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  • Pagination links broken - php/jquery

    - by ClarkSKent
    Hey, I'm still trying to get my pagination links to load properly dynamically. But I can't seem to find a solution to this one problem. vote down star Hi everyone, I am still trying to figure out how to fix my pagination script to work properly. the problem I am having is when I click any of the pagination number links to go the next page, the new content does not load. literally nothing happens and when looking at the console in Firebug, nothing is sent or loaded. I have on the main page 3 links to filter the content and display it. When any of these links are clicked the results are loaded and displayed along with the associated pagination numbers for that specific content. I believe the problem is coming from the sql query in generate_pagination.php (seen below). When I hard code the sql category part it works, but is not dynamic at all. This is why I'm calling $ids=$_GET['ids']; and trying to put that into the category section but then the numbers don't display at all. If I echo out the $ids variable and click on a filter it does display the correct name/id, so I don't know why this doesn't work Here is the main page so you can see how I am including and starting the function(I'm new to php): <?php include_once('generate_pagination.php'); ?> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script> <script type="text/javascript" src="jquery_pagination.js"></script> <div id="loading" ></div> <div id="content" data-page="1"></div> <ul id="pagination"> <?php //Pagination Numbers for($i=1; $i<=$pages; $i++) { echo '<li class="page_numbers" id="page'.$i.'">'.$i.'</li>'; } ?> </ul> <br /> <br /> <a href="#" class="category" id="marketing">Marketing</a> <a href="#" class="category" id="automotive">Automotive</a> <a href="#" class="category" id="sports">Sports</a> Here is the generate pagination where the problem seems to occur: <?php $ids=$_GET['ids']; include_once('config.php'); $per_page = 3; //Calculating no of pages $sql = "SELECT COUNT(*) FROM explore WHERE category='$ids'"; $result = mysql_query($sql); $count = mysql_fetch_row($result); $pages = ceil($count[0]/$per_page); ?> I thought I might as well post the jquery script if someone wants to see: $(document).ready(function(){ //Display Loading Image function Display_Load() { $("#loading").fadeIn(900,0); $("#loading").html("<img src='bigLoader.gif' />"); } //Hide Loading Image function Hide_Load() { $("#loading").fadeOut('slow'); }; //Default Starting Page Results $("#pagination li:first").css({'color' : '#FF0084'}).css({'border' : 'none'}); Display_Load(); $("#content").load("pagination_data.php?page=1", Hide_Load()); // Editing below. // Sort content Marketing $("a.category").click(function() { Display_Load(); var this_id = $(this).attr('id'); $.get("pagination.php", { category: this.id }, function(data){ //Load your results into the page var pageNum = $('#content').attr('data-page'); $("#pagination").load('generate_pagination.php?category=' + pageNum +'&ids='+ this_id ); $("#content").load("filter_marketing.php?page=" + pageNum +'&id='+ this_id, Hide_Load()); }); }); //Pagination Click $("#pagination li").click(function(){ Display_Load(); //CSS Styles $("#pagination li") .css({'border' : 'solid #dddddd 1px'}) .css({'color' : '#0063DC'}); $(this) .css({'color' : '#FF0084'}) .css({'border' : 'none'}); //Loading Data var pageNum = $(this).attr("id").replace("page",""); $("#content").load("pagination_data.php?page=" + pageNum, function(){ $(this).attr('data-page', pageNum); Hide_Load(); }); }); }); If any could assist me on solving this problem that would be great, thanks.

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  • Pagination number elements do not work - jquery

    - by ClarkSKent
    Hello, I am trying to get my pagination links to work. It seems when I click any of the pagination number links to go the next page, the new content does not load. literally nothing happens and when looking at the console in Firebug, nothing is sent or loaded. I have on the main page 3 links to filter the content and display it. When any of these links are clicked the results are loaded and displayed along with the associated pagination numbers for that specific content. Here is the main page so you can see what the structure looks like for the jquery: <?php include_once('generate_pagination.php'); ?> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script> <script type="text/javascript" src="jquery_pagination.js"></script> <div id="loading" ></div> <div id="content" data-page="1"></div> <ul id="pagination"> <?php generate_pagination($sql) ?> </ul> <br /> <br /> <a href="#" class="category" id="marketing">Marketing</a> <a href="#" class="category" id="automotive">Automotive</a> <a href="#" class="category" id="sports">Sports</a> The jquery below is pretty simple and I don't think I need to explain what it does. I believe the problem may be with the $("#pagination li").click(function(){ since the li elements are the numbers that do not work when clicked. Even when I try to fadeOut or hide the content on click nothing happens. I'm pretty new to the jquery structure so I do not fully understand where the real problem is occurring, this is just from my observation. The jquery file looks like this: $(document).ready(function(){ //Display Loading Image function Display_Load() { $("#loading").fadeIn(900,0); $("#loading").html("<img src='bigLoader.gif' />"); } //Hide Loading Image function Hide_Load() { $("#loading").fadeOut('slow'); }; //Default Starting Page Results $("#pagination li:first").css({'color' : '#FF0084'}).css({'border' : 'none'}); Display_Load(); $("#content").load("pagination_data.php?page=1", Hide_Load()); //Pagination Click $("#pagination li").click(function(){ Display_Load(); //CSS Styles $("#pagination li") .css({'border' : 'solid #dddddd 1px'}) .css({'color' : '#0063DC'}); $(this) .css({'color' : '#FF0084'}) .css({'border' : 'none'}); //Loading Data var pageNum = this.id; $("#content").load("pagination_data.php?page=" + pageNum, function(){ $(this).attr('data-page', pageNum); Hide_Load(); }); }); // Editing below. // Sort content Marketing $("a.category").click(function() { Display_Load(); var this_id = $(this).attr('id'); $.get("pagination.php", { category: this.id }, function(data){ //Load your results into the page var pageNum = $('#content').attr('data-page'); $("#pagination").load('generate_pagination.php?category=' + pageNum +'&ids='+ this_id ); $("#content").load("filter_marketing.php?page=" + pageNum +'&id='+ this_id, Hide_Load()); }); }); }); If anyone could help me on this that would be great, Thanks. EDIT: Here are the innards of <ul id="pagination">: <?php function generate_pagination($sql) { include_once('config.php'); $per_page = 3; //Calculating no of pages $result = mysql_query($sql); $count = mysql_fetch_row($result); $pages = ceil($count[0]/$per_page); //Pagination Numbers for($i=1; $i<=$pages; $i++) { echo '<li class="page_numbers" id="'.$i.'">'.$i.'</li>'; } } $ids=$_GET['ids']; generate_pagination("SELECT COUNT(*) FROM explore WHERE category='$ids'"); ?>

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  • How to make sql query dynamic?

    - by ClarkSKent
    Hello, I am using this pagination class and was looking for a way to make the sql query more dynamic instead of having it hardcoded. I have a 3 <li> elements that I want to be filter buttons, meaning when a user clicks on one of these elements I want It to send the id so I can use it in a sql query. So for the $sql = "select * from explore where category='marketing'"; (as seen below). When the user clicks on the 'automotive' button it will change the category above to automotive. Any help on this would be highly appreciated, Thanks. This is what my main page looks like: <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script> <script type="text/javascript" src="jquery_page.js"></script> <?php //Include the PS_Pagination class include('ps_pagination.php'); //Connect to mysql db $conn = mysql_connect('localhost', 'root', 'root'); mysql_select_db('ajax_demo',$conn); $sql = "select * from explore where category='marketing'"; //Create a PS_Pagination object $pager = new PS_Pagination($conn, $sql, 3, 11, 'param1=valu1&param2=value2'); //The paginate() function returns a mysql //result set for the current page $rs = $pager->paginate(); //Loop through the result set while($row = mysql_fetch_assoc($rs)) { echo "<table width='800px'>"; echo "<tr>"; echo"<td>"; echo $row['id']; echo"</td>"; echo"<td>"; echo $row['site_description']; echo"</td>"; echo"<td>"; echo $row['site_price']; echo"</td>"; echo "</tr>"; echo "</table>"; } echo "<ul id='pagination'>"; echo "<li>"; //Display the navigation echo $pager->renderFullNav(); echo "</li>"; echo "</ul>"; echo "<ul id='filter'>"; echo "<li id='marketing'>"; echo "Marketing"; echo "</li>"; echo "<li id='automotive'>"; echo "Automotive"; echo "</li>"; echo "<li id='sports'>"; echo "Sports"; echo "</li>"; echo "</ul>"; ?>

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  • This pagination script is doodoo - I need a better one!

    - by ClarkSKent
    Hello, I was looking at the pagination script (posted below) and found it to be gros,s and not very good at all especially when trying to customize it. This is what the main page looks like: <?php include('config.php'); $per_page = 9; //Calculating no of pages $sql = "select * from messages"; $result = mysql_query($sql); $count = mysql_num_rows($result); $pages = ceil($count/$per_page) ?> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/ libs/jquery/1.3.0/jquery.min.js"></script> <script type="text/javascript" src="jquery_pagination.js"></script> <div id="loading" ></div> <div id="content" ></div> <ul id="pagination"> <?php //Pagination Numbers for($i=1; $i<=$pages; $i++) { echo '<li id="'.$i.'">'.$i.'</li>'; } ?> </ul> The top part of the code gets the results from the mysql db and than uses this information to display the numbers in the body of this page. I am trying to put something like this on a separate page like count_page.php and then just include it. I guess my question is, if there is a better way of doing the above with better structure. A better way to go through the db and count the results and display the appropriate numbers. The above seems messy. Thanks for any help or suggestions on this.

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  • How do I add a filter button to this pagination?

    - by ClarkSKent
    Hey, I want to add a button(link), that when clicked will filter the pagination results. I'm new to php (and programming in general) and would like to add a button like 'Automotive' and when clicked it updates the 2 mysql queries in my pagination script, seen here: As you can see, the category automotive is hardcoded in, I want it to be dynamic, so when a link is clicked it places whatever the id or class is in the category part of the query. 1: $record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='automotive'")); 2: $get = mysql_query("SELECT * FROM explore WHERE category='automotive' LIMIT $start, $per_page"); This is the entire current php pagination script that I am using: <?php //connecting to the database $error = "Could not connect to the database"; mysql_connect('localhost','root','root') or die($error); mysql_select_db('ajax_demo') or die($error); //max displayed per page $per_page = 2; //get start variable $start = $_GET['start']; //count records $record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='automotive'")); //count max pages $max_pages = $record_count / $per_page; //may come out as decimal if (!$start) $start = 0; //display data $get = mysql_query("SELECT * FROM explore WHERE category='automotive' LIMIT $start, $per_page"); while ($row = mysql_fetch_assoc($get)) { // get data $name = $row['id']; $age = $row['site_name']; echo $name." (".$age.")<br />"; } //setup prev and next variables $prev = $start - $per_page; $next = $start + $per_page; //show prev button if (!($start<=0)) echo "<a href='pagi_test.php?start=$prev'>Prev</a> "; //show page numbers //set variable for first page $i=1; for ($x=0;$x<$record_count;$x=$x+$per_page) { if ($start!=$x) echo " <a href='pagi_test.php?start=$x'>$i</a> "; else echo " <a href='pagi_test.php?start=$x'><b>$i</b></a> "; $i++; } //show next button if (!($start>=$record_count-$per_page)) echo " <a href='pagi_test.php?start=$next'>Next</a>"; ?>

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  • Converting PHP pagination to jQuery?

    - by ClarkSKent
    Hey, I have been trying to get this pagination class that I am using to be more ajaxy - meaning when I click on the page number like page [2] the data loads, but I want to load in the data without going to a different page (HTTP request in the background, with no page reloads). Being new to both php and jquery, I am a little unsure on how to achieve this result, especially while using a php class. This is what the main page looks like by the way: <?php $categoryId=$_GET['category']; echo $categoryId; ?> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script> <script type="text/javascript" src="jquery_page.js"></script> <?php //Include the PS_Pagination class include('ps_pagination.php'); //Connect to mysql db $conn = mysql_connect('localhost', 'root', 'root'); mysql_select_db('ajax_demo',$conn); $sql = "select * from explore where category='$categoryId'"; //Create a PS_Pagination object $pager = new PS_Pagination($conn, $sql, 3, 11, 'param1=value1&param2=value2'); //The paginate() function returns a mysql //result set for the current page $rs = $pager->paginate(); //Loop through the result set echo "<table width='800px'>"; while($row = mysql_fetch_assoc($rs)) { echo "<tr>"; echo"<td>"; echo $row['id']; echo"</td>"; echo"<td>"; echo $row['site_description']; echo"</td>"; echo"<td>"; echo $row['site_price']; echo"</td>"; echo "</tr>"; } echo "</table>"; echo "<ul id='pagination'>"; echo "<li>"; //Display the navigation echo $pager->renderFullNav(); echo "</li>"; echo "</ul>"; ?> <div id="loading" ></div> <div id="content" ></div> Would I need to do something with this part of the class?, as seen above: $pager = new PS_Pagination($conn, $sql, 3, 11, 'param1=value1&param2=value2'); Or this?: echo $pager->renderFullNav(); I don't no much about jquery,but i guess I would start it like: $("#pagination li").click(function() { Then load something maybe... I don't no. Any help on this would be great. Thanks.

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  • how to add multiple filters to pagination?

    - by ClarkSKent
    Hey, I am trying to allow multiple filters to be selected for my pagination script. So in my pagination, when a user clicks the 'marketing' button(link) it queries the database just for the category that = marketing. The same goes for the other 2 filter buttons as seen in the script below. (automotive, sports). The problem is, I want to be able to select multiple filters like only marketing and auomotive or automotive and sports, for example if I click the marketing filter and then the automotive, it would display the categories that equal marketing, and automotive. I have no idea how to accomplish this, so I have come to the experts to help me out. This is the script I am working on (I am very new to PHP and programming in general): <h3>Filter results by:</h3> <a href='pagi_test.php?category=marketing'>marketing</a> <a href='pagi_test.php?category=automotive'>automotive</a> <a href='pagi_test.php?category=sports'>sports</a> <br /> <h3>Results:</h3> <?php //connecting to the database $error = "Could not connect to the database"; mysql_connect('localhost','root','root') or die($error); mysql_select_db('ajax_demo') or die($error); //max displayed per page $per_page = 3; //get start variable $start = $_GET['start']; $category = mysql_real_escape_string($_GET['category']); //count records $record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='$category'")); //count max pages $max_pages = $record_count / $per_page; //may come out as decimal if (!$start) $start = 0; //display data $get = mysql_query("SELECT * FROM explore WHERE category='$category' LIMIT $start, $per_page"); ?> <table width="800px"> <?php while ($row = mysql_fetch_assoc($get)) { // get data $id = $row['id']; $site_name = $row['site_name']; $site_description = $row['site_description']; ?> <tr> <td><?php echo $id; ?></td> <td><?php echo $site_name; ?></td> <td><?php echo $site_description; ?></td> </tr> <?php } //setup prev and next variables $prev = $start - $per_page; $next = $start + $per_page; //show prev button if (!($start<=0)) echo "<a href='pagi_test.php?category=$category&start=$prev'>Prev</a> "; //show page numbers //set variable for first page $i=1; for ($x=0;$x<$record_count;$x=$x+$per_page) { if ($start!=$x) echo " <a href='pagi_test.php?category=$category&start=$x'>$i</a> "; else echo " <a href='pagi_test.php?category=$category&start=$x'><b>$i</b></a> "; $i++; } //show next button if (!($start>=$record_count-$per_page)) echo " <a href='pagi_test.php?category=$category&start=$next'>Next</a>"; ?>

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  • Pagination links do not work properly - incorrect PHP function??

    - by ClarkSKent
    Hi everyone, I am still trying to figure out how to fix my pagination script to work properly. the problem I am having is when I click any of the pagination number links to go the next page, the new content does not load. literally nothing happens and when looking at the console in Firebug, nothing is sent or loaded. I have on the main page 3 links to filter the content and display it. When any of these links are clicked the results are loaded and displayed along with the associated pagination numbers for that specific content. I believe the problem is coming from the function(generate_pagination.php (seen below)). Here is the main page so you can how I am including and starting the function(I'm new to php): <?php include_once('generate_pagination.php'); ?> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script> <script type="text/javascript" src="jquery_pagination.js"></script> <div id="loading" ></div> <div id="content" data-page="1"></div> <ul id="pagination"> <?php generate_pagination($sql) ?> </ul> <br /> <br /> <a href="#" class="category" id="marketing">Marketing</a> <a href="#" class="category" id="automotive">Automotive</a> <a href="#" class="category" id="sports">Sports</a> This is as mentioned above, where i think the problem persists since I know nothing of the function formats and how to properly incorporate them: <?php function generate_pagination($sql) { include_once('config.php'); $per_page = 3; //Calculating no of pages $result = mysql_query($sql); $count = mysql_fetch_row($result); $pages = ceil($count[0]/$per_page); //Pagination Numbers for($i=1; $i<=$pages; $i++) { echo '<li class="page_numbers" id="'.$i.'">'.$i.'</li>'; } } $ids=$_GET['ids']; generate_pagination("SELECT COUNT(*) FROM explore WHERE category='$ids'"); ?> I thought I might as well through in the jquery if someone wants to see: $(document).ready(function(){ //Display Loading Image function Display_Load() { $("#loading").fadeIn(900,0); $("#loading").html("<img src='bigLoader.gif' />"); } //Hide Loading Image function Hide_Load() { $("#loading").fadeOut('slow'); }; //Default Starting Page Results $("#pagination li:first").css({'color' : '#FF0084'}).css({'border' : 'none'}); Display_Load(); $("#content").load("pagination_data.php?page=1", Hide_Load()); //Pagination Click $("#pagination li").click(function(){ Display_Load(); //CSS Styles $("#pagination li") .css({'border' : 'solid #dddddd 1px'}) .css({'color' : '#0063DC'}); $(this) .css({'color' : '#FF0084'}) .css({'border' : 'none'}); //Loading Data var pageNum = this.id; $("#content").load("pagination_data.php?page=" + pageNum, function(){ $(this).attr('data-page', pageNum); Hide_Load(); }); }); // Editing below. // Sort content Marketing $("a.category").click(function() { Display_Load(); var this_id = $(this).attr('id'); $.get("pagination.php", { category: this.id }, function(data){ //Load your results into the page var pageNum = $('#content').attr('data-page'); $("#pagination").load('generate_pagination.php?category=' + pageNum +'&ids='+ this_id ); $("#content").load("filter_marketing.php?page=" + pageNum +'&id='+ this_id, Hide_Load()); }); }); }); Any help would be appreciated on getting the function to work properly. Thank you.

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  • How to Load In Content with jQuery?

    - by ClarkSKent
    Hello, I am trying to add ajax functionality to my pagination so the content loads in the same page instead of the user having to navigate to another page when clicking the page links. I should mention that I am using this php pagination class. Being new to jquery, I am unsure of how to properly do this with the pagination class. This is what the main page looks like: <?php $categoryId=$_GET['category']; echo $categoryId; ?> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script> <script type="text/javascript" src="jquery_page.js"></script> <?php //Include the PS_Pagination class include('ps_pagination.php'); //Connect to mysql db $conn = mysql_connect('localhost', 'root', 'root'); mysql_select_db('ajax_demo',$conn); $sql = "select * from explore where category='$categoryId'"; //Create a PS_Pagination object $pager = new PS_Pagination($conn, $sql, 3, 11, 'param1=value1&param2=value2'); //The paginate() function returns a mysql //result set for the current page $rs = $pager->paginate(); //Loop through the result set echo "<table width='800px'>"; while($row = mysql_fetch_assoc($rs)) { echo "<tr>"; echo"<td>"; echo $row['id']; echo"</td>"; echo"<td>"; echo $row['site_description']; echo"</td>"; echo"<td>"; echo $row['site_price']; echo"</td>"; echo "</tr>"; } echo "</table>"; echo "<ul id='pagination'>"; echo "<li>"; //Display the navigation echo $pager->renderFullNav(); echo "</li>"; echo "</ul>"; ?> <div id="loading" ></div> <div id="content" ></div> <a href="#" class="category" id="marketing">Marketing</a> <a href="#" class="category" id="automotive">Automotive</a> <a href="#" class="category" id="sports">Sports</a> Any help on this would be great. Thanks.

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