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PHP - A Simple Website,pages,posts Views Counter Using Files
I made this for my site and decided to share with the community.
Depending on how you do the code and where, this can keep count of the views in different ways. Please read the somewhat directions have in the code, make a folder called counters, and be sure to include this below function file. I named the function file index.php and belongs in the counters folder you make. <?php /* name this file index.php and place it into a folder called counters the function will create a unique text file for each unique item, good for posts or pages and store all the files in the counters folder For the total website views,add this to your header file or top of your page <?php include('counters/index.php'); $website_view_url = "http://".$_SERVER['HTTP_HOST']; } $website_views = getViews("$website_view_url"); echo "<br />".$website_views."<br />"; ?> For the total pages or scripts views,add this to your header file or top of your page <?php include('counters/index.php'); $total_page_view_url = "http://".$_SERVER['HTTP_HOST'].$_SERVER['SCRIPT_NAME']; } $total_page_views = getViews("$total_page_view_url"); echo "<br />".$total_page_views."<br />"; ?> For entire urls including queries,add this to your header file or top of your page <?php include('counters/index.php'); $page_queries_view_url = "http://".$_SERVER['HTTP_HOST'].$_SERVER['SCRIPT_NAME']; if (!empty($_SERVER["QUERY_STRING"])) { $page_queries_view_url .= "?".$_SERVER['QUERY_STRING']; } $page_queries_views = getViews("$page_queries_view_url"); echo "<br />".$page_views."<br />"; ?> for usage in posts like by id: if do not want the pages counted and did not do the above code,include the below somewhere near the beginning of your page <?php include('counters/index.php');?> Then in the posts loop, you can associate your $row['id']; or some other unique value <?php $post_views = $row['id']; $post_views = getViews("$post_views"); echo "<br />Post Views ".$post_views."<br />"; ?> if would like to combine any the above or use them all, just be sure to just include include('counters/index.php'); only one time */ function getViews($views_count_value) { $views_count_value = md5("$views_count_value"); $views_count_file_name = "counters/$views_count_value.txt"; if (!file_exists($views_count_file_name)) { $views_count =0; $file_handle = fopen($views_count_file_name, 'w') or die("can't open file"); fwrite($file_handle, "$views_count"); fclose($file_handle); } $file_handle = fopen($views_count_file_name, "r"); $views_count = fread($file_handle, filesize("$views_count_file_name")); fclose($file_handle); if ($views_count <= 0){ $views_count =1; } else { ++$views_count; } $file_handle = fopen($views_count_file_name, "w+"); fwrite($file_handle, "$views_count"); fclose($file_handle); return("$views_count"); } ?> Similar TutorialsCurrently on my website i record the statistics, for my pages they include unique views, and total page views. at the moment every time a visitor loads a page a script updates a MySql Db. This seems very resource intensive, is there a better way to record statistics?? Hi all, I'm trying to create a PHP script for user's profile to display the amount of times they've been viewed. I'm looking to have this script increase on a unique view, and it should update the variable in the database. Profiles are accessed by the following link: userprofile.php?userid=X (where X is the ID, e.g. 1, 2, 1001, 345982, etc.). The database variable I'm looking to update is called ProfileViews. I began developing the script, which is as follows: Code: [Select] $_SESSION['Viewed'] = 0; if ($_SESSION['Viewed'] == 0) { $profileViewsQuery = mysql_query("SELECT ProfileViews FROM Users WHERE UserID='????'"); $getProfileViews = mysql_fetch_array($profileViewsQuery); $profileViews = $getProfileViews['ProfileViews']; $profileViews = $profileViews + 1; mysql_query("UPDATE Users SET ProfileViews='$profileViews' WHERE UserID='????'"); $_SESSION['Viewed'] = 1; } However, I'm stumped on a couple things. Could you possibly help me out? 1. How can I get the script to recognize the link accessed's ID? E.g. when a user goes to userprofile.php?userid=1001, how can I get the script to identify the ID to update should be 1001? This is where the "????" would be replaced in the code. 2. On page load, the variable is always going to be $_SESSION['Viewed'] = 0, which isn't going to produce unique hits. Do you have any recommendations how I could achieve unique hits using this method? Thanks very much for reading. I've made a page which lists information from a database like: 1 Andy 2 James 3 Bob But if James over takes Andy it goes 1 James 2 Andy 3 Bob. I have that working. However, I've added multi pages into my website [1, 2, 3, 4, 5] etc, however, when you go onto a new page, it starts back at number 1. It doesn't continue adding 1 on. Does anyone know how to make this work? Thanks in advance, Andy. need a code in php - show all posts from a irc channel on website if anyone knows how to do it, please let me know... i only need to display all the talking on mirc on my website. thank you This topic has been moved to PHP Applications. http://www.phpfreaks.com/forums/index.php?topic=354184.0 Hey all,
This code was given to me by my client who swears it works, yet I can't seem to get it to function.
<?php
$now = time(); // or your date as well
$your_date = strtotime("2016-06-01");
$datediff = $now - $your_date;
$referrals = number_format(162250 + (floor($datediff/(60*60*24)) * 527) + (87920 + (floor($datediff/(60*60*24)) * 45)));
?>
//javascript:
<script>
//vars from template
var referrals = "<?php echo $referrals;?>";
var currentdate = new Date();
jQuery("document").ready(function() {
jQuery ("#sp-menu > div > nav > ul > li:nth-child(1) > a").html("Home");
jQuery(".referrals").html(referrals);
jQuery(".current_date").html((currentdate.getMonth()+1) + "/" + currentdate.getDate() + "/" + currentdate.getFullYear());
jQuery(window).on("scroll", function() {
var scrollPos = jQuery(window).scrollTop();
if (scrollPos <= 0) {
jQuery(".counter").fadeIn();
jQuery(".sec-nav").fadeIn();
} else {
jQuery(".counter").fadeOut();
jQuery(".sec-nav").fadeOut();
}
});
});
</script>
I may not be a PHP wizard, but a view counter seems pretty damn simple and easy to implement. Unless you leave it in my hands... It's a pretty simple problem, so I'm hoping it'll be a simple solution. Every time you load a game on my website, at the top of the php "playgame.php" is a function call: $objGlobal->update_gameplay($gameid); In my functions.class.php is the function declaration: function update_gameplay() { $args = func_get_args(); $getsql = mysql_query("select * from games where gameid = $args[0]"); if($getsql) { $resultgame = mysql_fetch_array($getsql, MYSQL_ASSOC); $timesplayed = $resultgame['timesplayed'] + 1; $updatesql = "update games set timesplayed = '$timesplayed', last_played = now() where gameid = $args[0]"; if(mysql_query($updatesql)) return true; else return false; } } Yet, when I load the playgame page, it adds 1,2 or 3!!! So this means the function is being called 1, 2 or 3 times, but how can that be if it is at the VERY top of the page? Try it yourself on my website, www.gpstudios.com. Click on a game and reload. For some reason, the newer games always add 1, but other older/more popular games add 1, 2 or 3 I've searched for the string "update" and the string "timesplayed" in every file using a program called FileSeek and I can say for certain that the function must be being called multiple times for some reason. What's weirder is that my game reviews (http://www.gpstudios.com/reviews.php) are also having their view counter incremented by 1 or 2 on a page load. Please help me solve this quick. Thanks, -Tom Hi, (Just wanna say thanks to the guy who helped me earlier as i cannot find the topic) I have a game, it has 3 frames, menu, header & the main content, it WAS fine until i deleted a 4th frame, now when i click a link from the menu, instead of showing the link in the 'Main' content area, it takes you to the actual page with no menu, header etc. Thanks in advance! hello , i have created web application in which i have divided the all parts of page like - main header part - contains the <!DOCTYPE to <body> part other header part - contains the logo,slideshow of the header of page then the main part and at the end footer part i have created on folder which contains the detail of page content info referenced from the other pages from main content.. here i am facing problem as the style sheet , javascripts and the other header part of file is not references well.. how to do it? please guide me with correct solution awaiting for better response... I'm am somewhat new to PHP and am trying to set up a website for my cousin's wedding. Her idea is to have the guests sign in with a user/pass that she provides, and once they sign in, they will be taken to a page that has their name on it (i.e. "Mr. and Mrs. So and So, you are invited...). I have come to the conclusion that I will need to make an image for each guest's name (she wants to use a font for their names that nobody will have on their computer) so what I need to know is: How do I link each user name to their own personalized webpage, where the image of their name on the next page will change based on what username is entered? The php code I have right now is this (i'm sorry it's so long, I just don't want to leave anything out that might be important): $LOGIN_INFORMATION = array( 'steve' => 'password', 'rick' => 'password', 'tom'=> 'password' ); // request login? true - show login and password boxes, false - password box only define('USE_USERNAME', true); // User will be redirected to this page after logout define('LOGOUT_URL', 'http://www.example.com/'); // time out after NN minutes of inactivity. Set to 0 to not timeout define('TIMEOUT_MINUTES', 0); // This parameter is only useful when TIMEOUT_MINUTES is not zero // true - timeout time from last activity, false - timeout time from login define('TIMEOUT_CHECK_ACTIVITY', true); ################################################################## # SETTINGS END ################################################################## /////////////////////////////////////////////////////// // do not change code below /////////////////////////////////////////////////////// // show usage example if(isset($_GET['help'])) { die('Include following code into every page you would like to protect, at the very beginning (first line):<br><?php include("' . str_replace('\\','\\\\',__FILE__) . '"); ?>'); } // timeout in seconds $timeout = (TIMEOUT_MINUTES == 0 ? 0 : time() + TIMEOUT_MINUTES * 60); // logout? if(isset($_GET['logout'])) { setcookie("verify", '', $timeout, '/'); // clear password; header('Location: ' . LOGOUT_URL); exit(); } if(!function_exists('showLoginPasswordProtect')) { // show login form function showLoginPasswordProtect($error_msg) { ?> <html> <head> <title>Please enter password to access this page</title> <META HTTP-EQUIV="CACHE-CONTROL" CONTENT="NO-CACHE"> <META HTTP-EQUIV="PRAGMA" CONTENT="NO-CACHE"> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <style type="text/css"> body,td,th { font-family: Verdana, Geneva, sans-serif; font-size: 10px; color: #666; } body { background-color: #FFFFFB; } </style> </head> <body> <div align="center"> <style> input { border: 1px solid black; } </style> <div style="width:600px; margin-left:auto; margin-right:auto; text-align:center"> <form method="post"> <h4>Please sign in using the information provided on the invitation</h4> <font color="red"><?php echo $error_msg; ?></font><br /> <?php if (USE_USERNAME) echo 'Login:<br /><input type="input" name="access_login" /><br />Password:<br />'; ?> <input type="password" name="access_password" /><p></p><input type="submit" name="Submit" value="Submit" /> </form> <br /> <a style="font-size:9px; color: #B0B0B0; font-family: Verdana, Arial;" href="http://www.zubrag.com/scripts/password-protect.php" title="Download Password Protector">Powered by Password Protect</a> </div> </body> </html> <?php // stop at this point die(); } } // user provided password if (isset($_POST['access_password'])) { $login = isset($_POST['access_login']) ? $_POST['access_login'] : ''; $pass = $_POST['access_password']; if (!USE_USERNAME && !in_array($pass, $LOGIN_INFORMATION) || (USE_USERNAME && ( !array_key_exists($login, $LOGIN_INFORMATION) || $LOGIN_INFORMATION[$login] != $pass ) ) ) { showLoginPasswordProtect("Incorrect password."); } else { // set cookie if password was validated setcookie("verify", md5($login.'%'.$pass), $timeout, '/'); // Some programs (like Form1 Bilder) check $_POST array to see if parameters passed // So need to clear password protector variables unset($_POST['access_login']); unset($_POST['access_password']); unset($_POST['Submit']); } } else { // check if password cookie is set if (!isset($_COOKIE['verify'])) { showLoginPasswordProtect(""); } // check if cookie is good $found = false; foreach($LOGIN_INFORMATION as $key=>$val) { $lp = (USE_USERNAME ? $key : '') .'%'.$val; if ($_COOKIE['verify'] == md5($lp)) { $found = true; // prolong timeout if (TIMEOUT_CHECK_ACTIVITY) { setcookie("verify", md5($lp), $timeout, '/'); } break; } } if (!$found) { showLoginPasswordProtect(""); } } ?> ************************************************************** THANK YOU SO MUCH for any help!! Sam I got a question regarding a news website content that i want to make ! my question is how do i call my contents without using a lot of page? i explain let's say i got 10 news how do i put this 10 news in different pages without using 10 pages ? ex: you can see some links having a number like this http://bbc.uk/news/murder_case-12 then the next page got http://bbc.uk/news/finance-13 the title and the number id change but the page news doesnt change thanks for your answer. Hello everyone: I wanted to see how I can make a simple login page (user name and password) that redirects to a page(s) if the login is correct. Also, I wanted to put protection on the page(s) that will send the user back to the login page if the credentials are nor correct. I would imagine the username/password would be stored in a database table (Admins), and the correct login info would be stored in a session ..? I am use to doing this with ASP, but never PHP. I want to make sure I understand how to do this properly and securely so I can use this as a model for other systems. In ASP I would do a protected page like this: a_login_check.asp Code: [Select] <% if session("admin_user_name") = "" then session.abandon response.redirect "login.asp" end if %> Protected-Page.asp Code: [Select] <!-- #include file="include/a_check_login.asp" --> <html> ... CONTENT ... </html> And of course there is the login page itself ... (I thought it would be nice to add a "Forgot Password" link on the login page, but if that is too complicated I can do that later .. or is it easy ??) Anyway, can someone point-out to me how to do this. I would appreciate it! I'm working with a downloaded open-source CMS on localhost, and trying to learn. Is there a tool that will list the files (php and otherwise) that are accessed (in order) as the CMS does various normal operations? I want to see the order and the repetition of files that are opened, even if it is something minor. Thanks. Hello, I never dealt with such a feature before but I found a small script online that I am editing that works just fine. The problem is that it includes all of the directories in the path of my site in the zip. So if the path of my webhost is: /home/bob/files/ the zip will create a folder such as /home (inside that folder) /bob (inside that folder) /files (then the zipped files I wanted). I just want the zip files and I did store them out of the www/public_html directory to prevent people from downloading the files directly from the web but is there no way to make it not include all of those directories in the zip and only show the files? If so I will post the code that I have. Thanks I have an xml file thats about 20MB, I want to parse it all into an array. SimpleXML seems to have trouble processing it, I just get "Fatal error: Balloc() failed to allocate memory". I've googled this topic and can't seem to find a straight forward solution. There must be someone here who has had to do this before. Is there an easy way to parse this? Ok i want it to make it that everything some clicks on a view the view goes up by one so i did this Code: [Select] <?php session_start(); $views = $_SESSION['views']++; ?> and echo the views like this Views<?php echo $views; ?> but it updates countines to add even if the video is a new video Hello,
I try to get website speed of some website, but i can read only ''domain.com'' i can't read website files like css , js ... why ? i use proxies for this job.
here is the php code:
$options = array(
'useragent' => "Firefox (+http://www.firefox.org)", // who am i
'connecttimeout' => 120, // timeout on connect
'timeout' => 120, // timeout on response
'redirect' => 10, // stop after 10 redirects
'referer' => "http://www.google.com",
'proxyhost' =>'85.25.8.14:80'
);
$response = http_get("http://solve-ict.com/wp-content/themes/ict%20theme/js/jquery-1.7.1.min.js", $options , $info);
but it works fine with http://domain.com/ , but with files css or js it gives 404, using some free proxy servers available
?
Thanks.
Hello folks, good day! I'm having trouble here selecting my desired data. I hope somebody will help me. Im having two tables. tbl_a seId course year 1 IT 1999 2 HE 1992 3 RT 1990 tbl_b id seId status 1 1 NO 2 2 YES I want to select A.*,B.* from tbl_a AS A, tbl_b AS B having status = NO and status IS NULL my desired output is to see the following. course year status IT 1999 NO RT 1990 is it possible? thank you in advance. hii i want to create a script where i want to store unique page views per 24 hrs using cookie.. i had create a code but rectify me if i'm wrong... 1) i wan to store a cookie so i set this code on top to create a cookie on views browser.. Quote <php setcookie("user", time()+84600); ?> 2) Now i want to find that if the cookie was set before, then the views will not update, if not then it will increase by 1. Quote <php if (isset($_cookie["user"])) echo "Total no of views" . $views . "; else echo" $views=$views+1; echo " $views"; $k=mysql_query("UPDATE `images` SET views = '$views' WHERE id = '$userid'") ?> but the problem i'm facing is that where to put that "$views" so that it will increase and will show to the user... now suppose every thing works well then how the cookie will understand that the views will be updated in specific users database... please guys i really need help... If some other possible ways are there then please refer... Thanks in advance.. Hiii i got a image hosting script but the problem i'm not able to solve is that in the show image page, the page views counter is not showing unique views. If we refresh the page 100 times from same ip, it shows (views=100)... Can any one help me with it... The views are stored in the database of the respective users... can anyone help me to get unique views per 24 hrs per ip.... The code is... Quote //UPDATE VIEWS COUNT AND LAST ACCESS DATE $views = $row['views'] + 1; $access = date("y-m-d"); $r = mysql_query("UPDATE `images` SET views = '$views', access = '$access' WHERE id = '$id' "); $imguserid = getUserId($id); $own = false; if ($imguserid != -1) { if ($userid == $imguserid) $own = true; $r = mysql_query("SELECT username FROM `members` WHERE id = '$imguserid'"); $row1 = mysql_fetch_row($r); $username = $row1[0]; } else $username = "Anonymous"; echo "<center>"; echo "<br><LABEL id='title'>Views:</LABEL> $views"; echo "<br><LABEL id='title'>Date Added:</LABEL> {$row['date']}"; if needed i can supply the full script to the person who can help me... You can check the demo in [ http://torrentz.0fees.net/wussa/show-image.php?id=c7693c02c3ad2fafe5f5d3f00263a68a ] The full page is attached below.... |
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