I am making a module and trying to figure out how to get the Search engine friendly url to articles from this module
this is the helper class today
public function getItems($amount)
{
$db = &JFactory::getDBO();
$query = 'SELECT * FROM `#__content`, `#__content_frontpage` WHERE `#__content_frontpage`.content_id = `#__content`.id AND `#__content`.state = 1 ORDER BY `#__content`.publish_up DESC LIMIT ' . $amount . '';
$db->setQuery($query);
$items = ($items = $db->loadObjectList())?$items:array();
return $items;
} //end getItems
And this is the default.php to display stuff
<ul class="frontpage_news">
<?php foreach ($items as $item) { ?>
<li>
<div class="frontpage_date"><?php echo JText::sprintf('DATE_FRONTNEWS', $item->publish_up); ?></div>
<div id="ffTitle" class="frontpage_title"><?php echo JText::sprintf('TITLE_FRONTNEWS', $item->title); ?></div>
<div id="ffRead" class="frontpage_readmore"><?php echo JText::sprintf('READ_MORE_FRONTNEWS'); ?></div>
</li>
<?php } ?>
</ul>
So how do I get the correct link to each article displayed in SEF format?
Thanks for any help!
For Joomla 1.5:
echo JRoute::_(ContentHelperRoute::getArticleRoute($article_id_and_alias, $category_id_and_alias, $section_id));
For Joomla 1.6/1.7:
echo JRoute::_(ContentHelperRoute::getArticleRoute($article_id_and_alias, $category_id));
Related
Here Is my question: What I am wanting To do is Take Results from a mysql table and turn them into a menu and a drop down menu
HERE IS A QUICK EXAMPLE:
if you see in my mysql table i have page_name and parent, So the example is:
page_name and if i have row 1 the page_name is 'Home' now it's parent is 'none' right but on id number 39 the page_name is 'Contact Us' and the Parent Is 'Far Far Away 123' so if the parent is equal to 'none' then it will show at the top of the menu not the drop down if it has a parent it will show under that parent like this:
Home | the ben page | The Brock Page | Far Far Away 123 | dsfk
Contact Us
You see Contact Us is under Far Far Away Because the parent Is Far Far Away 123
here is my table:
Here is my code That I am trying but it is not working for some reason:
<ul>
<?php
$sql = "SELECT * FROM pages ORDER by item_order";
$result = mysqli_query($db, $sql);
confirm_query($result);
while ($links = mysqli_fetch_assoc($result)) {
if($links['parent'] !== "none") {
?>
<li id = "<?php echo $links['id']; ?>"><a href="
<?php
echo "page.php?id=" . $links['id'] . "\" title=\"" . $links['page_title'] . "\"";
?>>
<?php
echo $links['page_name'];
?>
</a>
<?php
if($links['parent'] !== "none") {
$child = "";
$sql = "SELECT * FROM pages";
$result = mysqli_query($db, $sql);
while($row = mysqli_fetch_assoc($result)) {
if($row['parent'] !== "none") {
$child = $row['page_name'];
}
}
echo "<ul id=\"sub_menu\" class=\"sub_navagation" . $links['id'] . "\">";
echo "<li>";
echo $child;
echo "<li>";
echo "</ul>";
}
?>
</li>
<?php
}
}
?>
</ul>
CSS:
#sub_menu {
display: none;
}
#sub_menu:hover {
display: block;
}
Ok if as you can see i have the parent row in the MYSQL table and on id number 39 i want the 'Far Far Away123' to be the parent of Contact Us and i want to show it when i hover over 'Far Far Away123'
My suggestion is to build out an array of all the results. Then run through that array (instead of multiple database queries).
I added a function build_dropdown() that will take the page name and run through the array of pages to see if there are any items with a parent matching. If so, we make an array of those items and run through them to build the dropdown menu. If not, it does nothing and moves on to the next menu item.
<?php
function build_dropdown ($parent, $pages){
foreach($pages as $page){
if($page['parent'] == $parent){
$items = $page;
} // END if
} // END foreach
if(is_array($items)){ // If a sub
echo '<ul id="sub_menu" class="sub_navagation'. $item['id'] .'">';
foreach($items as $item){
echo '<li>'.$item['name'].'<li>';
} // END foreach
echo '</ul>';
} // END if
}
$sql = "SELECT * FROM pages ORDER by item_order";
$result = mysqli_query($db, $sql);
confirm_query($result);
while ($row = mysqli_fetch_assoc($result)) {
$pages[] = $row; // Add each row to $pages array to use later
}
foreach($pages as $key => $page){
if($page['parent'] == 'none'){ ?>
<li id = "<?php echo $page['id']; ?>">
<a href="page.php?id=<?php echo $page['id']; ?>" title="<?php echo $page['page_title']; ?>">
<?php echo $page['page_name']; ?>
</a>
<?php
build_dropdown($page['page_name'], $pages); // If there are child items then build them out
?>
</li>
<?php
} // END if
} // END foreach
?>
I suggest you will need to JOIN your table to basically query it again to get the parent value, and add that to your markup.
SELECT *
FROM Pages
LEFT JOIN Pages p2 on page_name = p2.parent
(note: the syntax above may not be right, but I wanted to give you an idea of where I would start).
I hope somebody can help me, because i got an menu that is auto generated via my MySQL db.
Because i got the menu to work inside the website and with that i mean it works with "test.dk/about" but the a href is empty when it's going out of the website like "http://google.com"...
btw it's just a very simple UL LI menu no dropdown or something.
Here is my script
static function build_menu()
{
$result = mysql_query("SELECT * FROM menu");
$menu = array();
while ($row = mysql_fetch_assoc($result)) {
if ($row["is_external"]) {
$url = $row["url"];
} else if (empty($row["is_external"])) {
$url = get_page_url($row["page_id"]);
}
$menu[] = array("name" => $row["name"], "page_id" => $row["page_id"], "is_external" => $row["url"], "url" => $url);
}
return $menu;
}
static function get_page_url($page_id)
{
$result = mysql_query("SELECT view_id FROM page WHERE id = '$page_id'");
$result = mysql_fetch_assoc($result);
$view_id = $result["view_id"];
$result = mysql_query("SELECT listen_path FROM view WHERE id = '$view_id'");
$result = mysql_fetch_assoc($result);
$listen_path = $result["listen_path"];
return $listen_path;
}
static function render()
{
$result = mysql_query("SELECT * FROM menu"); ?>
<div class="menu">
<ul><?php while ($item = mysql_fetch_assoc($result)) { ?>
<li><?php echo $item["name"] ?>
</li> <?php } ?></ul></div><?php
}
How can i fix it, so it works both internal and external?
<div class="menu"> <ul> <li>Homepage</li> <li>About</li> <li>Develop</li> <li>Support</li>
This should be <li>Support</li>; </ul> </div>
You only check for an external link in the function build_menu(), but this function isn't called anywhere from your render() function.
The render() function only calls get_page_url() which doesn't distinguish between internal and external links.
Href parameter of external URL must start with protocol declaration, so with "http://" in your case.
So change your code in condition inside the function "build_menu", if the URL is external, add "http://" to it, something like this:
$url = 'http://'.$row["url"];
I got it work after a while!
I simply just created an If else statement in the render function
static function render(){
$menu_items = self::get();
?><div class="menu"><ul><?php while ($item = mysql_fetch_assoc($menu_items)) { ?>
<li><a href="<?php
if(empty($item["is_external"]))
{
echo self::get_page_url($item["page_id"]);
}
else if($item["is_external"] = 1)
{
echo $item["url"];
}
?>"><?php echo $item["name"] ?></a>
</li> <?php } ?></ul></div><?php
}
I am new w/ OPP and big pardon if my question maybe too simple :)
Table category, navigation, etc contains multiple rows (category : samsung, apple, etc; and navigation : about us, terms, etc) and both stand as Menu in all pages (home, product,etc)
My old php code and work good is below
<div id="categories">
<ul>
<?
$mydbcategories = new myDBC();
$resultcategories = $mydbcategories->runQuery("SELECT * FROM `category`");
while ($rowcategories = $mydbcategories->runFetchArray($resultcategories)) {
echo '<li>'.$rowcategories[title].'</li>';
}
?>
</ul>
</div>
<div id="navigation">
<ul>
<?
$mydbnavigation = new myDBC();
$resultnavigation = $mydbnavigation->runQuery("SELECT * FROM `navigation`");
while ($rownavigation = $mydbnavigation->runFetchArray($resultnavigation)) { echo '<li>'.$rownavigation [title].'</li>';
}
?>
</ul>
</div>
I would like to implement OOP PHP and create class then store in classes.php
<?
class Menu{
var $title;
var $url;
function setMenu($db){
$mydbMenu= new myDBC();
$resultmenu = $mydbMenu->runQuery("SELECT * FROM `$db`");
$resultmenurows = mysqli_num_rows($resultmenu);
while ($rowmenu = $mydbMenu->runFetchArray($resultmenu)){
$this->title = $rowmenu[title];
$this->url = $rowmenu[url];
}
}
function getTitle() { return $this->title;}
function getUrl() { return $this->url;}
}
?>
Then i'm edit my old code with new one below;
<div id="categories">
<ul>
<?
$catmenu = new Menu();
while ($catmenu ->setMenu('category')) {
echo '<li>'.$catmenu->getTitle().'</li>';
}
?>
</ul>
</div>
<div id="navigation">
<ul>
<?
$navmenu = new Menu();
while ($navmenu ->setMenu('category')) {
echo '<li>'.$navmenu ->getTitle().'</li>';
}
?>
</ul>
</div>
I tested and error maybe because there are multiple rows (from table) in the setMenu func.
How can i return this multiple rows ? should i use array ?
Please help me to solve this and any reply really very appreciate
You are coding PHP4 OOP style, this is very outdated. Don't use var, use public, protected, private.
$this->title = $rowmenu[title] in here, title is used as a constant (no quotes), proper: $this->title = $rowmenu['title'], same with $rowcategories[title]
"SELECT * FROM $db" is this correct? Or do you mean SELECT * FROM menu WHERE xxx='" . $db . "', do you catch errors if the lookup fails?
You should also look at PHP design patterns and code style to improve!
Try following PHP code
<?
class Menu {
var $title;
var $url;
function setMenu($db) {
$mydbMenu = new myDBC();
$resultmenu = $mydbMenu->runQuery("SELECT * FROM `$db`");
$resultmenurows = mysqli_num_rows($resultmenu);
$this->title = array();
$this->url = array();
while ($rowmenu = $mydbMenu->runFetchArray($resultmenu)) {
$this->title[] = $rowmenu['title'];
$this->url[] = $rowmenu['url'];
}
}
function getTitle($ind) {
return $this->title[$ind];
}
function getUrl($ind) {
return $this->url[$ind];
}
}
?>
And HTML
<div id="categories">
<ul>
<?
$catmenu = new Menu();
$catmenu->setMenu('category');
$i = 0;
while ($catmenu->getTitle($i)) {
echo '<li>' . $catmenu->getTitle($i) . '</li>';
$i++;
}
?>
</ul>
</div>
<div id="navigation">
<ul>
<?
$navmenu = new Menu();
$navmenu->setMenu('navigation');
while ($navmenu->getTitle($i)) {
echo '<li>' . $navmenu->getTitle($i) . '</li>';
$i++;
}
?>
</ul>
</div>
I have a problem that concerns blog posts and displaying the tag words from another table.
I seem to be able to pull the info out of the tables fine, however when I try to display the posts and the tags, I get one tag per post. In other words if I have 7 tags for a post, I get 7 iteration's of that post each with one tag instead of 1 post with 7 tags.
My Controller ( do have a question about the $this->db->get(posts, tags) is that correct
$this->db->order_by('posts.id', 'DESC');
$where = "publish";
$this->db->where('status', $where);
$this->db->join('tags', 'tags.post_id = posts.id');
$this->db->limit('7');
$query = $this->db->get('posts', 'tags');
if($query->result())
$data = array();
{
$data['blog'] = $query->result();
}
$data['title'] = 'LemonRose';
$data['content'] = 'home/home_content';
$this->load->view('template1', $data);
The view.
$limit = 5; // how many posts should we show in full?
$i = 1; // count
foreach ($blog as $row):
$permalink = "http://".$_SERVER['HTTP_HOST'].$_SERVER['PHP_SELF'].$_SERVER['QUERY_STRING'];
$url = CompressURL ("$permalink");
$description = $row->title . $row->post;
$twittermsg = substr($description, 0, 110);
$twittermsg .= "...more " . $url;
if ($i < $limit) // we are under our limit
{ ?>
<div class="titlebox">
<div class="title"><? echo ucwords($row->title); ?></div>
<span><? echo $row->date, nbs(10), $row->author; ?></span>
</div>
<div class="clear"></div>
<? $str = str_word_count($row->post, 0);
if ($str >= 500) {
$row->post = html_entity_decode($row->post);
$row->post = $this->typography->auto_typography($row->post); // display?>
<div class="split"> <? echo $row->post = word_limiter($row->post, 480); ?>
<div class="tags"><? echo $row->tag; ?></div>*** These 3 lines seem to be where I am confused and getting the wrong display
<p><h3>More <?php echo anchor("main/blog_view/$row->id", ucwords($row->title)); ?> </h3></p>
<p>Trackback URL: <? echo base_url() . "trackbacks/track/$row->id"; ?></p>
<!-- tweet me -->
<?echo anchor("http://twitter.com/home?status=$twittermsg", 'Tweet'); ?>
This is my first attempt with join and I have very little experience getting the display with implode, if that is the right way to go.
Thank you in advance.
Try
<div class="tags"><? echo implode(', ', $row->tag); ?></div>
and remove the 2 rows before this one.
I am relatively new to The Zend framework having only been working with it probably two months at the most. I am now trying to get a list of results from a query I run, send it through the zend paginator class and display 4 results per page, however it does not seem to be recognising when it has got 4 results, I believe there is an error in my code but I can not for the life of me see it, I was hoping someone here will be able to pick up on it and help me and out and probably tell me it is something stupid that I have missed. Thanks here is the code I have written.
This the query in my model
public function getAllNews()
{
$db = Zend_Registry::get('db');
$sql = "SELECT * FROM $this->_name WHERE flag = 1 ORDER BY created_at";
$query = $db->query($sql);
while ($row = $query->fetchAll()) {
$result[] = $row;
}
return $result;
}
This is code in my controller
function preDispatch()
{
$this->_helper->layout->setLayout('latestnews');
Zend_Loader::loadClass('newsArticles');
$allNews = new newsArticles();
$this->view->allNews = $allNews->getAllnews();
//die (var_dump($this->view->allNews));
$data = $this->view->allNews;
// Instantiate the Zend Paginator and give it the Zend_Db_Select instance Argument ($selection)
$paginator = Zend_Paginator::factory($data);
// Set parameters for paginator
$paginator->setCurrentPageNumber($this->_getParam("page")); // Note: For this to work of course, your URL must be something like this: http://localhost:8888/index/index/page/1 <- meaning we are currently on page one, and pass that value into the "setCurrentPageNumber"
$paginator->setItemCountPerPage(1);
$paginator->setPageRange(4);
// Make paginator available in your views
$this->view->paginator = $paginator;
//die(var_dump($data));
}
Below is the view that builds the paginator,
<?php if ($this->pageCount): ?>
<div class="paginationControl">
<!-- Previous page link -->
<?php if (isset($this->previous)): ?>
< Previous |
<?php else: ?>
<span class="disabled">< Previous</span> |
<?php endif; ?>
<!-- Numbered page links -->
<?php foreach ($this->pagesInRange as $page): ?>
<?php if ($page != $this->current): ?>
<?= $page; ?> |
<?php else: ?>
<?= $page; ?> |
<?php endif; ?>
<?php endforeach; ?>
<!-- Next page link -->
<?php if (isset($this->next)): ?>
Next >
<?php else: ?>
<span class="disabled">Next ></span>
<?php endif; ?>
</div>
<?php endif; ?>
And finally the code the makes up my view
<div id="rightColumn">
<h3>Latest News</h3>
<?php if (count($this->paginator)): ?>
<ul>
<?php foreach ($this->paginator as $item): ?>
<?php
foreach ($item as $k => $v)
{
echo "<li>" . $v['title'] . "</li>";
}
?>
<?php endforeach; ?>
</ul>
<?php endif; ?>
<?= $this->paginationControl($this->paginator, 'Sliding', '/latestnews/partial_pagination_control.phtml'); ?>
</div>
I will be greatful for any help you can give me.
Thanks
Sico
$paginator should be set to 4 as you want to show only 4 records at a time :
$paginator->setItemCountPerPage(4);
So I think the root of your problem is in your model class with the getAllNews function. The Zend_Db fetchAll function retrieves an associative array of all the records matching your query. You are returning an array of one element that is an array of all the rows retrieved by your select statement.
Try this:
public function getAllNews()
{
$db = Zend_Registry::get('db');
$sql = "SELECT * FROM $this->_name WHERE flag = 1 ORDER BY created_at";
return $db->fetchAll($sql);
}
You sould not pass values to paginator if you want it to limit your query.
There is a code will work for you
$usersTable = $this->getTable();
$registerSelect = $usersTable->select()->where('status = ?', 'OK')->order('lastLogin DESC');
$paginator = new Zend_Paginator(new Zend_Paginator_Adapter_DbTableSelect($registerSelect));
$paginator->setCurrentPageNumber($page);
$paginator->setItemCountPerPage($limit);