PHP - 24 Hour Page Show
Hi
I want to make something like this - My Sites index.php will be avail avail to user after he has clicked in a link that will come after every 24 Hour in my site. Means when a user first enters the site it will come and clicking in there the site will be avail avail. again after 24 Hour it will come again. But i am not getting how to do it. So need help SaKIB Similar TutorialsHey guys only thing i can find online is examples of showing how much time is left until a predetermind time. but what I need is how many minutes are left until the next hour hits anyone have any ideas on this? Im wanting to do it in UNIX time Quesion: Show each movie in the database on its own page, and give the user links in a "page 1, Page 2, Page 3" - type navigation system. Hint: Use LIMIT to control which movie is on which page. I have provided 3 files: 1st: configure DB, 2nd: insert data, 3rd: my code for the question. I would appreciate the help. I am a noob by the way. First set up everything for DB: <?php //connect to MySQL $db = mysql_connect('localhost', 'root', '000') or die ('Unable to connect. Check your connection parameters.'); //create the main database if it doesn't already exist $query = 'CREATE DATABASE IF NOT EXISTS moviesite'; mysql_query($query, $db) or die(mysql_error($db)); //make sure our recently created database is the active one mysql_select_db('moviesite', $db) or die(mysql_error($db)); //create the movie table $query = 'CREATE TABLE movie ( movie_id INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, movie_name VARCHAR(255) NOT NULL, movie_type TINYINT NOT NULL DEFAULT 0, movie_year SMALLINT UNSIGNED NOT NULL DEFAULT 0, movie_leadactor INTEGER UNSIGNED NOT NULL DEFAULT 0, movie_director INTEGER UNSIGNED NOT NULL DEFAULT 0, PRIMARY KEY (movie_id), KEY movie_type (movie_type, movie_year) ) ENGINE=MyISAM'; mysql_query($query, $db) or die (mysql_error($db)); //create the movietype table $query = 'CREATE TABLE movietype ( movietype_id TINYINT UNSIGNED NOT NULL AUTO_INCREMENT, movietype_label VARCHAR(100) NOT NULL, PRIMARY KEY (movietype_id) ) ENGINE=MyISAM'; mysql_query($query, $db) or die(mysql_error($db)); //create the people table $query = 'CREATE TABLE people ( people_id INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, people_fullname VARCHAR(255) NOT NULL, people_isactor TINYINT(1) UNSIGNED NOT NULL DEFAULT 0, people_isdirector TINYINT(1) UNSIGNED NOT NULL DEFAULT 0, PRIMARY KEY (people_id) ) ENGINE=MyISAM'; mysql_query($query, $db) or die(mysql_error($db)); echo 'Movie database successfully created!'; ?> ******************************************************************** *********************************************************************** second file to load info into DB: <?php // connect to MySQL $db = mysql_connect('localhost', 'root', '000') or die ('Unable to connect. Check your connection parameters.'); //make sure you're using the correct database mysql_select_db('moviesite', $db) or die(mysql_error($db)); // insert data into the movie table $query = 'INSERT INTO movie (movie_id, movie_name, movie_type, movie_year, movie_leadactor, movie_director) VALUES (1, "Bruce Almighty", 5, 2003, 1, 2), (2, "Office Space", 5, 1999, 5, 6), (3, "Grand Canyon", 2, 1991, 4, 3)'; mysql_query($query, $db) or die(mysql_error($db)); // insert data into the movietype table $query = 'INSERT INTO movietype (movietype_id, movietype_label) VALUES (1,"Sci Fi"), (2, "Drama"), (3, "Adventure"), (4, "War"), (5, "Comedy"), (6, "Horror"), (7, "Action"), (8, "Kids")'; mysql_query($query, $db) or die(mysql_error($db)); // insert data into the people table $query = 'INSERT INTO people (people_id, people_fullname, people_isactor, people_isdirector) VALUES (1, "Jim Carrey", 1, 0), (2, "Tom Shadyac", 0, 1), (3, "Lawrence Kasdan", 0, 1), (4, "Kevin Kline", 1, 0), (5, "Ron Livingston", 1, 0), (6, "Mike Judge", 0, 1)'; mysql_query($query, $db) or die(mysql_error($db)); echo 'Data inserted successfully!'; ?> ************************************************************** **************************************************************** MY CODE FOR THE QUESTION: <?php $db = mysql_connect('localhost', 'root', '000') or die ('Unable to connect. Check your connection parameters.'); mysql_select_db('moviesite', $db) or die(mysql_error($db)); //get our starting point for the query from the URL if (isset($_GET['offset'])) { $offset = $_GET['offset']; } else { $offset = 0; } //get the movie $query = 'SELECT movie_name, movie_year FROM movie ORDER BY movie_name LIMIT ' . $offset . ' , 1'; $result = mysql_query($query, $db) or die(mysql_error($db)); $row = mysql_fetch_assoc($result); ?> <html> <head> <title><?php echo $row['movie_name']; ?></title> </head> <body> <table border = "1"> <tr> <th>Movie Name</th> <th>Year</th> </tr><tr> <td><?php echo $row['movie_name']; ?></td> <td><?php echo $row['movie_year']; ?></td> </tr> </table> <p> <a href="page.php?offset=0">Page 1</a>, <a href="page.php?offset=1">Page 2</a>, <a href="page.php?offset=2">Page 3</a> </p> </body> </html> I want to show a cookie of a referral's username on a sign up page. The link is like this, www.mysite.com/signup?ref=johnsmith. The cookie doesn't show if I go to that url page. But it does show up once I reload the page. So I'm wondering if it's possible to show the cookie the first time around, instead of reloading the page? Here is my code. // This is in the header $url_ref_name = (!empty($_GET['ref']) ? $_GET['ref'] : null); if(!empty($url_ref_name)) { $number_of_days = 365; $date_of_expiry = time() + 60 * 60 * 24 * $number_of_days; setcookie( "ref", $url_ref_name, $date_of_expiry,"/"); } else if(empty($url_ref_name)) { if(isset($_COOKIE['ref'])) { $user_cookie = $_COOKIE['ref']; } } else {} // This is for the sign up form if(isset($_COOKIE['ref'])) { $user_cookie = $_COOKIE['ref']; ?> <fieldset> <label>Referred By</label> <div id="ref-one"><span><?php if(!empty($user_cookie)){echo $user_cookie;} ?></span></div> <input type="hidden" name="ref" value="<?php if(!empty($user_cookie)){echo $user_cookie;} ?>" maxlength="20" placeholder="Referrer's username" readonly onfocus="this.removeAttribute('readonly');" /> </fieldset> <?php }
Hi, I would like to do the following but not sure how. If the you/user is on index.php of http://www.domain.com/ show one page If not show another How would I do this? Thanks I am trying to display date and time below code display date correctly but when it come to display time it display 3 hour behind. eg. when it is 12:47 PM in my system it display 9:47 PM Code: [Select] <?php $todaydate=date("j - F - Y g:i:s a"); echo $todaydate; ?> $time="19:00:00"; how to do i output one hour later from the variable above? Hi, i'm trying to create some detailed statistics about customer activity, i have entries in my mysql db when the customer has been active for the last time and want to create some statistics about that, basically a "online last 24 hours" but from specific countries. Now i've tried this: Code: [Select] $time = date('Y-m-d H:i:s'); $time24 = date("Y-m-d H:i:s", time()-((60*60)*24)); $query = "SELECT COUNT(*) as Anzahl FROM customers WHERE country = 'de' AND time BETWEEN '$time' AND '$time24' "; to get the current date and select all customers that have been available from the current date minus 24 hours, what's my mistake here, as this doesn't seem to work! Thanks Hi All, I want to be able to show selected pages of my website as PDF using a 'PDF' button much like this site does: http://www.westmeon.org.uk/index.php?option=com_content&task=blogsection&id=8&Itemid=35 I presume that I will need to install some sort of software on my server (which runs the latest PHP, MySQL etc.) but after hours of searching online I cannot find a simple way of doing this. Does anyone have any suggestions or pointers for how I can do this? FYI my website is written in PHP drawing data from a MySQL database. Regards, Neil So I am trying to check if the current time is one hour before a variable time: Code: [Select] $date_game=$dt->format('Y n j'.$pieces[2]); echo $date_game; echo date('Y n j H'); if (date('Y n j H') < $date_game) { echo "The time is before the stored time"; } This displays 2012 1 21 17:30 2012 1 21 16 i.e $pieces[2] = 17:30. and $dt formatted Y n j = 2012 1 21. The current Y n j H is 2012 1 21 16. I want to know if it is more than one hour until the date/time stored in $date_game. At the moment it just tells me that it is before that time. Can I do something like Code: [Select] $data_game - 1->format('H'); or something? Thanks guys Hi,
I have a blog which records the amount of views on each article. I now want to be able to work out the average number of views per hour.
How can I work out the number of hours passed from a datetime format?
From there I can just do views divided by hours passed.
Thanks in advance!
Hello, I have a very simple table... DAY (Y-m-d) / TIME (00:00:00) / POWER (INT) I am using a Jquery datepicker to select the date on the page, and that POST to the PHP file. I am trying to select values from Mysql to make 3 HighCharts Graphs using the selected day of the datepicker. I have started with the DAY graph PHP to get the values for each hour of a 24 hour day. I need to get the values for each hour... 23,12,15,35 etc... , and then I need for the 31 days of a month for the month graph, and 12 months of the year for the month graph all in the same way so the HighCharts can use the values to make the chart (3 php files, one for each graph) Here is the PHP I have for the 1 day that should get the 24 individual hour data, but it does not seem to work... <?php $choice = (isset($_POST['choice'])) ? date("Y-m-d",strtotime($_POST['choice'])) : date("Y-m-d"); $con = mysql_connect("localhost","root","mackie1604"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("inverters", $con); $sql = "SELECT HOUR(time), COUNT(power) FROM feed WHERE time = '".$choice."' GROUP BY HOUR(time) ORDER BY HOUR(time) "; $res = mysql_query($sql) or die('sql='.$sql."\n".mysql_error()); $row = mysql_fetch_assoc($res); echo $row['choice'].'<br />'; ?> What have a written wrong in the sql query ??? Alan Hello, I am trying to load an external image from a NOAA site directory. This image refreshes every three hours, starting with 0 and then 3,6,9,12.....21 in a 24 hour calendar, and are stored for days in advance. The date/time is in the file name. Originally I had written the script to refresh on the hour, and then realized the tri-hourly spacing. This is what I have: Code: [Select] <?php date_default_timezone_set('Etc/GMT-1'); $today = date("YmdH"); $pattern = '/WaveHeight_'.$today.'_mic.png'; $base_url = 'http://www.crh.noaa.gov/images/greatlakes/ndfd/MIC/dynamic2'; print '<a href="http://www.crh.noaa.gov/greatlakes/?c=map&l=lm&p=a"><img src="'.$base_url.$pattern.'" width= "237" hieght= "314" " ></a>'; ?> I am not fluent in php, but I assume I need a "If H=1,4,7,10,13,16,19,22; Then H+2 ; along with If H=2,5,8,11,14,17,20,23; Then H+1" to keep the images relatively current. I just can't quite wrap my head around how to express this. Any help is appreciated. Thank you. Hi all, I have a loop that is modifying a DateTime object. It goes through X number of times and adds 1 day to the date time object and stores the modified object in an array. I use ->add(new DateInterval('P1D')); For some reason it is losing an hour each time that I add a day. I start out at 19:00 and each iteration through the loop adds a day, but loses an hour. 2010-09-19 19:00:00 2010-09-20 18:00:00 2010-09-21 17:00:00 I can correct for this by doing ->add(new DateInterval('P1DT1H'));, but adding 25 hours instead of 24 hours doesn't make any sense to me. I am using PHP Version 5.3.1. Any ideas? Thanks, W Hi guys, I have a PHP Script were user would enter a username and password to get access, but the problem I keep having is every half n hour or so user have to sign in again cause the keeps logging them out for some reason. Please I need your help and advice to make this stop. Hi, This may be something for JavaScript but I would like to know if and how it's possible to show who is currently active/viewing the page. Users are logged into the system with their own account. The purpose of this is for a CRM where more than one person may be editing the same record, so undesired overwrites might occur which is what I'd like to avoid with this "other user editing this record" notification. This topic has been moved to PHP Freelancing. http://www.phpfreaks.com/forums/index.php?topic=332844.0 My site allows users to vote once per hour, and it checks the next time they vote if it's been 3600 seconds (1 hour) since their last vote. But, it always returns the value 0. $grab = mysql_fetch_assoc($has_voted); $calc = strtotime($grab['date']) - strtotime(date("M-d-Y")); if($calc > 3600) { mysql_query("DELETE FROM votes WHERE ip = '". $_SERVER['REMOTE_ADDR'] ."'"); header("location:votes.php?id=". $id .""); } else { echo echo "You've voted within the last hour. Please wait ". $calc ." seconds."; } Hi, I am creating a date string using the following: $curDate = date("Y-m-d H:i:s"); However, what I want to do is take one hour off this so that $curDate now becomes the current date and time minus one hour. Thanks for your help. I insert a time value using the following code
$ti= ( !empty($_POST['time']) ) ? "'{$_POST['time']}'" : 'NULL';
I want to insert a second time value into a differen column which would be the same time minus 1 hour
something along the lines of :
$ti2= ( !empty($_POST['time']) ) ? "'{$_POST['time'] -1 hour}'" : 'NULL';
What would be the correct way to do it
I know the php is basic and not completed yet, but I am working on that. My next step I want the result to show on index.php when a form is submitted to calculator2.php. I just cant seem to get it. Please help! index.php Code: [Select] <div class="post"> <h2 class="title">Calculator</h2> <hr /> <form method="post" action="calculator2.php"> Fireplace Front Width: <input type="text" name="fw"> <br /> Fireplace Back Width: <input type="text" name="bw"> <br /> Fireplace Depth: <input type="text" name="fd"> <br /> <input type="submit" name="Submit" value="Submit"> </form> Pounds Of Glass Needed: <?php echo $res1; ?> </div> calculator2.php Code: [Select] <?php $frontwidth = $_POST['fw']; $backwidth = $_POST['bw']; $firedepth = $_POST['fd']; $x = $frontwidth + $backwidth+ $firedepth; $y = ($x / 3) * .6667; $res1 = $y *2; echo $res1; ?> |