PHP - Display Dates Between From Date To Date
Hi...
Good day! I have table that has a field from_date and to_date. Now I just want to know if how can I display as table format the dates between from_date to_date. Like this from_date: 2011-12-16 to_date: 2011-12-31 I want to display it: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 // table format. Thank you Similar TutorialsI would like to know the best practice to achieve the following: I have a list of dates related to live events for performing artists. When someone views the web page that contains a section to display the dates, I would only want to show the dates from today into the future and not show any dates from the past. What is the best way to accomplish this? Thanks in advance... 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.
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"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 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 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 have two text fields in a form. Both accepts dates. I'd like to check whether the second date provided is exactly one year in the future from the first provided date.
I can easily do this in PHP but since I need to do the check once the user leaves the second field, I have to resort to JavaScript, using onblur or something.
There are two issues I have.
1. Make sure that the strings provided are indeed dates (I would use strtodate() in PHP for this)
2. Regardless of month provided, making sure that the second date is exactly a year in the future of the already provided year
I've only captured the values and passed them to Date(). However, it can only handle european style, and if I enter 1981-06-16, the value returned is Jun 15 1981 17:00:00 which is obviously wrong.
Here is the simple broken code I have
var text1 = document.getElementById("text1"); var start_date = new Date(text1.value); var text2 = document.getElementById("text2"); var end_date = new Date(text2.value); alert(start_date);I need some guidance. I'm trying to wrap my head around how i should go about doing this. I have two dates...2011-05-03 and 2011-05-08. I want to write a function that creates an array of the days that consist of that date range. So say something like Code: [Select] <?php $start = "2011-05-03"; $end = "2011-05-08"; function get_days($start, $end) { //code to get the days } echo get_days($start, $end); //hopefully produce something like... $day['1'] = "2011-05-03"; $day['2'] = "2011-05-04"; $day['3'] = "2011-05-05"; $day['4'] = "2011-05-06"; $day['5'] = "2011-05-07"; $day['6'] = "2011-05-08"; anybody know any idea how to do something like this? Pretty sure i'm going to need the mktime() function to account for ranges going across months and years. I have got a date field in my mysql database with events, and as you know date in MYSQL uses yyyy/mm/dd but I would first like to turn this around to dd/mm/yyyy but also turn this into for example "Tuesday - May 1990" would anyone like to point me into the right direction? Much appreciated! I have a query that fetches all the dates in a date range then displays them in a Date:HH:MM;SS format. However when displayed sometimes there are missing Hours as I don't have data for them and I would like to display it. So right now I have something like this. Date Column: Revenue_Column: 07/29/2010 00:00:00 $250.00 07/29/2010 01:00:00 $150.00 07/29/2010 03:00:00 $350.00 07/29/2010 04:00:00 $450.00 As you can see 02:00:00 is missing, how can I use php or possibly mysql to fill in that gap? I have tried creating an hours table and joining them and grouping by date however that doesn't seem to work. Any ideas? I'm trying to do the following PHP. I have written it in English if (today's date) => date1 AND <= date2 then {display image1} elseif (today's date) => date3 AND <= date4 then {display image2} else {display image 13} There are 24 fixed dates and 13 fixed images. I have tried using combinations of strtotime(), replacing the date value with variables, hard coding the dates within the program. I don't seem to be able to get any combination to work properly. I'm sure it's just a syntax error but I can't see it. When I've searched the web all the answers I've found relate to dates within databases but as I only have 24 dates it seems a bit of overkill. I would appreciate any pointers to the correct method I might be able to use. The dates need only a day and month as I would like this to repeat year after year. <?php $today = strtotime(date('d-m')); if (strtotime($today) >= strtotime('28-10') && strtotime($today) <= strtotime('24-11')) {echo "<div>image1</div>";} elseif (strtotime($today) >= strtotime('25-11') && strtotime($today) <= strtotime('22-12')) {echo "<div>image2</div>" ;} elseif (strtotime($today) >= strtotime('23-12') && strtotime($today) <= strtotime('24-02')) {echo "<div>image3</div>" ;} else {echo "<div>image13</div>";} ?> The result I get from this is image1 appears on the web page but I would expect image3 as today is 22-01 Thank you in advance. Andrew I have date stored in database in any of the given forms 2020-06-01, 2020-05-01 or 2019-04-01 I want to compare the old date with current date 2020-06-14 And the result should be in days. Any help please? PS: I want to do it on php side. but if its possible to do on database side (I am using myslq) please share both ways🙂 Edited June 14, 2020 by 684425Hey guys, How would I go about subtracting Today from a previous day to find the difference? For example, I want to subtract TODAY from a previous date in my database, to determine if the difference is greater than 1 day. Any ideas? I tried doing the subraction in TIMESTAMPS, but when I convert the date back to Y-m-d H:i:s, I got some weird year and time. 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. 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. 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? (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 the following table: ID picName description date 1 Us Time.jpg A couple watching the sunset at Ballintoy Harbour 2011-02-04 2 Pic.jpg This is the house on Castle Island taken in fog. 2011-02-03 3 catblack.jpg Pic of a cat 2011-02-05 I am trying to display th picName that belongs to the highest date but only know how to display all pictures. Can anybody help me? Here is what I have: <?php include("db.inc"); $cxn = mysqli_connect($host,$user,$password,$dbname) or die ("couldn't connect to server"); /* Select */ $query = "SELECT * FROM pictures"; $result = mysqli_query($cxn,$query) or die ("Couldn't execute query."); while($row = mysqli_fetch_assoc($result)) { echo "<tr><td colspan=2> </td> <td><img src='./images/{$row['picName']}' border='0' width='500' height='400' /><p></td></tr>\n"; } ?> 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. Hi Guys.. How can I change a date on the fly ? Everything is UTC on my server. How can I change a date to something else on the fly? Ie: $timezone = "cet"; $datetime = "2011-09-04 19:53:00"; echo $datetime($timezone); So I can give it a datetime and have it echo the datetime as if it were in the other timezone? Thanks Graham 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. |