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PHP - Please Can Someone Show Me How To Code A Simple Addition All On The Same Page?
Hello People.
Please can someone point me to where I can learn to program the following all one one page? Here is a screenshot of what I'm after. Idea being that you put hours and minutes into the top row. Then hours and minutes into the next row. Click the button and get the answer at the bottom. The answer to 1 hour 40 minutes added to 1 hour 40 minutes should obviously be 3 hours 20 minutes. Here's how I would approach it in another language, but am brand new to php hence the help required. Make the following variables. Hour1 Hour2 Minute1 Minute2 Total_Hours Total_Minutes Combined_Time I would then write something like Total_minutes = minute1 + Minute2 Total_hours = Hour1 + hour2 combined time = Total_hours * 60 + (Total_Minutes) // reset the total hours Total_hours = 0 Loop here.... while combined_time >60 Total_hours = Total_hours + 1 combined_time = combined time -60 re-do the loop Now in the boxes at the bottom put bottom hour box = total hours bottom minute box = combined_time (remaining minutes) here's the code I have so far. Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Untitled Document</title> </head> <body> <table width="527" border="0" cellspacing="5" cellpadding="5" bgcolor="#c0c5c9"> <p>Time Calculator</p> <p>Put your hours and minutes in the top boxes. Click the add time button<br /> to get the answer </p> <table width="401" border="1" cellspacing="5" cellpadding="5"> <tr> <td width="109"><input type="int" name="hour01" size="10" maxlength="10" /> Hrs</td> <td width="251"><input type="int" name="min01" size="10" maxlength="10" /> Mins</td> </tr> <tr> <td><input type="int" name="hour02" size="10" maxlength="10" /> Hrs</td> <td><input type="int" name="min02" size="10" maxlength="10" /> Mins</td> </tr> <tr> <td><input type="submit" name="submit" value="Add Time" /></td> <td><input name="hour_answer" type="int" size="10" maxlength="10" /> Hrs <input type="int" name="minute_answer" size="10" maxlength="10" /> Mins</td> </tr> </table> <tr></tr> </body> </html> Similar TutorialsHey, guys!
Why the output of my code in both "echo" is only the number "2"?
<?php $x = 4; $y = 2; //Addition. echo "Addtion: " . $x + $y . "<br>"; echo 'Addtion: ' . $x + $y . '<br>'; ?>Thanks! Hi there, Im working on my little project and I would appreciate your help. I have only basic knowledge of php, mostly I just copy some scripts that could be useful for me. Im trying to find some simple script that allows me to see the written text on web page no. 1 on webpage no. 2.. Something like send the form to email, except I dont want to send it on email, but different webpage. Something like different way of eshop, where you get your order shipped to email, but I wanna send this information to webpage. Is there such a script like Im describing? Probably is but I dont know how to search for this..
Thank You
Edited by Radim, 21 October 2014 - 07:15 AM. 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'm trying to build a small php script to help automate adding vacation time to keep track of employees available vacation. I can run the sql below, and it executes fine: Quote Update `vacation` set avail_vacation = `avail_vacation` - `used` where employee = "employee1"; I'm trying to build this into a php function. This is what I have so far: Code: [Select] <?php function sql_addition() { global $conn; global $_POST; $sql = "update `vacation` set `avail_vacation`= `avail_vacation` + `added` where employee = .sqlvalue(@$_POST["employee"])"; mysql_query($sql, $conn) or die(mysql_error()); } <html> <tr> <td class="hr"><?php echo htmlspecialchars("Add Hours")." " ?></td> <td class="dr"><input type="text" name="added" value="<?php echo sql_addition('"', '"', trim($row["added"])) ?>"></td> </tr> ?> </html> It's showing the error below where the field should be for submitting hours to add: Quote <input type="text" name="added" value="You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 Help would be greatly appreciated. Hi I'm having an issue with a piece of code, it pretty much works but the line to output the total weight only seems to be picking up the count and not multiplying it by the weight. Any ideas why this is? Thanks in advance Code: [Select] <?php mysql_connect("localhost", "*****", "******") or die(mysql_error()); mysql_select_db("inventory") or die(mysql_error()); $data = mysql_query("SELECT * FROM testDB ORDER BY uid DESC LIMIT 30") or die(mysql_error()); Print "<p>"; while($info = mysql_fetch_array( $data )) { $totalweight = ($info['count' * 'itemtype']); Print $info['count'] . " "; Print $info['itemtype'] . " "; Print $totalweight . " kgs <br />"; } Print "</p>"; ?> Am trying to get a current price, add it to what is in the database and update the table with the new value. Every time i do it, it won't do the addition but update the table with the current price. Please, what am i doing wrong?
$sql2 = mysql_query("SELECT * FROM pbudget WHERE project_name = '$project' ");
$budget = mysql_fetch_array($sql2);
$actual = $budget['actual'] ;
$actual = $budget['spent'] ;
$tot = ($spent + $amount);
$final = ($actual - $amount);
//$final = ($actual + $amount);
$r = mysql_query("UPDATE pbudget SET actual = '$final', spent ='$tot' WHERE project_name = '$project'") or die(mysql_error());
$que = mysql_query("UPDATE budget SET actual = '$final', spent ='$tot' WHERE category = '$cat'") or die(mysql_error());
hii everyone how to add the values of rows in a database.... I have added the code of my database table Code: [Select] CREATE TABLE `users` ( `id` int(11) NOT NULL AUTO_INCREMENT, `username` varchar(50) NOT NULL, `money` int(11) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM AUTO_INCREMENT=23 DEFAULT CHARSET=utf8; And i have some rows like (1,peter, $94) (2,John,$16) .... .... .... .... .... i want to add those values (i.e their money) and shows like [ total money = $110]... I have done few queries like Code: [Select] SELECT id,SUM(money) FROM usersand Code: [Select] mysql_query("SELECT * FROM `users` SUM `money` WHERE id='$id'");But nothing seems to work out... Any help guysss.... Thanks in advance... Heja I have a MySQL database which contains information of members and the current balance of there accounts (eg: $1000) it also shows there current gain (eg: 1.2%) what i have been doing is individually updating each clients percent and balance. What i'm trying to work out is how to add a percent gain for the week and both the percent and balance of the clients are updated according to the submitted gain for example: Client 1 has $1,000 and has a current gain of 1.5% Client 2 has $,2000 and has a current gain of 2.5% I want to update both clients by 1.5% and then both clients be updated like so: Client 1 would now have $1,010.50 and now has a gain of 3.0% Client 2 would now have $2,020.50 and now has a gain of 4.0% I was looking around and could find much information on the internet or exactly has this can be achieved so i started to work on a addition and try atleast add the % however it dont go accordingly, i suppose im very rusty at my php. I started working around something like this: Code: [Select] mysql_select_db('mem') or trigger_error("SQL", E_USER_ERROR); $a1 = mysql_query("SELECT profits FROM account"); $a= $a1; $b=$b1; /* html <input type="text" name="b1" id="b1"> the % i want to update*/ $add=($a+$b); $sql = "UPDATE mem SET balance = balance +'$add'"; if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } This obviously does not work but it could possibly give you a better understanding of maybe what im trying to do here. Any help would be appreciated. Thanks Hi, this is my first time posting here. I am just delving into PHP and I am learning about foreach loops. I have written code in Notepad++ EXACTLY the way I saw it in a tutorial video I watched (I wish I could show the tutorial video to you, but it is on Lynda.com and you have to pay to watch) I attached the file with my code. The example 1 code works just fine. The example 2 code is the one that is not working for some reason. However, it worked for the guy that wrote it in the video, so I am not sure where I am going wrong? *The comments in green are mainly for myself, I explain things to myself so that I don't forget what the code does forloops.php 1.74KB 2 downloads I would appreciate some help. Thank you!!! I am new to php and need help creating addition and multiplication tables using the html code we were given. I feel like I am on the right track, but I am getting a few errors and I cant figure them out. One of the errors is that it tells me I am not putting in a valid number for rows even though it is a positive number so it should work. Here is my code and all help is appreciated thanks in advance.
<html>
<head/>
<body>
<form method="POST" action="<?= $_SERVER['PHP_SELF'] ?>">
<table border="1">
<tr><td>Number of Rows:</td><td><input type="text" name="rows" /></td></tr>
<tr><td>Number of Columns:</td><td><select name="columns">
<option value="1">1</option>
<option value="2">2</option>
<option value="4">4</option>
<option value="8">8</option>
<option value="16">16</option>
</select>
</td></tr>
<tr><td>Operation:</td><td><input type="radio" name="operation" value="multiplication" checked="yes">Multiplication</input><br/>
<input type="radio" name="operation" value="addition">Addition</input>
</td></tr>
</tr><td colspan="2" align="center"><input type="submit" name="submit" value="Generate" /></td> </tr>
</table>
</form>
<?php
//check to see if num of rows is numberic
if (isset($_POST["rows"]) && is_numeric($_POST["rows"])){
//check to see if rows is a positive number
if($_POST["rows"] > 0){
if(isset($_POST) && $_POST['operation'] == "multiplication")
{
for($r = 0; $r < $_POST["rows"]; $r++){
echo '<tr>';
for($c = 0; $c < $_POST["columns"]; $c++){
echo '<td>' .$c*$r. '</td>';
echo '</tr>';
}}
}
else if (isset($_POST) && $_POST['operation'] == "addition")
{
for($r = 0; $r < $_POST["rows"]; $r++){
echo '<tr>';
for($c = 0; $c < $_POST["columns"]; $c++){
echo '<td>' .$c+$r. '</td>';
echo '</tr>';
}
}
}
}
else{
echo 'Invalid rows columns parameters';
exit();
}
}
else{
echo 'Invalid rows columns parameters';
exit();
}
?>
</body>
</html>
Hello, Here's the weird thing, some of the pages shows the results, and some pages just won't show results at all, and I'm not getting any errors here is the one that doesn't Code: [Select] <? function display_mptt($user) { global $db; // retrieve the left and right value of the $root node $sql2 = "SELECT * from mptt where id='25'"; $result2 = mysql_query($sql2 ,$db); if(!$row2 = mysql_fetch_array($result2)) echo mysql_error(); echo '<h1>Users List</h1>'; // start with an empty $right stack $right = array(); // now, retrieve all descendants of the $root node $sql = "SELECT * from mptt WHERE 'left' BETWEEN '".$row2['left']."' AND '".$row2['right']."' ORDER BY 'left' ASC"; $result = mysql_query($sql ,$db); // display each row while ($row = mysql_fetch_array($result)or die(mysql_error())) { // only check stack if there is one $count = mysql_num_rows($result); if (count($right)>0) { // check if we should remove a node from the stack while ($right[count($right)-1]<$row['right']) { array_pop($right); } } // display indented node title // add this node to the stack $var3 = '10'; echo "<table width='589' border='1'> <tr> <th>ID</th> <th>Name</th> </tr>"; echo "<tr><td><a href=\"user.php?id=".$row['id']."\">".$row['id']."</a></td>"; echo "<td>" . $row['title'] ." </td>"; echo "</tr>"; echo "</table>"; $right[] = $row['right']; } } display_mptt(1); ?> and here is the page that does Code: [Select] <? function display_mptt($user) { global $db; $id = $_GET['id']; // retrieve the left and right value of the $root node $sql2 = "SELECT * from mptt where id= ".$id.""; $result2 = mysql_query($sql2 ,$db); if(!$row2 = mysql_fetch_array($result2)) echo mysql_error(); echo '<h1>Your Tree</h1>'; // start with an empty $right stack $right = array(); // now, retrieve all descendants of the $root node $sql = "SELECT * from mptt WHERE `left` BETWEEN ".$row2['left']." AND ".$row2['right']." ORDER BY 'left' ASC"; $result = mysql_query($sql ,$db); // display each row while ($row = mysql_fetch_array($result)) { // only check stack if there is one if (count($right)>0) { // check if we should remove a node from the stack while ($right[count($right)-1]<$row['right']) { array_pop($right); } } // display indented node title echo str_repeat(' ',count($right)).$row['title']."<br>"; // add this node to the stack $right[] = $row['right']; } } display_mptt(1); ?> Hi all, The below example is a workable code, taken from tutor_profile.sql table Code: [Select] <?php $query = "SELECT tutor_id, religion_id FROM tutor_profile WHERE tutor_id = '" . $_GET['tutor_id'] . "'"; $data = mysqli_query($dbc, $query) or die(mysqli_error($dbc)); // The user row was found so display the user data if (mysqli_num_rows($data) == 1) { $row = mysqli_fetch_array($data); print_r($row); if ($row != NULL) { $religion_id = $row['religion_id']; $tutor_id = $row['tutor_id']; } else { echo '<p class="error">There was a problem accessing your profile.</p>'; } } <!--Religion--> <tr> <td class="label">Religion:</td> <td> <select id="religion_id" name="religion_id"> <option value="1" <?php if (!empty($religion_id) && $religion_id == '1') echo 'selected = "selected"'; ?>>Buddhism</option> <option value="2" <?php if (!empty($religion_id) && $religion_id == '2') echo 'selected = "selected"'; ?>>Christianity</option> <option value="3" <?php if (!empty($religion_id) && $religion_id == '3') echo 'selected = "selected"'; ?>>Hinduism</option> <option value="4" <?php if (!empty($religion_id) && $religion_id == '4') echo 'selected = "selected"'; ?>>Islam</option> <option value="5" <?php if (!empty($religion_id) && $religion_id == '5') echo 'selected = "selected"'; ?>>Taoism</option> <option value="6" <?php if (!empty($religion_id) && $religion_id == '6') echo 'selected = "selected"'; ?>>Others</option> </select> </td> </tr> ?> As you can see I have hard coded the names of the religion in html code example - <option value="3" <?php if (!empty($religion_id) && $religion_id == '3') echoselected = "selected"'; ?>>Hinduism</option> And if our record shows that the tutor has previously selected '3', it will reflect as 'hinduism' in his profile. View profile.jpg for example example - <option value="3" <?php if (!empty($religion_id) && $religion_id == '3') echo 'selected = "selected"'; ?>>Hinduism</option> In fact, these names can be found in another table called religion.sql, but I hard coded it anyway, without using loop (while function), since there are only 8 names Code: [Select] <!--Religion--> <tr> <td class="label">Religion:</td> <td> <select id="religion_id" name="religion_id"> <option value="1" <?php if (!empty($religion_id) && $religion_id == '1') echo 'selected = "selected"'; ?>>Buddhism</option> <option value="2" <?php if (!empty($religion_id) && $religion_id == '2') echo 'selected = "selected"'; ?>>Christianity</option> <option value="3" <?php if (!empty($religion_id) && $religion_id == '3') echo 'selected = "selected"'; ?>>Hinduism</option> <option value="4" <?php if (!empty($religion_id) && $religion_id == '4') echo 'selected = "selected"'; ?>>Islam</option> <option value="5" <?php if (!empty($religion_id) && $religion_id == '5') echo 'selected = "selected"'; ?>>Taoism</option> <option value="6" <?php if (!empty($religion_id) && $religion_id == '6') echo 'selected = "selected"'; ?>>Others</option> </select> </td> </tr> Currently I am facing an issue, I guess I will need to use looping, as there are 22 names in another table which I will need to call forth, tutor_educational_level.sql, and the number of names get more and more in other tables. My question is, how do I pull out the entire list of names into a drop down box and yet showing the selected name which the user has chosen, more elaboration can be seen in profile.jpg. In profile.jpg - as you can see the list of names are shown in the drop down box and the system is able to decipher the chosen name. Another Example 1) N level 2) O level 3) A level 4) University User selected '3', which is A level, and system would still show the list of educational_names in a drop down box,, but selecting A level as the one to appear. Example 1) N level 2) O level 3) A level (selected) 4) University It should have the same overall result as the religion which I have stated above, however this time round, it is using looping function (while) to retrieve the entire list of names, select and show the name which the user has chosen Below is my code, and I know it is wrong, but generally would like to relate my idea across. Code: [Select] <?php <!--Teaching Credentials--> <tr> <td class="label">Teaching Credentials:</td> <td> <?php echo '<select name="educational_level" id="educational_level">'; $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) or die(mysqli_error($dbc)); $query = "SELECT tp.educational_id, el.educational_name AS educational_name, el.educational_id AS list_educational_id " . "FROM tutor_profile AS tp " . "INNER JOIN * tutor_educational_level AS el USING (educational_id) " . "WHERE tp.tutor_id = '" . $_GET['tutor_id'] . "'"; $sql = mysqli_query($dbc, $query) or die(mysqli_error()); while($data = mysqli_fetch_array($sql)) { echo'<option value="'.$data['list_educational_id'].'">'.$data['educational_name'].'</option>'; if (!empty ($data['educational_id']) && ($data['educational_id']) == ($data['list_educational_id'])) { echo 'selected = "selected"'; } } echo '</select><br/>'; mysqli_close($dbc); ?> </td> </tr> ?> This topic has been moved to Other Libraries and Frameworks. http://www.phpfreaks.com/forums/index.php?topic=314194.0 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 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 We are building a site for users to post articles on certain topics and we want to show their number of followers on various social media platforms. Apart from them having the means to add it manually, there must be a method of syncing with their account, that they punch in, so that it dynamically updates our database with their number of followers. How is it done? 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. 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 |
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