I would like to display one image at a time on a webpage. I used a query to get a set of images from the database into an array. I have tried along these lines:
$my=& JFactory::getUser();
$db=& JFactory::getDBO();
$a=& $my->id;
$query="SELECT id FROM jos_phocagallery_categories WHERE accessuserid='$a'";
$db->setQuery($query);
$r=$db->loadResult();
$queryc="SELECT filename FROM jos_phocagallery WHERE catid='$r'";
$db->setQuery($queryc);
$results=$db->loadResultArray();
$image=current($results);
echo 'imgag srx="images/phocagallery/'.$image.'" />';
I changed echo line above because the line was not allowed
After this code I've tried to add an input button that calls "next($results)" and then replace $image, but it seems the whole query is beeing implemented from the beginning again. So I can't get the pointer of the array to step forward. Not even in an echo.
What Am I doing wrong?
Regards / Viktor
If you are reloading the page to get to the next image, then you are going to need to pass the ID or array key to the next page so it knows where you left off. The other option would be to step through the array without reloading the page, instead using some Javascript to load the next image and replace the current one when the button is clicked.
EDIT
Ok, with that info you will definitely need to use some Javascript to make it work. Even though Joomla comes with Mootools, I prefer jQuery partially due to the availability of plugins to do all sorts of stuff. I didn't look too hard, but this carousel image plugin could be very easily modified to do what you like by simply changing it to display one image at a time instead of three.
http://www.gmarwaha.com/jquery/jcarousellite/index.php
With a little Googling you might even be able to find an exact match.
Related
I am a beginner programmer/coder who is currently trying to get to grips with HTML and PHP. I currently have a locally hosted searchable database that (when used) brings up a list of the first twenty entries that correspond to your search terms, with buttons to send you to the next page, last page, etc. (you know... pages...) The search outputs information into the URL ([url]?q=Alphonse&f=Elric etc). I have two problems with this at the moment.
Problem A:
My URL contains information that is unused. If I don't put anything into the search term it simply comes out with "q=&x=&f=..." etc. This makes the URL absurdly long even on the most simple searches.
Can this be cleaned up through just php?
Would this method be different if I end up hosting this online?
Problem B:
The way my paging functions is to send the user to the following link '.$_SERVER['PHP_SELF'].'?'.$_SERVER['QUERY_STRING'].'&pn='.$nextPage.'. This outputs the current link but with "pn=1" at the end (Or whatever relevant page they click).
This method itself makes the URL quite messy. If they click through multiple pages, and perhaps go back and forth, the link ends up having "pn=1&pn=2&pn=3&pn=1...." etc at the end. I assume that this will be answered by the first query, but it is slightly different in that this is information that is actually present.
How do I remove this superfluous information, and just keep the (final) relevant one?
I am thinking that I can use parse_str to turn the URL into an array, then delete each entry of the array that are empty, then create a new string out of that array and make that the link the search/next page button goes to.
Does that sound like it would work? If so, how do I delete those specific array entries, and how would that array then be stored? Would the array lose those entries and calling a deleted entry "$array['1']" for example result in an error, or does deleting entries in an array move everything up one to fill the gap?
Apologies if I'm asking too many different things in one post here!
I need help in identifying how I should set up this non public website.
Basically there are around 2000 images referenced in a database.
Each user will be able to check or uncheck each image with the use of a checkbox.
Some users might have different images checked or unchecked.
I want the checkbox to process an AJAX request to a user specific XML (PHP generated from the db) which contains a boolean variable for each image entry.
The PHP then references the XML and highlights checkedboxes and disables the uncheckedboxes.
Again each user will have different references for each image.
Im not sure if the above is the correct method to use.
I want the page to dynamically load the first 20 images and if the checkbox is changed, instantly updated and refreshed without a page reload. Then I'll paginate to the next 20.
If I'm on the right track I'll attempt a demo and post an update.
Thanks,
Depends how you want to do it. You'll need to brush up on your javascript as well to track the on change event. If you're bring in the images with ajax to begin with I would create an object for each and attach the event to element. Then on click you could post the image_id or what have you to the server. Alternatively you could use an attribute on the checkbox/image like data-imageid="1"
JSON or XML, both will work, it just depends how you want to design it. Though from my experience I would use JSON instead.
The JSON array you would return might look something like {1:true,2:false,3:true} etc, so foreach key value pair you would either check it be it true or not on false.
Consider editing your tags for javascript or jquery to see if you get some better answers. Or I could elaborate further if I'm on the right track.
I didn't know how to ask professional so if topic is wrong, please correct it.
My problem is quite complicated. I was training on symfony webpage how to generate form and then send this data to another page.
But now i would like do something different. I would like create page xyzSuccess.php.
Then generate 29 links on this page. Each links would have its own number.
Each link would redirect to the same page. For example numberSuccess.php. After that, this site would give me data of the number which i clicked.
I will show the example :)
Page xyzSuccess.php have links, the third one is http://localhost:8080/web/number?nr=3
I click on it and i go to numberSuccess.php, the page give me number 3
After the page got number, also take from any datafile information under that number 3 ( i don't know, maybe any file wchich can store symfony / array somewhere?.)
I dont know how to do this, i don't want use any database eq. *sql
So far i created only page xyz.php which use php for loop. I see that action.class.php would work with it if i would use submit button, but i really dont have any idea how make this working with links. And... that problem with storing data :( maybe create file in lib/form ?
corrected the link to page (deleted success which used only in templates), guy under is right :)
First of all, you're talking about URL's like /web/numberSuccess... That probably means you've misconfigured the root. Also see this question, asked today.
Then it's common to give your actions names like number, and then in you're actions file you have the executeNumber action, which then renders the numberSuccess template. So you shouldn't include the Success in the action name/URL.
Then, your question:
What I would do, if you don't want to use a database:
Create a file links.txt in your /data directory, and on every line create an url, followed by a space and then the title of your link.
In your list action open the this file using file_get_contents(), explode it on the newline-character, and assign this array to the view.
In your view loop through the array, use the array index as your number, and explode each item on the first space, so you can display the title.
In the view action your open and explode the links file again, and now you can pointer directly to the index, explode again and you have the URL.
Hey guys I am really messing up with this.
This is I'm doing: If there are 100 users to fetch from Database, I'm using pagination and showing 10 users at a time. when user will click on next page, he will get next 10 users through ajax(called ajax on click) and so on. I'm showing 10 page-links right now with first, next last and previous links.
This is how flow will go: On a.php created links and called ajax function with every link, passing url(b.php) & target(where I will get result), with url also passing clicked pageno., this pageno. will go to b.php and next 10 users will be shown with the help of ajax.
This is the problem: Currently I'm showing 1-10 links with first and last links, unable to show next and previous links because to redirect to next or previous, I am not getting the current page number on a.php i.e i'm passing to b.php. also links are created in foreach loop.
I am trying hard to get this done, but no success yet.
waiting for valuable reply.
Ok, I think I understand...
I think what you need to do is store the current page number as a JavaScript variable. Use the function that is doing the Ajax to update this variable and also update the next/previous links.
Hope this helps.
Update:
On second thoughts, here's a better idea. Having your ajax call directly in your link means you have to put together a url and update it, which is awkward and annoying. This isn't ideal. Instead it is better to store the variables and have some set functions work with them in a set way. So, you could have something like:
<a href="javascript:void(0)" onclick="gotoNextPage()">
Rather than passing them the number of the page to go to, you can store the current page number in a javascript variable and do something like this:
gotoNextPage(){
// lets assume the current page number is stored in a
// variable defined outside of this function called curr_page
curr_page++; // increment the current
call_ajax('user_info.php?pageno='+curr_page);
}
This way the code for the link doesn't ever change.
You can still pass other variables (such as 'order', 'perpage' etc) to gotoNextPage() but you might find you don't need to.
A couple of things to bear in mind:
1. Using jQuery or something similar would probably make things easier for you.
2. Anyone without javascript enabled will not be able to use your site. Consider changing it to something like
<?php
echo '<a href="your_main_page.php?page_number=', ($curr_page + 1), '" onclick="gotoNextPage()">';
?>
This way it would work for via the ajax method but would also work (if not as smoothly and with a page reload) for people without javascript. It's a bit more work for you though...so hat's your choice!
Hope this makes sense. It's been a long day!
Further to my question yesterday (here), I am working on a webpage that has a section that shows 'live' order details.
The top half of my webpage has Spry Tabbed Panels. One of the panels contains an include call to a separate php page that I have created (getOpenOrders.php). This contains an SQL query to obtain all open orders and then puts the details into a table.
As a result, the table of open orders is shown in the Spry panel. What steps do I now need to take to have this refresh every 15 seconds?
Do you really want to call the database every 15 seconds for each user? isn't that an overload?
I'm not saying that your database will be overloaded, but, thats how you shouldn't do things!
Edited
you should show an image, or the link to that page in order to gt an appropriate answer, because it all depends in what are you doing in the table.
because I don't know, I will give you an answer on what probably is happening.
Because you said that you're new to the ajax world, let's make things simple, and not to complicate on the you should return a JSON object and use it to re populate your table. :)
So we will start with 2 buttons (Previous and Next) so the user can move the data that is showing (you probably don't want to give him/her 100 lines to see right?)
let's say that you have 2 pages, a showData.php and getTable.php, in the showData.php you will need to load jQuery (wonderful for this) and add a little code, but where the table is to be placed, just add a div tag with an id="myTable" because we will get the data from the getTable.php file.
getTable.php file has to output only the table html code with all the data in, without no html, body, etc... the idea is to add inside the div called myTable all the code generated by getTable.php
Let's imagine that getTable.php gets a page variable in the queryString, that will tell what page you should show (to use LIMIT in your MySQL or PostgreSQL database)
You can use jQuery plugin called datatables witch is one of my choices, check his example and how small code you need to write! just using jQuery and Datatables plugin.
The first description follows the jQuery.Load() to load the getTable.php and add as a child of the div and wold do this for the previous and next buttons, passing a querystring with the page that the user requested. It's to simple and you can see the website for that, if you prefer to use the DataTables plugin, then just follow their examples :)
if you, after all this need help, drop me a line.
<META HTTP-EQUIV=Refresh CONTENT="15; URL=<?php print $PHP_SELF ?>">
This should be in between the head tags.
-or-
header('Refresh: 15');
This should be before the head tag and directly after the html tag.
As said by balexandre, a different method should be used. One that does not require a database hit every 15 seconds for every single user that is connected to the site. But, there is your answer anyways.
Although, balexandre makes a very good point, if you do decide that you need a refresh, you could simply do something like this in your JavaScript:
window.onload = function( )
{
setTimeout( 'window.location.refresh( )', 1500 );
}
(I've not tested the above code, so syntax may need to be tweaked a little, but you get the idea)