bit ugly, but the links are available inside each <table> in the contents container. Rip 'em out and push them in an array. If you use the link itself as the key, there is no concern for duplicates as they'll just rewrite over the same key-value.
function get_page_links($document) {
$content = $document->getElementById('content');
$tables = $content->getElementsByTagName('table');
$links = array();
foreach ($tables as $table) {
if ($table->getAttribute('class') === 'itemlist-table') {
// table > tbody > tr > td > a
$link = $table->firstChild->firstChild->firstChild->firstChild->getAttribute('href');
// No duplicates because they just overwrite the same entry.
$links[$link] = "http://www.sayuri.co.jp{$link}";
}
}
return $links;
}
Perhaps also obvious, but these will break if this site changes their formatting. You'd be better off asking if they have a REST API or some such available for long term use, though I'm guessing you don't care as much if it's just a personal project for tinkering.
Hope it helps prod you in the right direction.
I'm having a problem here and I think people here can help me.
I have a file that generates an image, ler.php, and the file that loads the images through a while, carregar.php.
I need to block direct access to the images generated by ler.php, tried to make a system like this session:
carregar.php:
<?
$_session['a'] = 1;
while($a != 50) { echo "<img src='ler.php?imagem=$a'>"; $a++; }
$_session['a'] = 0;
?>
ler.php:
<? if($_session['a'] == 1) { //load image } ?>
The result is the only loading the first image.
I'm trying to now use the $_SERVER ["PHP_SELF"], placing the IF of ler.php, what happens is I load it through <img src=''> she identifies as carregar.php.
Who has the best solution?
I've tried several ways with $_SESSION and it seems to not really work.
I could suggest two ways:
Easy to implement, less protective: in ler.php check that $_SERVER["HTTP_REFERER"] refers to "carregar.php".
A little bit more complicated: in carregar.php generate an unique code for each image you're going to output and store it in $_SESSION. Then pass the code to ler.php as a GET parameter. In ler.php check if the code exists in $_SESSION object, then generate an image and remove the code from $_SESSION.
I have a hard time identifying the problem, but your use of the session is going to lead to unexpected results:
You are adding 50 (I guess...) image tags to a page and right after you have added these tags, you set the session variable to 0.
The browser only loads a few files from the same server at the same time, so when the script is done and the first image is loaded, the browser is going to request the next image but that will fail as you have set the session variable to 0 already.
The only way to reliably set the session variable to 0 after all images have loaded, is an ajax request from your page that checks and triggers after all images have completely loaded.
Good! I managed to resolve one way and improvised without using AJAX:
ler.php:
<?php
if(isset($_SERVER["HTTP_REFERER"])) {
$check2 = (strpos($_SERVER["HTTP_REFERER"], 'carregar.php') > 0) ? true : false;
if(print_r($check2) != 11) {
// Blank
}
} else {
if(isset($_SERVER["HTTP_REFERER"]))
{
// Load Image
}
if(!isset($_SERVER["HTTP_REFERER"]))
{
// Blank
}
?>
So, the image only can be loaded into my page. Maybe...
I've searched far and wide and every CMS tutorial out there either doesn't explain this at all or gives you a huge chunk of code without explaining how it works. Even on stack overflow I can't find anything close to the answer, though I'd be okay with eating my words if someone could point me to the answer.
I am using PHP and mysql for this project.
I am building a CMS. Its extremely simple and I understand every concept I think I'll need except how to dynamically generate pages and page links. The way I want to do it is by having a database table that stores the name of a page and the main content of the page. That's all. Then I'd just call a script to pull the main content of a page into whatever page I happen to call. No big deal, right? Wrong.
Here's the problem. If I were to do this then I'd have to create a file for every page I want to create that calls the script that pulls the content from the correct database row. So I could add all sorts of page names and contents into the table but I don't know how to call them without manually creating new files each time I want to link to a new page.
Ideally there'd be a script that creates links to pages based on the page name row of the DB table as the pages are created. But how do you get those links with the ?=pageName at the end? If I just knew how that worked then I could figure the rest out.
UPDATE
The second answer really confirmed everything I thought I had to do but there is one catch. My plan now is to split up all the code into a series of functions and either include or require them in different templates that will be used to format the way pages are displayed. I need one look for the home page and one other design for the rest of the pages. I'm thinking that I'll have a function that says if ID is 0 then call this page template.php else call this other template file.php. But how do I pass the required variables to these new files? Do I just include the index.PHP page in them?
Bill your actually on the right track. Almost all web software today does extensive URL processing. Traditionally you would have php pages on your web root and then utilize the query string in the URL to refine the page's output. You have already arrived at why this might not be desired. So the popular alternative is the Front Controller design pattern. Basically we funnel every request to your index.php page and then route the request to internal pages or apps outside the web root. This can get complicated fast and everybody seems to implement this pattern in unique ways.
We can utilize this pattern without the routing by simply putting our app in the index page. The script below shows an example of what your trying to do in the simplest of ways. We basically have one page with our script. We can request the virtual pages by changing the id query string in our url. For example www.demo.net/?id=0 can be utilized as an index to your site. This should be the same as www.demo.net without the 'id' query. Just keep solving those problems one by one even if you don't know what the problem is. Once you start looking at other peoples code, then you can start seeing how other people solved the same problems you have.
The solution below will get you started, but then what do you do when you want an admin page? How do you authenticate the user? Do you duplicate alot of the code for yet another page? If your serious about your CMS then your going to want to implement some kind of framework underneath it. A framework to process the url, route to your application, load configuration files, and probably manage your database connection. Yea it gets complicated, but not if you solve each problem one at a time. Utilize classes or functions to share code to start. At the very least include a common "bootstrap" file at the top of your page to initialize common functionality such as a database connection. Read Stack Overflow just to keep up with whats going on. You can learn alot of terminology and probably find some answers to questions you didn't even know you wanted to ask.
Below assume we have a table with the following fields:
page_id
page_name
page_title
page_body
<?php
//<--------Move outside of web root-------------->
define('DB_HOST', 'localhost');
define('DB_USER', 'cms');
define('DB_PASS', 'changeme');
define('DB_DB', 'cms');
define('DB_TABLE', 'cms_pages');
//<---------------------------------------------->
//Display errors for development testing
ini_set('display_errors','On');
//Get the requested page id
if(isset($_GET['id']))
{
$id = $_GET['id'];
}
else
{
//Make page id '0' an index page to catch all
$id = 0;
}
//Establish a connection to MySQL
$conn = mysql_connect(DB_HOST,DB_USER,DB_PASS) or die(mysql_error());
//Select the database we will be querying
mysql_select_db(DB_DB, $conn) or die(mysql_error());
//Lets just grab the whole table
$sql = "SELECT * FROM ".DB_TABLE;
$resultset = mysql_query($sql, $conn) or die(mysql_error());
//The Select Query succeeded, but returned 0 result.
if (mysql_num_rows($resultset)==0)
{
echo "<pre>Add some Pages to my CMS</pre>";
exit;
}
//This is our target array we need to fill with arrays of pages
$result = array();
//Convert result into an array of associative arrays
while($row = mysql_fetch_assoc($resultset))
{
$result[] = $row;
}
//We now have all the information needed to build our app
//Page name - Short name for buttons, etc.
$name = "";
//Page title - The page content title
$title = "";
//Page body - The content you have stored in a table
$body = "";
//Page navigation - Array of formatted links
$nav = array();
//Process all pages in one pass
foreach($result as $row)
{
//Logic to match the requested page id
if($row['page_id'] == $id)
{
//Requested Page
$name = $row['page_name'];
$title = $row['page_title'];
$body = $row['page_body'];
$page = "<b>$name</b>";
}
else
{
//Not the requested page
$page = $row['page_name'];
}
//Build the navigation array preformatted with list items
$url = "./?id=" . $row['page_id'];
$nav[] = "<li>$page</li>";
}
?>
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title>SimpleCMS | <?php echo $title; ?></title>
</head>
<body>
<div>
<div id="navigation" style="float:left;">
<ul>
<?php
foreach($nav as $item)
{
echo $item;
}
?>
</ul>
</div>
<div id="content"><?php echo $body;?></div>
</div>
</body>
</html>
I think you need to read about $_GET.
I also recommend a decent PHP book. Forget online tutorials; they are (for the most part) utterly useless.