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PHP - Updating Last Login Date
I'm making an amendment to my login page so that when a user logs in, the date_last_login field in the members table is updated to NOW(). Well, that's the idea.
The login works fine. Always has. It's just this update that isn't happening. No error message appears, but date_last_login remains 0000-00-00 00:00:00 . Any tips?
$sql_date_last_login = $pdo->exec("update members set date_last_login = now() where member_id = '".$SESSION['member_id']."'");
Cheers,
Dave
Similar TutorialsI am trying to update last logged in entry in the database upon succesful login. I may be way off in the logic here or I may be missing something simple. I don't get any errors and it logs in fine. Just does not update the lastvisit field in the database. Code: [Select] //record date of most recent login $result = mysql_query("SELECT username FROM users WHERE user_id ='".$_SESSION['userId'] . "'"); $dtCreated = date('Y-m-d'); mysql_query("UPDATE users SET lastvisit=('$dtCreated') WHERE username = $result"); As I am new to all this php, I need some help and/or ideas on how to code this little idea of mine: I have a database with a members table, I would like to create some Teams, and so I want too create a new table with the info for these teams. My problem is to set what team a member belongs too. My idea short Add a new column in the members area to hold the Id of the team this member belongs too. So if member 1000 sends a "join team link" too member 2000 the script needs to get the team id from member 1000 and add this id to member 2000. select teamid from members where id=1000 update teamid='same id as member 1000' where id='membersid' I hope this all makes sense, cause I am having a hard time trying to explain it. Hey, I'm using a script which allows you to click on a calendar to select the date to submit to the database. The date is submitted like this: 2014-02-08 Is there a really simple way to prevent rows showing if the date is in the past? Something like this: if($currentdate < 2014-02-08 || $currentdate == 2014-02-08) { } Thanks very much, Jack hi i need help an idea how can i separate members from admins since i dont know how to create login form i used tutorial ( http://www.youtube.com/watch?v=4oSCuEtxRK8 ) (its session login form only that i made it work other tutorials wre too old or something) how what i want to do is separate members and admins because admin need more rights to do now i have idea but dont know will it work like that what i want to do is create additional row in table named it flag and create 0 (inactive user) 1 (member) 2 (admin) will that work? and how can i create different navigation bars for users and admins? do you recommend that i use different folders to create it or just script based on session and flag? How to add the ability to login with username or email for login?
<?php
ob_start();
include('../header.php');
include_once("../db_connect.php");
session_start();
if(isset($_SESSION['user_id'])!="") {
header("Location: ../dashboard");
}
if (isset($_POST['login'])) {
$email = mysqli_real_escape_string($conn, $_POST['email']);
$password = mysqli_real_escape_string($conn, $_POST['password']);
$result = mysqli_query($conn, "SELECT * FROM users WHERE email = '" . $email. "' and pass = '" . md5($password). "'");
if ($row = mysqli_fetch_array($result)) {
$_SESSION['user_id'] = $row['uid'];
$_SESSION['user_name'] = $row['user'];
$_SESSION['user_email'] = $row['email'];
header("Location: ../dashboard");
} else {
$error_message = "Incorrect Email or Password!!!";
}
}
?>
Hi guys. What I want to create is really complicated. Well I have a login system that works with post on an external website. I have my own website, but they do not give me access to the database for security reasons, therefore I have to use their login system to verify my users. What their website does is that it has a post, with username and password. The POST website is lets say "https://www.example.com/login". If login is achieved (i.e. username and password are correct), it will redirect me to "https://www.example.com/login/success" else it will redirect me to "https://www.example.com/login/retry". So I want a PHP script that will do that post, and then according to the redirected website address it will return me TRUE for success, FALSE for not successful login. Any idea?? Thanks Hello guys, Is there on web any updated tutorial on how can I add Facebook login on my simple php login script? Hi guys, Can anyone assist me. I am trying to create a login for admin and user (if user not a member click register link) below is my code: But whenever I enter the value as: Username: admin Password:123 - I got an error message "That user does not exist!" Any suggestion and help would be appreciated. Thanks. login.php <?php //Assigned varibale $error_msg as empty //$error_msg = ""; session_start(); $error_msg = ""; if (isset($_POST['submit'])) { if ($a_username = "admin" && $a_password = "123") { //Define $_POST from form text feilds $username = $_POST['username']; $password = $_POST['password']; //Add some stripslashes $username = stripslashes($username); $password = stripslashes($password); //Check if usernmae and password is good, if it is it will start session if ($username == $a_username && $password == $a_password) { session_start(); $_SESSION['session_logged'] = 'true'; $_SESSION['session_username'] = $username; //Redirect to admin page header("Location: admin_area.php"); } } $username = (isset($_POST['username'])) ? $_POST['username'] : ''; $password = (isset($_POST['password'])) ? $_POST['password'] : ''; if($username && $password) { $connect = mysql_connect("localhost", "root", "") or die ("Couldn't connect!"); mysql_select_db("friendsdb") or die ("Couldn't find the DB"); $query = mysql_query ("SELECT * FROM `user` WHERE username = '$username'"); $numrows = mysql_num_rows($query); if ($numrows != 0){ while ($row = mysql_fetch_array($query)) { $dbusername = $row['username']; $dbpassword = $row['password']; } //Check to see if they are match! if ($username == $dbusername && md5($password) == $dbpassword) { header ("Location: user_area.php"); $_SESSION['username'] = $username; } else $error_msg = "Incorrect password!"; //code of login }else $error_msg = "That user does not exist!"; //echo $numrows; } else $error_msg = "Please enter a username and password!"; } ?> <!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=iso-8859-1" /> <title>Login Page</title> </head> <body> <br /> <?php require "header.php"; ?><br /> <div align="center"> <table width="200" border="1"> <?php // If $error_msg not equal to emtpy then display error message if($error_msg!="") echo "<div id=\"error_message\"style=\"color:red; \">$error_msg</div><br />";?> <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post"> <!--form action="login_a.php" method="post"--> Username: <input type="text" name="username" /><br /><br /> Password: <input type="password" name="password" /><br /><br /> <input type="submit" name = "submit" value="Log in" /> </form> <p> </p> Register a <a href="register.php">New User</a> </table> </div> </body> </html> Hello. I'm new to pHp and I would like to know how to get my $date_posted to read as March 12, 2012, instead of 2012-12-03. Here is the code: Code: [Select] <?php $sql = " SELECT id, title, date_posted, summary FROM blog_posts ORDER BY date_posted ASC LIMIT 10 "; $result = mysql_query($sql); while($row = mysql_fetch_assoc($result)) { $id = $row['id']; $title = $row['title']; $date_posted = $row['date_posted']; $summary = $row['summary']; echo "<h3>$title</h3>\n"; echo "<p>$date_posted</p>\n"; echo "<p>$summary</p>\n"; echo "<p><a href=\"post.php?id=$id\" title=\"Read More\">Read More...</a></p>\n"; } ?> I have tried the date() function but it always updates with the current time & date so I'm a little confused on how I get this to work. Alright, I have a Datetime field in my database which I'm trying to store information in. Here is my code to get my Datetime, however it's returning to me the wrong date. It's returning: 1969-12-31 19:00:00 $mysqldate = date( 'Y-m-d H:i:s', $phpdate ); $phpdate = strtotime( $mysqldate ); echo $mysqldate; Is there something wrong with it? I have tried a large number of "solutions" to this but everytime I use them I see 0000-00-00 in my date field instead of the date even though I echoed and can see that the date looks correct. Here's where I'm at: I have a drop down for the month (1-12) and date fields (1-31) as well as a text input field for the year. Using the POST array, I have combined them into the xxxx-xx-xx format that I am using in my field as a date field in mysql. <code> $date_value =$_POST['year'].'-'.$_POST['month'].'-'.$_POST['day']; echo $date_value; </code> This outputs 2012-5-7 in my test echo but 0000-00-00 in the database. I have tried unsuccessfully to use in a numberof suggested versions of: strtotime() mktime Any help would be extremely appreciated. I am aware that I need to validate this data and insure that it is a valid date. That I'm okay with. I would like some help on getting it into the database. (continuing from topic title) So if I set a date of July 7 2011 into my script, hard coded in, I would like the current date to be checked against the hard coded date, and return true if the current date is within a week leading up to the hard coded date. How could I go about doing this easily? I've been researching dates in php but I can't seem to work out the best way to achieve what I'm after. Cheers Denno Hi, I have a job listing website which displays the closing date of applications using: $expired_date (This displays a date such as 31st December 2019) I am trying to show a countdown/number of days left until the closing date. I have put this together, but I can't get it to show the number of days.
<?php
$expired_date = get_post_meta( $post->ID, '_job_expires', true );
$hide_expiration = get_post_meta( $post->ID, '_hide_expiration', true );
if(empty($hide_expiration )) {
if(!empty($expired_date)) { ?>
<span><?php echo date_i18n( get_option( 'date_format' ), strtotime( get_post_meta( $post->ID, '_job_expires', true ) ) ) ?></span>
<?php
$datetime1 = new DateTime($expired_date);
$datetime2 = date('d');
$interval = $datetime1->diff($datetime2);
echo $interval->d;
?>
<?php }
}
?>
Can anyone help me with what I have wrong? Many thanks Hi guys, I'm putting together a small event system where I want the user to add his own date and time into a textfield (I'll probably make this a series of drop-downs/a date picker later). This is then stored as a timestamp - "0000-00-00 00:00:00" which displays fine until I try to echo it out as a UK date in this format - jS F Y, which just gives today's date but not the inputted date. Here's the code I have right now: Code: [Select] $result = mysql_query("SELECT * FROM stuff.events ORDER BY eventdate ASC"); echo "<br />"; echo mysql_result($result, $i, 'eventvenue'); echo ", "; $dt = new DateTime($eventdate); echo $dt->format("jS F Y"); In my mysql table eventdate is set up as follows: field - eventdate type - timestamp length/values - blank default - current_timestamp collation - blank attributes - on update CURRENT_TIMESTAMP null - blank auto_increment - blank Any help as to why this could be happening would be much appreciated, thanks. I'm getting this Time Zone error. Perhaps it's a compatibility issue with PHP 5.3. Looked all over for an answer without finding one. Here is the error message Warning: date() [function.date]: It is not safe to rely on the system's timezone settings. You are *required* to use the date.timezone setting or the date_default_timezone_set() function. In case you used any of those methods and you are still getting this warning, you most likely misspelled the timezone identifier. We selected 'America/New_York' for 'EST/-5.0/no DST' instead in /blocked.php on line 41 12/02/12 Here is the code. Line 41 is near the bottom, the one with the d,m,y. Perhaps the echo date (d/m/y") needs to be changed. Appreciate any help! Code: [Select] <table border="3" width="16%" align="center" cellspacing="0" bgcolor="#FF6600" bordercolor="red" bordercolordark="red" bordercolorlight="red"> <tr> <td width="176"> <p align="center"><?php // shows IP Number on Page echo $ip; ?> </p> </td> </tr> </table> <p align="center"><?php // Show the user agent echo 'Your user agent is: <b>'.$_SERVER['HTTP_USER_AGENT'].'</b><br />';?></p> [b]<h1 align="center"><?php echo date("d/m/y");?></h1>[/b] </td> </tr> </table [,code] i have a table that shows payments made but want to the payments only showing from a set date(06/12/14) and before this date i dont want to show
this is my sql that doesnt seem to work and is showing dates before the specified date.
.
"SELECT * FROM payments2014, signup2014, editprop2014 WHERE signup2014.userid = payments2014.payment_userid AND editprop2014.prop_id = signup2014.prop_id AND signup2014.userid !='page1' AND signup2014.userid !='page6' AND signup2014.userid !='page4' AND payments2014.payment_transaction_status !='none' AND payments2014.payment_transaction_status !='CANCELLEDa' AND payments2014.payment_type !='deposit' AND payments2014.payment_paid_timestamp NOT LIKE '%2012%' AND payments2014.payment_paid_timestamp NOT LIKE '%2011%' AND payments2014.payment_paid_timestamp >= '06/12/14' ORDER BY payments2014.payment_id DESC"i have some other parts in the statment but this one that should be filtering is host_payments2014.payment_paid_timestamp >= '06/12/14'thanks in advance Hi there, I have a string '12/04/1990', that's in the format dd/mm/yyyy. I'm attempting to convert that string to a Date, and then insert that date into a MySQL DATE field. The problem is, every time I try to do so, I keep getting values like this in the database: 1970-01-01. Any ideas? Much appreciated. Hi, I am trying to convert a String date into numeric date using PHP function's, but haven't found such function. Had a look at date(), strtotime(), getdate(); e.g. Apr 1 2011 -> 04-01-2011 Could someone please shed some light on this? Regards, Abhishek Hi, Currently I am making a module for joomla. every article has an publish date, if the article was published in 7 days ago, it will displayed as "article in last week", My idea is to use today's date - publish date, if the result is greater than 7 and smaller than 14, the article will be displayed as "article in last week. Any one know how to write this code? Here is what I have got, but not working. <?php $todays_date = date("Y-m-d"); $result = mysql_query("select * from jos_content where $test between $todays_date-14 and $todays_date-7"); while($row = mysql_fetch_array($result)) { echo "$todays_date - $row[title]"; } ?> |
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