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PHP - Recently Viewed Items
Hi Guys,
I have a script that shows the user of my shop what they have looked at. I was wondering if someone can tell me how I can limit this so it only lists 10 and no more. Here is the code: Code: [Select] if(RECENTVIEWED ==1) { if(isset($_COOKIE['jimbeam'])){ $productUrl = SITE_URL."ecom/index.php?action=ecom.pdetails&mode="; $productUrl = $this->libFunc->m_safeUrl($productUrl); foreach ($_COOKIE['jimbeam'] as $name => $value) { $this->obDb->query="SELECT vTitle FROM ".PRODUCTS." WHERE vSeoTitle = '".$value."'"; $rsProd = $this->obDb->fetchQuery(); $this->obTpl->set_var("TPL_VAR_PRODUCTTITLE",stripslashes($rsProd[0]->vTitle)); $this->obTpl->set_var("TPL_VAR_PRODUCTURL",$productUrl.$value); $this->obTpl->parse("recent_blk","TPL_RECENT_BLK",true); } $this->obTpl->parse("mainrecent_blk","TPL_MAINRECENT_BLK"); } } Thanks, Jordan Similar Tutorialsam currently building the cart system of a product, now there is this part wherein, the non-logged-in OR logged-in user should also be able to see the items that he/she viewed, how to do that?., am not asking for code snippets , just give me some ideas/hints/strategies/tips that may help me get the big picture on how to do this thing and proceed coding. I want to be able to grab any new images uploaded and display them right away on the front page, at the moment I can grab the id of each but that won't update the gallery. Once 4 images are uploaded, the next one to be uploaded pushes the last out of the gallery.
At the moment, I only have 2 images on the db, but I want it ready to be used by a community
<form action="<? echo $_SERVER['PHP_SELF']; ?>" method="post"
enctype="multipart/form-data">
Upload:<br><br>
<input type="file" name="image"><br><br>
<input type="submit" name="submit" value="Upload">
</form>
<?php
if(isset($_POST['submit']))
{
mysql_connect("localhost","____","____");
mysql_select_db("moduni_images");
$imageName = mysql_real_escape_string($_FILES["image"]["name"]);
$imageData = mysql_real_escape_string(file_get_contents($_FILES["image"]["tmp_name"]));
$imageType = mysql_real_escape_string($_FILES["image"]["type"]);
if(substr($imageType,0,5) == "image")
{
mysql_query("INSERT INTO `images` VALUES('','$imageName','$imageData')");
echo "Image uploaded!";
}
else
{
echo '<br>O<font color="#8B0000">nly images are allowed!</font>';
}
}
?>
<?php
mysql_connect("localhost","____","____");
mysql_select_db("moduni_images");
if(isset($_GET['id']))
{
$id = mysql_real_escape_string($_GET['id']);
$query = mysql_query("SELECT * FROM `images` WHERE `id`='$id'");
while($row = mysql_fetch_assoc($query))
{
$imageData = $row["image"];
}
header("content-type: image/jpeg");
echo $imageData;
}
else
{
echo "Error!";
}
?>
<div id="user-gallery"> <h2>Gallery</h2> <div class="img"> <a href="scripts/show_image.php?id=2" data-lightbox="image-1"> <img src="scripts/show_image.php?id=2" width="125px" height="71px"> </a> </div> <div class="img"> <a href="scripts/show_image.php?id=2" data-lightbox="image-1"> <img src="scripts/show_image.php?id=2" width="125px" height="71px"> </a> </div> <div class="img"> <a href="scripts/show_image.php?id=2" data-lightbox="image-1"> <img src="scripts/show_image.php?id=2" width="125px" height="71px"> </a> </div> <div class="img"> <a href="scripts/show_image.php?id=2" data-lightbox="image-1"> <img src="scripts/show_image.php?id=2" width="125px" height="71px"> </a> </div> </div> How can I make it so that the things that the user recently typed in an input is not displayed as a drop down. I want nothing to be displayed. This topic has been moved to Other Libraries and Frameworks. http://www.phpfreaks.com/forums/index.php?topic=307986.0 I have a webpage where all the database items are displyes in a table format.The table also has a check box.Upon clicking the delete button i need to delete all the items whish has the checkbox checked. How will i do that hi i am working on a project in which i have to retrieve the articles which are the most visited by the users. i have serched the internet but not find any useful code snippets. can any one guide me how to solve this problem. So, you know how on forums (including this one) the icon of a topic/forum will be greyed out if your account viewed that topic's latest post, regardless of which computer you viewed it on? How exactly does it store that? I know that logically it would store it on the Database, but I'm curious as to the actual implementation. It seems to me that storing every single topic you viewed it and when would take up huge numbers of rows, even for a small number of people, like say 100. If you have a forum with 1000's of people, each viewing even 3 topics a day, you're record numbers would be huge, slowing down the database. And calling your last viewed time for each and ever topic? Wouldn't that slow down the code excecution heaps? If you have any input, I'd love to hear it. It's a question I've been wrestling with in the back of my mind for a few years now Hello everyone I wonder if you could help me. I want to create a function called <?php function mythings($what) Which would allow me to pass an ID number to, such as <?php mythings(5419); What I want to do with that function is store the ID numbers of the past 6 items I have viewed, but I want to store each item as a SESSION once it has been passed. So essentially I would end up with the following 6 SESSIONS <?php $_SESSION['item1']; $_SESSION['item2']; $_SESSION['item3']; $_SESSION['item4']; $_SESSION['item5']; $_SESSION['item6']; So the "item1" SESSION would be the latest item ID passed through the function But then if I then passed a new ID through the function, that new ID passed would then become $_SESSION['item1']; and the current SESSION $_SESSION['item1']; would become $_SESSION['item2']; and $_SESSION['item2']; would become $_SESSION['item3']; and so on. This means I can do a list such as <?php print "Item 1".$_SESSION['item1']."<br>"; print "Item 2".$_SESSION['item2']."<br>"; print "Item 3".$_SESSION['item3']."<br>"; print "Item 4".$_SESSION['item4']."<br>"; print "Item 5".$_SESSION['item5']."<br>"; print "Item 6".$_SESSION['item6']."<br>"; ?> Does that make much sense? Can anyone help. Thanks very much all John Hello all, Wondering if you can help me, I am wondering if it is possible to get PHP to show the last websites you viewed, just giving the names of the websites in a list. If so? is there any pointers someone can give me as to how to learn to achieve something like this..I'd really appreciate the help! Hello people i have a system that takes people to generated pages from the database, the user has a field to input a video and others watching videos will get redirecting to tht video in time, what i want to know is how can i tell if a user viewed the page so to stop them getting redirected to it again by my random video query? The link always is video.php?id=blabla the id changes every refresh, so i can call that id to check if there on the page but how can i tell if they have been on it before. This script ...
<?php
echo "Your IP Address: <br/>";
print_r($_SERVER['REMOTE_ADDR']);
?>
... prints:
::1I am looking to have the IP address from which the site is viewed printed on screen. What is the issue, here, with the script I have? I have been trying this on localhost and I am using XAMPP. This topic has been moved to PHP Applications. http://www.phpfreaks.com/forums/index.php?topic=327636.0 hey all, i have an issue, i have a directory site called hackingtoolz.tk and if you go to it, you will see a list of folders as it is just a directory the problem is how to i stop the banner from apearing in the directory. i have tried changing the chmod to 4000 and that just made the jpeg stay their and the banner screw up if you go to the site you will understand what i mean better hope you can help cheers nick <?php query_posts('meta_key=post_views_count&posts_per_page=3&orderby=meta_value_num& order=DESC'); if (have_posts()) : while (have_posts()) : the_post(); ?> <ul class="rpul"> <li> <div class="class="post-thumb wp-post-image" style=' '> <?php if ( has_post_thumbnail() ) : ?> <?php the_post_thumbnail('thumbnail') ?> <?php endif ?> </div><div class='cat-post-text'> <h1> <a href="<?php the_permalink(); ?>"><?php the_title(); ?>
so the user is reading a story, to finish reading he has to click a link that redirects them to the signup page. Code: [Select] <?php session_start(); $beginurl = $_SERVER['HTTP_REFERER']; $_SESSION['beginurl'] = $beginurl; echo $_SESSION['beginurl']; ?> <html> <head> </head> <body> <script type="text/javascript"><!-- location.replace("http://www.mysite.com/members/"); //--></script> </body> </html> When they get to the second page, they have to click a link that opens up a modal. this is the code that runs when they hit the register button Code: [Select] session_start(); $beginurl = $_SESSION['beginurl']; $beginurl= (isset($_SESSION['beginurl'])) ? $_SESSION['beginurl'] : 'Error'; if( $_SESSION['status'] ='authorized') $_SESSION['$makemodal'] = 0; //sends the user to the page upon successful password credential if(!isset($_SESSION['SESS_USERID'])||(trim($_SESSION['SESS_USERID']=='admin'))) { echo '<script language="javascript">'; echo "top.location.href = $beginurl"; echo '</script>'; exit(); } Am I passing this variable correctly? and I'm not sure if the top.location.href towards the bottom is correct either, right now after I hit the register button I'm redirected to a blank page where the url is, "http://www.mysite.com/function Error() { [native code]}" Hello, I was wondering how to create the 'This post is being viewed by user1, user2, user3 and ... guests' on a certain location of the forum/website. I was thinking about having all these usernames / guests stored in the mysql database, however this caused an issue for me, since I got no idea if you can/are allowed to place an array in a table. I am hoping someone out here knows what is the easiest/most efficient way of doing this. Thanks in advance. How can I select the last 5 db items? I was thinking a timestamp or some thing. What is the best way to capture items that someone wants to purchase when they don't have an account yet? And I prefer not using cookies.
Guys, i want to display ads after 4 items to be displayed. Iam fetching item using mysql database & iam showing 8 items per page. I used below code but its not working because i used the item id to specify the first number but the items is not listed as correct order Code: [Select] <?php // first number $first = $row['id']; // second number $second = 4; //checking if the first number is muliple of second. If true then display ads. If($first % $second == "0") { echo 'ads to display'; } It will work if i list items in correct order. Please let me know correct solution. Thanks in advance ! How can I check if an array contains two items that are the same? For example ("apples", "bananas", "oranges") returns false But ("apples", "bananas", "bananas") returns true |
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