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PHP - Add 4 Days To Today
Hi I am trying to add a field to a database that is 4 days from the date the record is added, but it is not adding a value
Code: [Select] $end_date=strtotime("+ 4 days"); $add_vehicle_sql=mysql_query("INSERT INTO `tbl_auction_lot`(`cust_id`,`reserve`,`make`,`model`,`spec`,`fuel`,`doors`,`mot_date`,`fns`,`fos`,`rns`,`ros`,`condition`,`reg_no`,`service_history`,`sale_type`,`status`,`keepers`,`gearbox`,`emissions`,`colour`,`date_first_reg`,`date_manufacture`,`bhp`,`engine_size`,`end_date`) VALUES ('$seller_id','$reserve','$make','$model','$body_style','$fuel_type','$no_of_doors','$mot','$fns','$fos','$rns','$ros','$vehicle_condition','$vrm','$service_history','auction','$status','$prev_keepers','$gearbox','$emissions','$colour','$date_reg','$date_man','$bhp','$engine_size','$end_date')") or die(mysql_error()); What am I doing wrong and what is there a better way to achieve the desired result. Similar TutorialsHi, 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 I'm looking for a simple little code to display today's date, month, day, year and countdown to 365 days. Can anyone please help. Hello, In my Mysql database, it has a datetime field. and I have created a $today = date('Y-m-d H:i:s'); <- today's date and time How do I write a query to run in PHP in order to get all today's items by comparing datetime field and $today? Thanks! Is there a way of getting today's date (in European format - day-month-year) into the body of an email sent via phpmailer? Many thanks. This topic has been moved to PHP Freelancing. http://www.phpfreaks.com/forums/index.php?topic=333553.0 is it possible to do something like Code: [Select] $today = date("Y-m-d"); $result = mysql_query("SELECT * FROM staff where date = '.$today.' "); also, is it normal for the first entry in the database not to be displayed? i have 6 entries in a table and only 2-6 are shown. when i changed the id for 1 to 7, it only displayed 3-7. Hello, i'm trying to get the number of users that registered today and the number of users that registered yersterday (seperate), i've got this field in mysql: 'registertime' which stores data in this format 2011-11-14 14:53:49 also i have the field 'time' in the same format that updates everytime the user logs in. Now previously i wanted to find out users online in last 24 hours, the code i used for that is this: Code: [Select] $query = "SELECT COUNT(*) as Anzahl FROM customers WHERE country = 'de' AND time BETWEEN DATE_SUB( NOW(), INTERVAL 24 HOUR ) and NOW() "; $queryerg = mysql_query($query) OR die(mysql_error()); while($row = mysql_fetch_array($queryerg)){ $customers_on_de_24 = $row[0]; } How could i edit that to select the count just by date and ignore the time (hours minutes seconds) ? As far as I can see, this code: Code: [Select] $this->_licence_expires = date('Y/m/d', mktime(0, 0, 0, date("m"), date("d")-1, date("Y"))); results in accounts expiring 1 year from today, right? How can I change it so that it will result in expiry 9 months from today? TIA still shows all records in the database... any idea how i go about this? Code: [Select] $date= date('y-m-d'); $query=mysql_query("SELECT * FROM listing WHERE date >= $date") or die (mysql_error()); Hi All, Newbie here.
I run a website using xmb forums software as a base login script to connect to my database and created my own crude and simple pages for adding and displaying the data learning as I went. Over the years with recent upgrades of php versions its now virtually useless and unsecure code. I archive a lot of news articles but have had trouble pasting text that includes apostrophe's and special characters, but managed to fix that using addslashes() now. Can someone suggest a tutorial or a way to rebuild my site from scratch or something I can use as a template to connect to my pages ?. I have to two variables $cronlast and $cronnow in my php code. I need to have the time 7am of previous day in $cronlast variable and 7am today in $cronnow variable. I need the time in UNIX timestamp format. So everyday when I print the variable I should have: $cronlast=7am Yesterday(in Unix Timestamp) $cronnow=7am Today(in Unix Timestamp) I would appreciate any help. Hi, I have a table that contains posts and each post has a datetime field. Im trying to work out how I can show entries from today, this week and this month so I can have a link that shows all posts from today or this month etc. Any one know how I can do this ? My current piece of code that pulls the data from the db looks like this : Code: [Select] $texts= mysql_query("SELECT * FROM submittedtexts Order by id DESC LIMIT " . (($page - 1) * 6) . ", 6"); I want to add a bit that acts like : WHERE date = today Hope that makes sense, Im a bit of a newbie Thanks in advance, Scott I'm trying to figure out if today's current date and time falls between Mon & Fri, but so far, I can't get it to work. Can anybody look at my code and see what I'm doing wrong? Code: [Select] <?php $day_start = date('D h:i a', strtotime("Mon 05:30")); $day_end = date('D h:i a', strtotime("Fri 10:00")); $day_current = date('D h:i a', strtotime("+1 hours")); if (($day_current > $day_start) && ($day_current < $day_end)) { echo "yup"; } else { echo "nope"; } ?> Thanks in advance This topic has been moved to PHP Regex. http://www.phpfreaks.com/forums/index.php?topic=320501.0 When I try to add 30 days: Code: [Select] $date = date("Y-m-d"); $date = strtotime(date("Y-m-d", strtotime($date)) . " +30 days"); echo $date; and I echo date I get 1330664400 How do I get it to echo out 3/1/2012? I know the answer lies in the strtotime but I can't figure it out. I know it's a simple problem for most of you... Hello, I am trying to put together a mysql query that will return the number of visitors for four days ago, what i am trying to do is plot the last seven days visitors on a graph in the format of day seven, day six, etc and need to find away to get a count for each of those days. At the moment i am thinking about running various queries with each returning the results for a specific day. The code below is supposed to get the count of visitors four days ago. Not between now and four days ago but just for the 24 hour period which covers day 4. The current code just returns a value of 0. Any help would be appreciated. Code: [Select] $result = mysql_query("SELECT ip FROM ip_stats WHERE date= date_sub(NOW(), interval 4 DAY)"); $num_rows = mysql_num_rows($result); echo "$num_rows"; Hi, I have db table that records the days (Sunday, Monday...) when the employee login to the system. Say that the employee logged in on Monday then Logged in on Wednesday so this means he was absent on Tuesday. I calculated the number of days to get the answer 2 (between Wednesday and Monday before logging in on Wednesday) how to get the name of the missing day "Tuesday"? code:
$curdate = date("Y-m-d");
$currday = date('l');
$lastdate = $row['indate'];
$lastdayin = $row['lastdayin'];
if ($lastdate !=Null)
{
$misseddates = strtotime($curdate) - strtotime($lastdate);
$misseddates = $misseddates / 86400;
echo $misseddates;
$misseddays = strtotime($lastdayin) - strtotime($currday);
$misseddays = $misseddays / 86400;
echo $misseddays;
}
I tried to get the name of day in the last step but it only calculates 5 in numbers how to get the name. I want the answer to be "Tuesday" as of the example. Thanks. I have a SQL row that has a date field: ex: 2010-11-01. When a car is sold there either is a 30 day warranty, a 60 day warranty, or 0 day warranty. What I'm trying to do is display when the vehicles warranty expires, based on the date it was sold, or when did it expire based on the same sold date pulled from the database. Example using last months date: 2010-10-01 60 day: "Expires 11-30-10" 30 day: "Expired 11-01-10" I can not seem to use the date function properly... Any help would be greatly appreciated. Hi there, for every case we have "gooo" as commen in all how this could be one code using anything like elseif or switch whatever... if($refere && (stripos($r, $refere) === false)){ echo "x1" }else{ echo"goooo" } And if($limited && ($line[hits] >= $limited)){ echo "x2" }else{ echo"goooo" } And if($pword && (isset($_POST['password']) && $_POST['password'] == $pword) ){ echo"goooo" }else{ echo"x3" } And if ($line[capt] == 1 && ($_SESSION["security_code"]) ){ echo"goooo" }else{ echo"x4" } thanks in advance |
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